# 7.3: Graphs of the Other Trigonometric Functions

Skills to Develop

• Analyze the graph of (y= an x).
• Graph variations of (y= an x).
• Analyze the graphs of (y=sec x) and (y=csc x).
• Graph variations of (y=sec x) and (y=csc x).
• Analyze the graph of (y=cot x).
• Graph variations of (y=cot x).

We know the tangent function can be used to find distances, such as the height of a building, mountain, or flagpole. But what if we want to measure repeated occurrences of distance? Imagine, for example, a police car parked next to a warehouse. The rotating light from the police car would travel across the wall of the warehouse in regular intervals. If the input is time, the output would be the distance the beam of light travels. The beam of light would repeat the distance at regular intervals. The tangent function can be used to approximate this distance. Asymptotes would be needed to illustrate the repeated cycles when the beam runs parallel to the wall because, seemingly, the beam of light could appear to extend forever. The graph of the tangent function would clearly illustrate the repeated intervals. In this section, we will explore the graphs of the tangent and other trigonometric functions.

## Analyzing the Graph of (y = an x)

We will begin with the graph of the tangent function, plotting points as we did for the sine and cosine functions. Recall that

[ an , x=dfrac{sin , x}{cos , x}. onumber]

The period of the tangent function is (pi) because the graph repeats itself on intervals of (kpi) where (k) is a constant. If we graph the tangent function on (−frac{pi}{2}) to (frac{pi}{2}), we can see the behavior of the graph on one complete cycle. If we look at any larger interval, we will see that the characteristics of the graph repeat.

We can determine whether tangent is an odd or even function by using the definition of tangent.

[egin{align*} an(-x)&= dfrac{sin(-x)}{cos(-x)} qquad ext{Definition of tangent} &= dfrac{-sin , x}{cos , x} qquad ext{Sine is an odd function, cosine is even} &= -dfrac{sin , x}{cos , x} qquad ext{The quotient of an odd and an even function is odd} &= - an , x qquad ext{Definition of tangent} end{align*}]

Therefore, tangent is an odd function. We can further analyze the graphical behavior of the tangent function by looking at values for some of the special angles, as listed in Table (PageIndex{1}).

 (x) ( an x) (−dfrac{pi}{2}) (−dfrac{pi}{3}) (−dfrac{pi}{4}) (−dfrac{pi}{6}) 0 (dfrac{pi}{6})﻿ (dfrac{pi}{4}) (dfrac{pi}{3}) (dfrac{pi}{2}) undefined (-sqrt{3}) (–1) (-dfrac{sqrt{3}}{3}) 0 (dfrac{sqrt{3}}{3}) 1 (sqrt{3}) undefined

These points will help us draw our graph, but we need to determine how the graph behaves where it is undefined. If we look more closely at values when (frac{pi}{3}

 (x) ( an x) 1.3 1.5 1.55 1.56 3.6 14.1 48.1 92.6

As (x) approaches (dfrac{pi}{2}), the outputs of the function get larger and larger. Because (y= an , x) is an odd function, we see the corresponding table of negative values in Table (PageIndex{3}).

 (x) ( an x) −1.3 −1.5 −1.55 −1.56 −3.6 −14.1 −48.1 −92.6

We can see that, as (x) approaches (−frac{pi}{2}), the outputs get smaller and smaller. Remember that there are some values of (x) for which (cos , x=0). For example, (cos left (frac{pi}{2} ight)=0) and (cos left (frac{3pi}{2} ight )=0). At these values, the tangent function is undefined, so the graph of (y= an , x) has discontinuities at (x=frac{pi}{2}) and (frac{3pi}{2}). At these values, the graph of the tangent has vertical asymptotes. Figure (PageIndex{1}) represents the graph of (y= an , x). The tangent is positive from (0) to (frac{pi}{2}) and from (pi) to (frac{3pi}{2}), corresponding to quadrants I and III of the unit circle. Figure (PageIndex{1}): Graph of the tangent function

## Graphing Variations of (y = an , x)

As with the sine and cosine functions, the tangent function can be described by a general equation.

[y=A an(Bx) onumber]

We can identify horizontal and vertical stretches and compressions using values of (A) and (B). The horizontal stretch can typically be determined from the period of the graph. With tangent graphs, it is often necessary to determine a vertical stretch using a point on the graph.

Because there are no maximum or minimum values of a tangent function, the term amplitude cannot be interpreted as it is for the sine and cosine functions. Instead, we will use the phrase stretching/compressing factor when referring to the constant (A).

FEATURES OF THE GRAPH OF (y = A an(Bx))

• The stretching factor is (|A|).
• The period is (P=dfrac{pi}{|B|}).
• The domain is all real numbers (x),where (x≠dfrac{pi}{2| B |}+dfrac{π}{| B |}k) such that (k) is an integer.
• The range is ((−infty,infty)).
• The asymptotes occur at (x=dfrac{pi}{2| B |}+dfrac{π}{| B |}k) where (k) is an integer.
• (y=A an(Bx)) is an odd function.

### Graphing One Period of a Stretched or Compressed Tangent Function

We can use what we know about the properties of the tangent function to quickly sketch a graph of any stretched and/or compressed tangent function of the form (f(x)=A an(Bx)). We focus on a single period of the function including the origin, because the periodic property enables us to extend the graph to the rest of the function’s domain if we wish. Our limited domain is then the interval (left (−dfrac{P}{2},dfrac{P}{2} ight )) and the graph has vertical asymptotes at (pm dfrac{P}{2}) where (P=dfrac{pi}{B}). On (left (−dfrac{pi}{2},dfrac{pi}{2} ight )), the graph will come up from the left asymptote at (x=−dfrac{pi}{2}), cross through the origin, and continue to increase as it approaches the right asymptote at (x=dfrac{pi}{2}). To make the function approach the asymptotes at the correct rate, we also need to set the vertical scale by actually evaluating the function for at least one point that the graph will pass through. For example, we can use

[f left (dfrac{P}{4} ight )=A an left (Bdfrac{P}{4} ight )=A an left (Bdfrac{pi}{4B} ight )=A onumber]

because ( an left (dfrac{pi}{4} ight )=1).

Howto: Given the function (f(x)=A an(Bx)), graph one period.

1. Identify the stretching factor, (| A |).
2. Identify B and determine the period, (P=dfrac{pi}{| B |}).
3. Draw vertical asymptotes at (x=−dfrac{P}{2}) and (x=dfrac{P}{2}).
4. For (A>0), the graph approaches the left asymptote at negative output values and the right asymptote at positive output values (reverse for (A<0)).
5. Plot reference points at (left (dfrac{P}{4},A ight )), ((0,0)), and (left (−dfrac{P}{4},−A ight )), and draw the graph through these points.

