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Back to topObjectives Sine, Cosine, and the other ratios Beyond the Unit Circle Practice Problems

Section 5.2: The Other Circular Functions

Learning Objectives

In this section you will:

  • Discover the connection between sine, cosine, and the other trigonometric ratios.
  • Evaluate the values of all trig ratios for special unit circle angles.
  • Evaluate the values of trig ratios beyond the unit circle.

A wheelchair ramp that meets the standards of the Americans with Disabilities Act must make an angle with the ground whose tangent is $\frac{1}{12}$ or less, regardless of its length. A tangent represents a ratio, so this means that for every 1 inch of rise, the ramp must have 12 inches of run. Trigonometric functions allow us to specify the shapes and proportions of objects independent of exact dimensions. We have already defined the sine and cosine functions of an angle. Though sine and cosine are the trigonometric functions most often used, there are four others. Together they make up the set of six trigonometric functions. In this section, we will investigate the remaining functions.


Connecting Sine and Cosine to the other Trigonometric Functions

We can also define the remaining functions in terms of the unit circle with a point (x,y) corresponding to an angle of $\theta$, as shown below:

A unit circle with a reference triangle in quadrant I. The legs are labeled x and y, and the hypotenuse is labeled r = 1.

From right triangle trigonometry, we know that $$\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$$ which in the case of our triangle here, means $$\tan(\theta) = \frac{y}{x}.$$

In a similar way, we can connect all of our trigonometric functions to the coordinates of the unit circle.

The Circular Functions

Suppose an angle $\theta$ is graphed in standard position. Let $P(x,y)$ be the point of intersection of the terminal side of $P$ and the Unit Circle.

  • The sine of $\theta$, denoted $\sin(\theta)$, is defined by $\sin(\theta) = y$.


  • The cosine of $\theta$, denoted $\cos(\theta)$, is defined by $\cos(\theta) = x$.


  • The tangent of $\theta$, denoted $\tan(\theta)$, is defined by $\tan(\theta) = \dfrac{y}{x}$, provided $x \neq 0$.


  • The secant of $\theta$, denoted $\sec(\theta)$, is defined by $\sec(\theta) = \dfrac{1}{x}$, provided $x \neq 0$.


  • The cosecant of $\theta$, denoted $\csc(\theta)$, is defined by $\csc(\theta) = \dfrac{1}{y}$, provided $y \neq 0$.


  • The cotangent of $\theta$, denoted $\cot(\theta)$, is defined by $\cot(\theta) = \dfrac{x}{y}$, provided $y \neq 0$.



In Section 5.1, we saw that we can define the $x$ coordinate of that point on the unit circle as the $\cos(\theta)$, and the $y$ coordinate as $\sin(\theta)$. Now, consider the tangent ratio. It then follows that $$\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}.$$

Similarly we can show that $\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}$. These facts are true for any value of $\theta$; therefore, we call them identities.


The Quotient Identities

  • $\tan(\theta) = \dfrac{\sin(\theta)}{\cos(\theta)}$, provided $\cos(\theta) \neq 0$; if $\cos(\theta) = 0$, $\tan(\theta)$ is undefined.


  • $\cot(\theta) = \dfrac{\cos(\theta)}{\sin(\theta)}$, provided $\sin(\theta) \neq 0$; if $\sin(\theta) = 0$, $\cot(\theta)$ is undefined.

We had also defined the cosecant and secant functions by their relationship to sine and cosine. We'll now call those definitions the reciprocal identities.


The Reciprocal Identities

  • $\sec(\theta) = \dfrac{1}{\cos(\theta)}$, provided $\cos(\theta) \neq 0$; if $\cos(\theta) = 0$, $\sec(\theta)$ is undefined.


  • $\csc(\theta) = \dfrac{1}{\sin(\theta)}$, provided $\sin(\theta) \neq 0$; if $\sin(\theta) = 0$, $\csc(\theta)$ is undefined.



Find $\tan\left(\frac{\pi}{6}\right)$ and $\sec\left(\frac{\pi}{6}\right)$.


Since $\frac{\pi}{6}$ is a special angle, we know (or should know soon after spending some time practicing) that $$\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$$ and $$\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$$ We also know from the quotient identities that $$\tan\left(\frac{\pi}{6}\right) = \frac{\sin\left(\frac{\pi}{6}\right)}{\cos\left(\frac{\pi}{6}\right)}.$$ So substituting in our values for the sine and cosine, we have: $$\tan\left(\frac{\pi}{6}\right) = \frac{1/2}{\sqrt{3}/2}$$ $$=\frac{1}{2}\cdot\frac{2}{\sqrt{3}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$$


Now, since the reciprocal identities tell us that $$\sec\left(\frac{\pi}{6}\right) = \frac{1}{\cos\left(\frac{\pi}{6}\right)}$$ we can substitute in our value for cosine to get $$\sec\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}{2}}$$ $$=\frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$$



Hence our final answer is $\tan\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{3}$ and $\sec\left(\frac{\pi}{6}\right) =\frac{2\sqrt{3}}{3}$.



Find the exact value of $\csc\left(\frac{14\pi}{3}\right)$ and $\cot\left(\frac{14\pi}{3}\right)$


$\frac{14\pi}{3}$ is simply too big for us to deal with right now, so let's find a smaller coterminal angle: $$\frac{14\pi}{3} - 2\pi$$ $$=\frac{14\pi}{3} - \frac{6\pi}{3}$$ $$= \frac{8\pi}{3}$$ Still too big! Let's go smaller... $$\frac{8\pi}{3} - \frac{6\pi}{3}$$ $$= \frac{2\pi}{3}$$ That's an angle we recognize from our common unit circle values! Since this is coterminal with our original angle, $$\cos\left(\frac{14\pi}{3}\right) = \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}$$ $$\sin\left(\frac{14\pi}{3}\right) = \sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2}$$


Now we turn to the question at hand. Using reciprocal identities, we have $$