Example (PageIndex{1}): Sketching a Compressed Tangent

Sketch a graph of one period of the function (y=0.5 an left (dfrac{pi}{2}x ight )).

Solution

First, we identify (A) and (B). Figure (PageIndex{2})

Because (A=0.5) and (B=dfrac{pi}{2}), we can find the stretching/compressing factor and period. The period is (dfrac{pi}{dfrac{pi}{2}}=2), so the asymptotes are at (x=±1). At a quarter period from the origin, we have

[egin{align*} f(0.5)&= 0.5 an left (dfrac{0.5pi}{2} ight ) &= 0.5 an left (dfrac{pi}{4} ight ) &= 0.5 end{align*}]

This means the curve must pass through the points ((0.5,0.5)), ((0,0)),and ((−0.5,−0.5)). The only inflection point is at the origin. Figure (PageIndex{3}) shows the graph of one period of the function. Figure (PageIndex{3}) (PageIndex{1})

Sketch a graph of (f(x)=3 an left (dfrac{pi}{6}x ight )). Figure (PageIndex{4})

### Graphing One Period of a Shifted Tangent Function

Now that we can graph a tangent function that is stretched or compressed, we will add a vertical and/or horizontal (or phase) shift. In this case, we add (C) and (D) to the general form of the tangent function.

[f(x)=A an(Bx−C)+D onumber]

The graph of a transformed tangent function is different from the basic tangent function ( an x) in several ways:

Howto: Given the function (y=A an(Bx−C)+D), sketch the graph of one period.

1. Express the function given in the form (y=A an(Bx−C)+D).
2. Identify the stretching/compressing factor, (| A |).
3. Identify (B) and determine the period, (P=dfrac{pi}{|B|}).
4. Identify (C) and determine the phase shift, (dfrac{C}{B}).
5. Draw the graph of (y=A an(Bx)) shifted to the right by (dfrac{C}{B}) and up by (D).
6. Sketch the vertical asymptotes, which occur at (x=dfrac{C}{B}+dfrac{pi}{2| B |}k),where (k) is an odd integer.
7. Plot any three reference points and draw the graph through these points.

Example (PageIndex{2}): Graphing One Period of a Shifted Tangent Function

Graph one period of the function (y=−2 an(pi x+pi)−1).

Solution

• Step 1. The function is already written in the form (y=A an(Bx−C)+D).
• Step 2.(A=−2), so the stretching factor is (|A|=2).
• Step 3. (B=pi), so the period is (P=dfrac{pi}{| B |}=dfrac{pi}{pi}=1).
• Step 4. (C=−pi), so the phase shift is (CB=dfrac{−pi}{pi}=−1).
• Step 5-7. The asymptotes are at (x=−dfrac{3}{2}) and (x=−dfrac{1}{2}) and the three recommended reference points are ((−1.25,1)), ((−1,−1)), and ((−0.75,−3)). The graph is shown in Figure (PageIndex{5}). Figure (PageIndex{5})

Analysis

Note that this is a decreasing function because (A<0).

Exercise (PageIndex{2})

How would the graph in Example (PageIndex{2}) look different if we made (A=2) instead of (−2)?

It would be reflected across the line (y=−1), becoming an increasing function.

Howto: Given the graph of a tangent function, identify horizontal and vertical stretches.

1. Find the period (P) from the spacing between successive vertical asymptotes or x-intercepts.
2. Write (f(x)=A an left (dfrac{pi}{P}x ight )).
3. Determine a convenient point ((x,f(x))) on the given graph and use it to determine (A).

Example (PageIndex{3}): Identifying the Graph of a Stretched Tangent

Find a formula for the function graphed in Figure (PageIndex{6}). Figure (PageIndex{6}): A stretched tangent function

Solution

The graph has the shape of a tangent function.

• Step 1. One cycle extends from (–4) to (4), so the period is (P=8). Since (P=dfrac{pi}{| B |}), we have (B=dfrac{π}{P}=dfrac{pi}{8}).
• Step 2. The equation must have the form (f(x)=A an left (dfrac{pi}{8}x ight )).
• Step 3. To find the vertical stretch (A),we can use the point ((2,2)). [egin{align*} 2&=A an left (dfrac{pi}{8}cdot 2 ight ) &=A an left (dfrac{pi}{4} ight ) end{align*}]

Because ( an left (dfrac{pi}{4} ight )=1), (A=2).

This function would have a formula (f(x)=2 an left (dfrac{pi}{8}x ight )).

Exercise (PageIndex{3})

Find a formula for the function in Figure (PageIndex{7}). Figure (PageIndex{7})

(g(x)=4 an(2x))

## Analyzing the Graphs of (y = sec x) and (y = csc x)

The secant was defined by the reciprocal identity (sec , x=dfrac{1}{cos x}). Notice that the function is undefined when the cosine is (0), leading to vertical asymptotes at (dfrac{pi}{2}), (dfrac{3pi}{2}) etc. Because the cosine is never more than (1) in absolute value, the secant, being the reciprocal, will never be less than (1) in absolute value.

We can graph (y=sec x) by observing the graph of the cosine function because these two functions are reciprocals of one another. See Figure (PageIndex{8}). The graph of the cosine is shown as a dashed orange wave so we can see the relationship. Where the graph of the cosine function decreases, the graph of the secant function increases. Where the graph of the cosine function increases, the graph of the secant function decreases. When the cosine function is zero, the secant is undefined.

The secant graph has vertical asymptotes at each value of (x) where the cosine graph crosses the (x)-axis; we show these in the graph below with dashed vertical lines, but will not show all the asymptotes explicitly on all later graphs involving the secant and cosecant.

Note that, because cosine is an even function, secant is also an even function. That is, (sec(−x)=sec x). Figure (PageIndex{8}): Graph of the secant function, (f(x)=sec x=dfrac{1}{cos x})

As we did for the tangent function, we will again refer to the constant (| A |) as the stretching factor, not the amplitude.

FEATURES OF THE GRAPH OF (y = A sec(Bx))

• The stretching factor is (| A |).
• The period is (dfrac{2pi}{| B |}).
• The domain is (x≠dfrac{pi}{2| B |}k), where (k) is an odd integer.
• The range is ((−∞,−|A|]∪[|A|,∞)).
• The vertical asymptotes occur at (x=dfrac{pi}{2| B |}k), where (k) is an odd integer.
• There is no amplitude.
• (y=Asec(Bx)) is an even function because cosine is an even function.

Similar to the secant, the cosecant is defined by the reciprocal identity (csc x=dfrac{1}{sin x}). Notice that the function is undefined when the sine is (0), leading to a vertical asymptote in the graph at (0), (pi), etc. Since the sine is never more than (1) in absolute value, the cosecant, being the reciprocal, will never be less than (1) in absolute value.

We can graph (y=csc x) by observing the graph of the sine function because these two functions are reciprocals of one another. See Figure (PageIndex{7}). The graph of sine is shown as a dashed orange wave so we can see the relationship. Where the graph of the sine function decreases, the graph of the cosecant function increases. Where the graph of the sine function increases, the graph of the cosecant function decreases.

The cosecant graph has vertical asymptotes at each value of (x) where the sine graph crosses the (x)-axis; we show these in the graph below with dashed vertical lines.

Note that, since sine is an odd function, the cosecant function is also an odd function. That is, (csc(−x)=−csc x).

The graph of cosecant, which is shown in Figure (PageIndex{9}), is similar to the graph of secant. Figure (PageIndex{9}): The graph of the cosecant function, (f(x)=csc x=frac{1}{sin x})

## Graphing Variations of (y = sec x) and (y= csc x)

For shifted, compressed, and/or stretched versions of the secant and cosecant functions, we can follow similar methods to those we used for tangent and cotangent. That is, we locate the vertical asymptotes and also evaluate the functions for a few points (specifically the local extrema). If we want to graph only a single period, we can choose the interval for the period in more than one way. The procedure for secant is very similar, because the cofunction identity means that the secant graph is the same as the cosecant graph shifted half a period to the left. Vertical and phase shifts may be applied to the cosecant function in the same way as for the secant and other functions.The equations become the following.

[y=Asec(Bx−C)+D]

[y=Acsc(Bx−C)+D]

FEATURES OF THE GRAPH OF (y = Asec(Bx−C)+D)

• The stretching factor is (|A|).
• The period is (dfrac{2pi}{|B|}).
• The domain is (x≠dfrac{C}{B}+dfrac{pi}{2| B |}k),where (k) is an odd integer.
• The range is ((−∞,−|A|]∪[|A|,∞)).
• The vertical asymptotes occur at (x=dfrac{C}{B}+dfrac{π}{2| B |}k),where (k) is an odd integer.
• There is no amplitude.
• (y=Asec(Bx)) is an even function because cosine is an even function.

HOWTO: Given a function of the form (y=Asec(Bx)), graph one period

1. Express the function given in the form (y=Asec(Bx)).
2. Identify the stretching/compressing factor, (|A|).
3. Identify (B) and determine the period, (P=dfrac{2pi}{| B |}).
4. Sketch the graph of (y=Acos(Bx)).
5. Use the reciprocal relationship between (y=cos , x) and (y=sec , x) to draw the graph of (y=Asec(Bx)).
6. Sketch the asymptotes.
7. Plot any two reference points and draw the graph through these points.

Q&A: Do the vertical shift and stretch/compression affect the secant’s range?

Yes. The range of (f(x)=Asec(Bx−C)+D) is ((−∞,−|A|+D]∪[|A|+D,∞)).

Howto: Given a function of the form (f(x)=Asec(Bx−C)+D), graph one period.

1. Express the function given in the form (y=A sec(Bx−C)+D).
2. Identify the stretching/compressing factor, (| A |).
3. Identify (B) and determine the period, (dfrac{2pi}{|B|}).
4. Identify (C) and determine the phase shift, (dfrac{C}{B}).
5. Draw the graph of (y=A sec(Bx)). but shift it to the right by (dfrac{C}{B}) and up by (D).
6. Sketch the vertical asymptotes, which occur at (x=dfrac{C}{B}+dfrac{pi}{2| B |}k), where (k) is an odd integer.

Example (PageIndex{5}): Graphing a Variation of the Secant Function

Graph one period of (y=4sec left (dfrac{pi}{3x}−dfrac{pi}{2} ight )+1).

Solution

• Step 1. Express the function given in the form (y=4sec left (dfrac{pi}{3x}−dfrac{pi}{2} ight )+1).
• Step 2. The stretching/compressing factor is (| A |=4).
• Step 3. The period is

[egin{align*} dfrac{2pi}{|B|}&= dfrac{2pi}{dfrac{pi}{3}} &= 2pi cdot dfrac{3}{pi} &= 6 end{align*}]

• Step 4. The phase shift is

[egin{align*} dfrac{C}{B}&= dfrac{dfrac{pi}{2}}{dfrac{pi}{3}} &= dfrac{pi}{2}cdot dfrac{3}{pi} &= 1.5 end{align*}]

• Step 5. Draw the graph of (y=Asec(Bx)), but shift it to the right by (dfrac{C}{B}=1.5) and up by (D=6).
• Step 6. Sketch the vertical asymptotes, which occur at (x=0), (x=3), and (x=6). There is a local minimum at ((1.5,5)) and a local maximum at ((4.5,−3)). Figure (PageIndex{12}) shows the graph. Figure (PageIndex{12})

Exercise (PageIndex{5})

Graph one period of (f(x)=−6sec(4x+2)−8). Figure (PageIndex{13})

Q&A: The domain of (csc , x) was given to be all (x) such that (x≠kpi) for any integer (k). Would the domain of (y=Acsc(Bx−C)+D) be (x≠dfrac{C+kpi}{B})?

Yes. The excluded points of the domain follow the vertical asymptotes. Their locations show the horizontal shift and compression or expansion implied by the transformation to the original function’s input.

Howto: Given a function of the form (y=Acsc(Bx)), graph one period.

1. Express the function given in the form (y=Acsc(Bx)).
2. (|A|).
3. Identify (B) and determine the period, (P=dfrac{2pi}{| B |}).
4. Draw the graph of (y=Asin(Bx)).
5. Use the reciprocal relationship between (y=sin , x) and (y=csc , x) to draw the graph of (y=Acsc(Bx)).
6. Sketch the asymptotes.
7. Plot any two reference points and draw the graph through these points.

Example (PageIndex{6}): Graphing a Variation of the Cosecant Function

Graph one period of (f(x)=−3csc(4x)).

Solution

• Step 1. The given function is already written in the general form, (y=Acsc(Bx)).
• Step 2. (| A |=| −3 |=3),so the stretching factor is (3).
• Step 3. (B=4),so (P=dfrac{2pi}{4}=dfrac{pi}{2}). The period is (dfrac{pi}{2}) units.
• Step 4. Sketch the graph of the function (g(x)=−3sin(4x)).
• Step 5. Use the reciprocal relationship of the sine and cosecant functions to draw the cosecant function.
• Steps 6–7. Sketch three asymptotes at (x=0), (x=dfrac{pi}{4}), and (x=dfrac{pi}{2}). We can use two reference points, the local maximum at (left (dfrac{pi}{8},−3 ight )) and the local minimum at (left (dfrac{3pi}{8},3 ight )). Figure (PageIndex{14}) shows the graph. Figure (PageIndex{14})

Exercise (PageIndex{6})

Graph one period of (f(x)=0.5csc(2x)). Figure (PageIndex{15})

Howto: Given a function of the form (f(x)=A csc(Bx−C)+D), graph one period

1. Express the function given in the form (y=Acsc(Bx−C)+D).
2. Identify the stretching/compressing factor, (|A|).
3. Identify (B) and determine the period, (dfrac{2pi}{| B |}).
4. Identify (C) and determine the phase shift, (dfrac{C}{B}).
5. Draw the graph of (y=Acsc(Bx)) but shift it to the right by and up by (D).
6. Sketch the vertical asymptotes, which occur at (x=dfrac{C}{B}+dfrac{pi}{| B |}k),where (k) is an integer.

Example (PageIndex{7}): Graphing a Vertically Stretched, Horizontally Compressed, and Vertically Shifted Cosecant

Sketch a graph of (y=2csc left (dfrac{pi}{2}x ight )+1). What are the domain and range of this function?

Solution

• Step 1. Express the function given in the form (y=2csc left (dfrac{pi}{2}x ight )+1).
• Step 2. Identify the stretching/compressing factor, (| A |=2).
• Step 3. The period is (dfrac{2pi}{| B |}=dfrac{2pi}{dfrac{pi}{2}}=2pi⋅dfrac{2}{pi}=4).
• Step 4. The phase shift is (dfrac{0}{dfrac{pi}{2}}=0).
• Step 5. Draw the graph of (y=Acsc(Bx)) but shift it up (D=1).
• Step 6. Sketch the vertical asymptotes, which occur at (x=0), (x=2), (x=4).

The graph for this function is shown in Figure (PageIndex{16}). Figure (PageIndex{16}): A transformed cosecant function

Analysis

The vertical asymptotes shown on the graph mark off one period of the function, and the local extrema in this interval are shown by dots. Notice how the graph of the transformed cosecant relates to the graph of (f(x)=2sin left (frac{pi}{2}x ight )+1),shown as the orange dashed wave.

Exercise (PageIndex{7})

Given the graph of (f(x)=2cos left (frac{pi}{2}x ight )+1) shown in Figure (PageIndex{17}), sketch the graph of (g(x)=2sec left (dfrac{pi}{2}x ight )+1) on the same axes. Figure (PageIndex{17}) Figure (PageIndex{18})

## Analyzing the Graph of (y = cot x)

The last trigonometric function we need to explore is cotangent. The cotangent is defined by the reciprocal identity (cot , x=dfrac{1}{ an x}). Notice that the function is undefined when the tangent function is (0), leading to a vertical asymptote in the graph at (0), (pi), etc. Since the output of the tangent function is all real numbers, the output of the cotangent function is also all real numbers.

We can graph (y=cot x) by observing the graph of the tangent function because these two functions are reciprocals of one another. See Figure (PageIndex{19}). Where the graph of the tangent function decreases, the graph of the cotangent function increases. Where the graph of the tangent function increases, the graph of the cotangent function decreases.

The cotangent graph has vertical asymptotes at each value of (x) where ( an x=0); we show these in the graph below with dashed lines. Since the cotangent is the reciprocal of the tangent, (cot x) has vertical asymptotes at all values of (x) where ( an x=0), and (cot x=0) at all values of (x) where ( an x) has its vertical asymptotes. Figure (PageIndex{19}): The cotangent function

## Graphing Variations of (y =cot x)

We can transform the graph of the cotangent in much the same way as we did for the tangent. The equation becomes the following.

[y=Acot(Bx−C)+D]

PROPERTIES OF THE GRAPH OF (y = A cot(Bx-c)+D)

• The stretching factor is (| A |).
• The period is (dfrac{pi}{|B|})
• The domain is (x≠dfrac{C}{B}+dfrac{pi}{| B |}k),where (k) is an integer.
• The range is ((−∞,−|A|]∪[|A|,∞)).
• The vertical asymptotes occur at (x=dfrac{C}{B}+dfrac{pi}{| B |}k),where (k) is an integer.
• There is no amplitude.
• (y=Acot(Bx)) is an odd function because it is the quotient of even and odd functions (cosine and sine, respectively)

Example (PageIndex{8}): Graphing Variations of the Cotangent Function

Determine the stretching factor, period, and phase shift of (y=3cot(4x)), and then sketch a graph.

Solution

• Step 1. Expressing the function in the form (f(x)=Acot(Bx)) gives (f(x)=3cot(4x)).
• Step 2. The stretching factor is (|A|=3).
• Step 3. The period is (P=dfrac{pi}{4}).
• Step 4. Sketch the graph of (y=3 an(4x)).
• Step 5. Plot two reference points. Two such points are (left (dfrac{pi}{16},3 ight )) and (left (dfrac{3pi}{16},−3 ight )).
• Step 6. Use the reciprocal relationship to draw (y=3cot(4x)).
• Step 7. Sketch the asymptotes, (x=0), (x=dfrac{pi}{4}).

The orange graph in Figure (PageIndex{20}) shows (y=3 an(4x)) and the blue graph shows (y=3cot(4x)). Figure (PageIndex{20})

Howto: Given a modified cotangent function of the form (f(x)=Acot(Bx−C)+D), graph one period.

1. Express the function in the form (f(x)=Acot(Bx−C)+D).
2. Identify the stretching factor, (| A |).
3. Identify the period, (P=dfrac{pi}{|B|}).
4. Identify the phase shift, (dfrac{C}{B}).
5. Draw the graph of (y=A an(Bx)) shifted to the right by (dfrac{C}{B}) and up by (D).
6. Sketch the asymptotes (x=dfrac{C}{B}+dfrac{pi}{| B |}k),where (k) is an integer.
7. Plot any three reference points and draw the graph through these points.

Example (PageIndex{9}): Graphing a Modified Cotangent

Sketch a graph of one period of the function (f(x)=4cot left (dfrac{pi}{8}x−dfrac{pi}{2} ight )−2).

Solution

• Step 1. The function is already written in the general form (f(x)=Acot(Bx−C)+D).
• Step 2. (A=4),so the stretching factor is (4).
• Step 3. (B=dfrac{pi}{8}), so the period is (P=dfrac{pi}{| B |}=dfrac{pi}{dfrac{pi}{8}}=8).
• Step 4. (C=dfrac{pi}{2}),so the phase shift is (CB=dfrac{dfrac{pi}{2}}{dfrac{pi}{8}}=4).
• Step 5. We draw (f(x)=4 an left (dfrac{pi}{8}x−dfrac{pi}{2} ight )−2).
• Step 6-7. Three points we can use to guide the graph are ((6,2)), ((8,−2)), and ((10,−6)). We use the reciprocal relationship of tangent and cotangent to draw (f(x)=4cot left (dfrac{pi}{8}x−dfrac{pi}{2} ight )−2).
• Step 8. The vertical asymptotes are (x=4) and (x=12).

The graph is shown in Figure (PageIndex{21}). Figure (PageIndex{21}): One period of a modified cotangent function

## Using the Graphs of Trigonometric Functions to Solve Real-World Problems

Many real-world scenarios represent periodic functions and may be modeled by trigonometric functions. As an example, let’s return to the scenario from the section opener. Have you ever observed the beam formed by the rotating light on a police car and wondered about the movement of the light beam itself across the wall? The periodic behavior of the distance the light shines as a function of time is obvious, but how do we determine the distance? We can use the tangent function.

Example (PageIndex{10}): Using Trigonometric Functions to Solve Real-World Scenarios

Suppose the function (y=5 an(dfrac{pi}{4}t)) marks the distance in the movement of a light beam from the top of a police car across a wall where (t) is the time in seconds and (y) is the distance in feet from a point on the wall directly across from the police car.

1. Find and interpret the stretching factor and period.
2. Graph on the interval ([0,5]).
3. Evaluate (f(1)) and discuss the function’s value at that input.

Solution

1. We know from the general form of (y=A an(Bt)) that (| A |) is the stretching factor and (dfrac{pi}{B}) is the period. Figure (PageIndex{22})

We see that the stretching factor is (5). This means that the beam of light will have moved (5) ft after half the period.

The period is (dfrac{pi}{ frac{pi}{4}}=dfrac{pi}{1}⋅dfrac{4}{pi}=4). This means that every (4) seconds, the beam of light sweeps the wall. The distance from the spot across from the police car grows larger as the police car approaches.

1. To graph the function, we draw an asymptote at (t=2) and use the stretching factor and period. See Figure (PageIndex{23}) Figure (PageIndex{23})

1. period: (f(1)=5 an(frac{pi}{4}(1))=5(1)=5); after (1) second, the beam of has moved (5) ft from the spot across from the police car.

## Key Equations

 Shifted, compressed, and/or stretched tangent function (y=A an(Bx−C)+D) Shifted, compressed, and/or stretched secant function (y=A sec(Bx−C)+D) Shifted, compressed, and/or stretched cosecant function (y=A csc(Bx−C)+D) Shifted, compressed, and/or stretched cotangent function (y=A cot(Bx−C)+D)

## Key Concepts

• The tangent function has period (π).
• (f( x )=A an( Bx−C )+D) is a tangent with vertical and/or horizontal stretch/compression and shift. See Example (PageIndex{1}), Example (PageIndex{2}), and Example (PageIndex{3}).
• The secant and cosecant are both periodic functions with a period of (2pi). (f( x )=Asec( Bx−C )+D) gives a shifted, compressed, and/or stretched secant function graph. See Example (PageIndex{4}) and Example (PageIndex{5}).
• (f( x )=Acsc( Bx−C )+D) gives a shifted, compressed, and/or stretched cosecant function graph. See Example (PageIndex{6}) and Example (PageIndex{7}).
• The cotangent function has period (pi) and vertical asymptotes at (0,±pi,±2pi),....
• The range of cotangent is (( −∞,∞ )), and the function is decreasing at each point in its range.
• The cotangent is zero at (±dfrac{pi}{2},±dfrac{3pi}{2}),....
• (f(x)=Acot(Bx−C)+D) is a cotangent with vertical and/or horizontal stretch/compression and shift. See Example (PageIndex{8}) and Example (PageIndex{9}).
• Real-world scenarios can be solved using graphs of trigonometric functions. See Example (PageIndex{10}).

## 7.3 Unit Circle

Looking for a thrill? Then consider a ride on the Singapore Flyer, the world’s tallest Ferris wheel. Located in Singapore, the Ferris wheel soars to a height of 541 feet—a little more than a tenth of a mile! Described as an observation wheel, riders enjoy spectacular views as they travel from the ground to the peak and down again in a repeating pattern. In this section, we will examine this type of revolving motion around a circle. To do so, we need to define the type of circle first, and then place that circle on a coordinate system. Then we can discuss circular motion in terms of the coordinate pairs.

### Finding Trigonometric Functions Using the Unit Circle

We have already defined the trigonometric functions in terms of right triangles. In this section, we will redefine them in terms of the unit circle. Recall that a unit circle is a circle centered at the origin with radius 1, as shown in Figure 2. The angle (in radians) that t t intercepts forms an arc of length s . s . Using the formula s = r t , s = r t , and knowing that r = 1 , r = 1 , we see that for a unit circle, s = t . s = t .

The x- and y-axes divide the coordinate plane into four quarters called quadrants. We label these quadrants to mimic the direction a positive angle would sweep. The four quadrants are labeled I, II, III, and IV.

### Try It #1

#### Finding Sines and Cosines of Angles on an Axis

For quadrantral angles, the corresponding point on the unit circle falls on the x- or y-axis. In that case, we can easily calculate cosine and sine from the values of x x and y . y .

### Example 2

#### Solution

We can then use our definitions of cosine and sine.

### Try It #2

Find cosine and sine of the angle π . π .

#### The Pythagorean Identity

Now that we can define sine and cosine, we will learn how they relate to each other and the unit circle. Recall that the equation for the unit circle is x 2 + y 2 = 1. x 2 + y 2 = 1. Because x = cos t x = cos t and y = sin t , y = sin t , we can substitute for x x and y y to get cos 2 t + sin 2 t = 1. cos 2 t + sin 2 t = 1. This equation, cos 2 t + sin 2 t = 1 , cos 2 t + sin 2 t = 1 , is known as the Pythagorean Identity . See Figure 7.

We can use the Pythagorean Identity to find the cosine of an angle if we know the sine, or vice versa. However, because the equation yields two solutions, we need additional knowledge of the angle to choose the solution with the correct sign. If we know the quadrant where the angle is, we can easily choose the correct solution.

### Pythagorean Identity

The Pythagorean Identity states that, for any real number t , t ,

### Example 3

#### Solution

Substituting the known value for sine into the Pythagorean Identity,

Because the angle is in the second quadrant, we know the x-value is a negative real number, so the cosine is also negative.

### Finding Sines and Cosines of Special Angles

We have already learned some properties of the special angles, such as the conversion from radians to degrees, and we found their sines and cosines using right triangles. We can also calculate sines and cosines of the special angles using the Pythagorean Identity.

#### Finding Sines and Cosines of 45° 45° Angles

From the Pythagorean Theorem we get

We can then substitute y = x . y = x .

Next we combine like terms.

If we then rationalize the denominators, we get

#### Finding Sines and Cosines of 30° 30° and 60° 60° Angles

Next, we will find the cosine and sine at an angle of 30° , 30° , or π 6 . π 6 . First, we will draw a triangle inside a circle with one side at an angle of 30° , 30° , and another at an angle of −30° , −30° , as shown in Figure 11. If the resulting two right triangles are combined into one large triangle, notice that all three angles of this larger triangle will be 60° , 60° , as shown in Figure 12.

Because all the angles are equal, the sides are also equal. The vertical line has length 2 y , 2 y , and since the sides are all equal, we can also conclude that r = 2 y r = 2 y or y = 1 2 r . y = 1 2 r . Since sin t = y , sin t = y ,

Using the Pythagorean Identity, we can find the cosine value.

From the Pythagorean Theorem, we get

We have now found the cosine and sine values for all of the most commonly encountered angles in the first quadrant of the unit circle. Table 1 summarizes these values.

Figure 14 shows the common angles in the first quadrant of the unit circle.

#### Using a Calculator to Find Sine and Cosine

To find the cosine and sine of angles other than the special angles, we turn to a computer or calculator. Be aware: Most calculators can be set into “degree” or “radian” mode, which tells the calculator the units for the input value. When we evaluate cos ( 30 ) cos ( 30 ) on our calculator, it will evaluate it as the cosine of 30 degrees if the calculator is in degree mode, or the cosine of 30 radians if the calculator is in radian mode.

Given an angle in radians, use a graphing calculator to find the cosine.

1. If the calculator has degree mode and radian mode, set it to radian mode.
2. Press the COS key.
3. Enter the radian value of the angle and press the close-parentheses key ")".
4. Press ENTER.

### Example 4

#### Solution

Enter the following keystrokes:

#### Analysis

We can find the cosine or sine of an angle in degrees directly on a calculator with degree mode. For calculators or software that use only radian mode, we can find the sine of 20° , 20° , for example, by including the conversion factor to radians as part of the input:

### Identifying the Domain and Range of Sine and Cosine Functions

Now that we can find the sine and cosine of an angle, we need to discuss their domains and ranges. What are the domains of the sine and cosine functions? That is, what are the smallest and largest numbers that can be inputs of the functions? Because angles smaller than 0 0 and angles larger than 2 π 2 π can still be graphed on the unit circle and have real values of x , y , x , y , and r , r , there is no lower or upper limit to the angles that can be inputs to the sine and cosine functions. The input to the sine and cosine functions is the rotation from the positive x-axis, and that may be any real number.

What are the ranges of the sine and cosine functions? What are the least and greatest possible values for their output? We can see the answers by examining the unit circle, as shown in Figure 15. The bounds of the x-coordinate are [ −1 , 1 ] . [ −1 , 1 ] . The bounds of the y-coordinate are also [ −1 , 1 ] . [ −1 , 1 ] . Therefore, the range of both the sine and cosine functions is [ −1 , 1 ] . [ −1 , 1 ] .

### Finding Reference Angles

We have discussed finding the sine and cosine for angles in the first quadrant, but what if our angle is in another quadrant? For any given angle in the first quadrant, there is an angle in the second quadrant with the same sine value. Because the sine value is the y-coordinate on the unit circle, the other angle with the same sine will share the same y-value, but have the opposite x-value. Therefore, its cosine value will be the opposite of the first angle’s cosine value.

Likewise, there will be an angle in the fourth quadrant with the same cosine as the original angle. The angle with the same cosine will share the same x-value but will have the opposite y-value. Therefore, its sine value will be the opposite of the original angle’s sine value.

Recall that an angle’s reference angle is the acute angle, t , t , formed by the terminal side of the angle t t and the horizontal axis. A reference angle is always an angle between 0 0 and 90° , 90° , or 0 0 and π 2 π 2 radians. As we can see from Figure 17, for any angle in quadrants II, III, or IV, there is a reference angle in quadrant I.

1. An angle in the first quadrant is its own reference angle.
2. For an angle in the second or third quadrant, the reference angle is | π − t | | π − t | or | 180° − t | . | 180° − t | .
3. For an angle in the fourth quadrant, the reference angle is 2 π − t 2 π − t or 360° − t . 360° − t .
4. If an angle is less than 0 0 or greater than 2 π , 2 π , add or subtract 2 π 2 π as many times as needed to find an equivalent angle between 0 0 and 2 π . 2 π .

### Example 5

#### Finding a Reference Angle

Find the reference angle of 225° 225° as shown in Figure 18.

#### Solution

Find the reference angle of 5 π 3 . 5 π 3 .

### Using Reference Angles

Now let’s take a moment to reconsider the Ferris wheel introduced at the beginning of this section. Suppose a rider snaps a photograph while stopped twenty feet above ground level. The rider then rotates three-quarters of the way around the circle. What is the rider’s new elevation? To answer questions such as this one, we need to evaluate the sine or cosine functions at angles that are greater than 90 degrees or at a negative angle. Reference angles make it possible to evaluate trigonometric functions for angles outside the first quadrant. They can also be used to find ( x , y ) ( x , y ) coordinates for those angles. We will use the reference angle of the angle of rotation combined with the quadrant in which the terminal side of the angle lies.

#### Using Reference Angles to Evaluate Trigonometric Functions

We can find the cosine and sine of any angle in any quadrant if we know the cosine or sine of its reference angle. The absolute values of the cosine and sine of an angle are the same as those of the reference angle. The sign depends on the quadrant of the original angle. The cosine will be positive or negative depending on the sign of the x-values in that quadrant. The sine will be positive or negative depending on the sign of the y-values in that quadrant.

### Using Reference Angles to Find Cosine and Sine

Angles have cosines and sines with the same absolute value as their reference angles. The sign (positive or negative) can be determined from the quadrant of the angle.

Given an angle in standard position, find the reference angle, and the cosine and sine of the original angle.

## 7.3: Graphs of the Other Trigonometric Functions

page 531:

problems 5, 12, 17, 29, 33

Possible Classroom Examples:

• Graph the function on the interval . What is the period? What is the phase shift? What is the domain? What is the range? Are there any asymptotes? If yes, what are they? Is the function even, odd or neither?
• Graph the function on the interval . What is the period? What is the phase shift? What is the domain? What is the range? Are there any asymptotes? If yes, what are they? Is the function even, odd or neither?
• Graph the function on the interval . What is the period? What is the phase shift? What is the domain? What is the range? Are there any asymptotes? If yes, what are they? Is the function even, odd or neither?
• Graph the function on the interval . What is the period? What is the phase shift? What is the domain? What is the range? Are there any asymptotes? If yes, what are they? Is the function even, odd or neither?
• Graph .
• Graph .
• Graph .
• Graph .
• Graph .

## Examples

If sin A  =  9/15, find the other trigonometric ratios

sin  θ  =  Opposite side/hypotenuse side

Opposite side  =  9, Hypotenuse side  =  15

(Hypotenuse side) 2   =  (Opposite side) 2 + (Adjacent side) 2

cos A  =  Adjacent  side/hypotenuse side   =  12/15

cosec A   =  Hypotenuse side/Opposite side  =  15/9

sec A  =  H ypotenuse side/Adjacent side   =  15/12

If cos A  =  15/17, find the other trigonometric ratios

cos  θ  =  Adjacent side/hypotenuse side

Adjacent side  =  15, Hypotenuse side  =  17

(Hypotenuse side) 2   =  (Opposite side) 2  + (Adjacent side) 2

sin A  =  Opposite  side/hypotenuse side   =  8/17

cosec A   =  Hypotenuse side/Opposite side  =  17/8

sec A  =  H ypotenuse side/Adjacent side   =  17/15

If sec  θ   =  17/8, find the other trigonometric ratios

sec  θ  =  Hypotenuse side/Adjacent side

Hypotenuse side  =  17, Adjacent side  =  8

(Hypotenuse side) 2   =  (Opposite side) 2  + (Adjacent side) 2

cosec  θ   =  H ypotenuse side/Opposite side   =  17/15 Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.

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## 7.3: Graphs of the Other Trigonometric Functions

### Author’s word

These math lessons has been written especially to meet the requirements of higher grade students. I’ve tried my best to present the work in a clear, simple and easy style so that students may not face any difficulty. Each lesson has solved examples and practice problems with answers.

While adding new topics is an ongoing process, efforts has been made to put the concepts in a logical sequence. In spite of my best efforts to make these lessons error free, some typing errors might have gone unnoticed. I shall be grateful to generous fellows if same are brought to my notice.

Worthy suggestions for improvement of these math lessons are always welcome.

• All Basic Trigonometric functions
• All Trigonometric Identities and Formulas
• Applications of Right triangle Trigonometry
• Co-terminal Angles
• De Moivre’s Theorem and nth Roots
• Degree, Radians and their Conversions
• Evaluating Trigonometric functions
• Ferris Wheel problems (applications of trigonometric functions)
• Graphing Sine and Cosine functions(stretching & shrinking)
• Graphing Sine and Cosine functions ( vertical & Horizontal Translation)
• Graphs of other Trigonometric functions (tanx , cotx, secx, cscx)
• Law of Cosines and its Applications
• Law of Sines and its Applications
• Reference Angle
• SOHCAHTOA rule and word problems.
• Solving Trigonometric equations
• Verify Trigonometric Identities
• Applications of exponential functions (Compound interest)
• Applications of exponential and logarithmic functions (Population and bacteria growth)
• Applications of exponential functions (Intensity of earthquakes and sound loudness)
• Domain and Range of log function
• Exponential and Logarithmic functions
• Log Functions and their Inverse
• Logistic Functions and their graphs
• Simplify Exponential expressions
• Simplify logarithmic expressions
• Solving Exponential Equations
• Solving Logarithmic equations
• Transforming and Graphing logarithmic functions

### Ferris Wheel problems (applications of trigonometric functions)

Ferris Wheel (applications of trigonometric functions)

One of the most common applications of trigonometric functions is, Ferris wheel, since the up and down motion of a rider follows the shape of sine or cosine graph. Equations used :

Y = aSin(bx-c)+d or

Y = aCos(bx-c)+d

Formula used :

vertical shift d=(max+min)/2

Example1. A Ferris wheel has a diameter of 30 m with its center 18 m above the ground. It makes one complete rotation every 60 seconds. Assuming rider starts at the lowest point, find the trigonometric function for this situation and graph the function.

Amplitude – radius of the wheel makes the amplitude so amplitude(a) = 30/2 =15.

Period– Wheel complete one rotation in 60 seconds so period is 60 sec. Using period we can find b value as,

Phase shift– There is no phase shift for this cosine function so no c value.

Vertical shift– Centre of wheel is 18m above the ground which makes the mid line, so d= 18.

Lowest point would be 18-15=3m and highest point would be 18+15= 33m above the ground. So the rider will start from 3m and reach to a height of 33 m in half the period (30 sec) and come back to lowest point (3m) again in 60 secs. So its graph would look like this. As the graph start from lowest point and the pattern is upside down so we put a negative sign in front of cos. Summing up all the parameters above we get trigonometric function as,

Example2. A water wheel on a paddle boat has a radius of 2 m. The wheel rotates every 30 secs and bottom 0.6m of wheel is submerged in water.

1. Considering the water surface as x axis , determine the cosine equation of the graph starting from a point at the top of wheel.
2. Graph the height of a point on the wheel relative to the surface of water, starting from highest point.
3. How long is the point on wheel under water.

Radius of wheel gives the amplitude so a= 2

Since radius is 2 and bottom point is -0.6, mid line will be at 2-0.6 =1.4m

Combining all the above parameters and considering top point as starting point , we get the cosine equation as, c) To find the time for wheel under water we need to find x intercepts , intersection of cosine function with y=0(water surface) using graphic calculator. We get the intersection points as x=11.2s and x=18.8s So the total time for wheel underwater is 18.8-11.2 = 6 seconds.

Example3. The earliest sunset occurs at 5:34 PM on Dec. 21 and latest at 11:45 PM on June 21.

1. Write cosine equation of the graph.
2. Draw the graph approximating the sunrise time during the year.
3. What is the sunset time on April 6.
4. The sunset time is earlier than 8PM for what percentage of the year.

First of all we need to convert time from hour/min to decimal hour form. 5PM is equal to 17 hours and 34 min. are equal to 34/60=0.57 so we get 5:34PM equal to 17.57 hrs.

Same way we get 11:45PM equal to 23.75 hrs. because 11PM is equal to 12+11=23hrs and 45 min = 45/60 =0.75hrs.

Using these maximum and minimum values we get amplitude

For these type of problems, period is taken as 365 days. so,

Starting the graph on Jan1, max. value occurs on 21 June so

c = 31+28+31+30+31+21=172 days

d = (23.75+17.57)/2 = 20.66

Combining all above parameters , we get cosine function as,

This equation can also be written as,

considering 21 Dec. as lowest point. 3) To find sunset time on April6, we find what day of year it is.

So we plug in x as 96 into the equation found in part a.

Converting back 21.46 into decimal hour form we get,

So sunset time on April 6 is 9:28PM.

4) Sunset is earlier than 8PM for days 0 to 68 and then again days 276 to 365. So that total number of days are 157 which are 43 % of the year.

Example 4: The following table gives the average recorded monthly temperature throughout the year. Write the cosine equation for the graph corresponding to the table given above.

Amplitude, a = [22-(-17)]/2 =39/2 = 19.5

Period = 12 months, here months are used instead of days.

Since the maximum temp. occur in the month of July which is the 7 th month so there is a phase shift of 7.

Vertical shift d =[22+(-17)]/2 = 5/2 =2.5

Combining all the parameters above, we get the final equation as,

Where x represents number of months and y represents approximate temperature.

Practice problems:

1) A Ferris wheel with radius 40 ft complete one revolution every 60 seconds. The lowest point of wheel is 5 m above the ground.

• Draw the graph of the situation, starting with a person getting on the bottom of the wheel at t=0 seconds.
• Determine an equation representing the path of the person on Ferris wheel.
• Determine how high the person will be after riding for 40 seconds.
• When the person first reach 50 ft.

2) The bottom of a windmill is 8m above the ground, and the top is 22m above the ground. The wheel rotates once every 5 seconds.

3)The average temp. for Regina is hottest at 27 on July 28, and coolest at -16 on January 10.

• Draw the graph and write the cosine equation for the graph.
• The average temp. is higher than 23 for how many days.

4)The latest sunrise occurs at 9:10 AM on Dec 21. The earliest occurs at 3:43 AM on June 21. Write the cosine equation for the graph.

## 7.3: Graphs of the Other Trigonometric Functions

The next trig function is the tangent, but that's difficult to show on the unit circle. So let's take a closer look at the sine and cosines graphs, keeping in mind that tan(&theta) = sin(&theta)/cos(&theta) .

The tangent will be zero wherever its numerator (the sine) is zero. This happens at 0 , &pi , 2&pi , 3&pi , etc, and at &ndash&pi , &ndash2&pi , &ndash3&pi , etc. Let's just consider the region from &ndash&pi to 2&pi , for now. So the tangent will be zero (that is, it will cross the x-axis) at &ndash&pi , 0 , &pi , and 2&pi .

The tangent will be undefined wherever its denominator (the cosine) is zero. Thinking back to when you learned about graphing rational functions, a zero in the denominator means you'll have a vertical asymptote. So the tangent will have vertical asymptotes wherever the cosine is zero: at &ndash&pi/2 , &pi/2 , and 3&pi/2 . Let's put dots for the zeroes and dashed vertical lines for the asymptotes:

Now we can use what we know about sine, cosine, and asymptotes to fill in the rest of the tangent's graph: We know that the graph will never touch or cross the vertical asymptotes we know that, between a zero and an asymptote, the graph will either be below the axis (and slide down the asymptote to negative infinity) or else be above the axis (and skinny up the asymptote to positive infinity). Between zero and &pi/2 , sine and cosine are both positive. This means that the tangent, being their quotient, is positive, so the graph slides up the asymptote: Copyright © Elizabeth Stapel 2010-2011 All Rights Reserved

Between &pi/2 and &pi , sine is positive but cosine is negative. These opposite signs mean that the tangent quotient will be negative, so it will come up the asymptote from below, to meet the x -axis at x = &pi :

Since sine and cosine are periodic, then tangent has to be, as well. A quick check of the signs tells us how to fill in the rest of the graph:

• &ndash&pi to &ndash&pi/2 : sine is negative and cosine is negative, so tangent is positive
• &ndash&pi/2 to 0 : sine is negative but cosine is positive, so tangent is negative
• &pi to 3&pi/2 : sine is negative and cosine is negative, so tangent is positive
• 3&pi/2 to 2&pi : sine is negative but cosine is positive, so tangent is negative

Now we can complete our graph:

The Tangent Graph

As you can see, the tangent has a period of &pi , with each period separated by a vertical asymptote. The concept of "amplitude" doesn't really apply.

For graphing, draw in the zeroes at x = 0 , &pi , 2&pi , etc, and dash in the vertical asymptotes midway between each zero. Then draw in the curve. You can plot a few more points if you like, but you don't generally gain much from doing so.

If you prefer memorizing graphs, then memorize the above. But I always had trouble keeping straight anything much past sine and cosine, so I used the reasoning demonstrated above to figure out the tangent (and the other trig) graphs. As long as you know your sines and cosines very well, you'll be able to figure out everything else.

## Frequency

It is the number of times something happens per unit of time.

Example – there is the sine function which is repeated 4 times between 0 and 1 –

Thus, the frequency is 4. The frequency and period are related to each other –
Frequency = 1 / period
Period = 1 / frequency

Example – 3 sin (100 (t + 0.01))

Here the period is 0.02π, thus the frequency will be 1 / 0.02π = 50π.

Thus, now you are clear with all the terms described and explained above, with examples.

## Trigonometric functions

This online calculator computes the values of elementary trigonometric functions, such as sin, cos, tg, ctg, sec, cosec for an angle, which can be set in degrees, radians, or grads.

Trigonometric functions are the set of elementary functions that relates the angles of a triangle to the lengths of the sides of the triangle. They're also called circular functions. See picture.

The trigonometric functions are:
sin — sine
cos — cosine
tg — tangent
ctg — cotangent
sec — secant
cosec — cosecant
versin — versine (versed sine)
vercos — vercosine (versed cosine)
haversin — haversed sine
exsec — exsecant
excsc — excosecant

To compute these functions, enter the angle value in the Angle field and get the table of results. The angle can be entered in degrees, radians, grads, minutes, or seconds.

Find the points at which f is discontinuous. At which of these points f is continuous from the right, from the left, or neither? Sketch the graph of f. First let us check the continuity at the point x  =  -1

By applying the limit, we get

By applying the limit, we get

So, the function is not continuous at x = -1.

Now let us check the continuity at the point x  =  1

By applying the limit, we get

By applying the limit, we get

So, the function is not continuous at x = 1.

To find a t which of these points f is continuous from the right, from the left, or neither, we have to draw the number line. Applying the limit, we get

Applying the limit, we get

Applying the limit, we get   Find the points at which f is discontinuous. At which of these points f is continuous from the right, from the left, or neither? Sketch the graph of f. First let us check the continuity at the point x  =  0

By applying the limit, we get

By applying the limit, we get

So, the function is not continuous at x = 0.

To find at which of these points f is continuous from the right, from the left, or neither, we have to draw the number line. Applying the limit, we get

Applying the limit, we get  After having gone through the stuff given above, we hope that the students would have understood, " How to Sketch the Graph and Find Continuity of Functions"

Apart from the stuff given in " How to Sketch the Graph and Find Continuity of Functions" ,   if you need any other stuff in math, please use our google custom search here.

If you have any feedback about our math content, please mail us :

You can also visit the following web pages on different stuff in math.

## Trigonometric Function and the Unit Circle The unit circle is often used to define a trigonometric function, like the versine.

A unit circle has a radius of 1, centered at the origin (0, 0) of the Cartesian plane. Many trigonometric functions are defined in terms of the unit circle, including the sine function, cosine function and tangent function. That’s why trig functions are sometimes called “circular” functions.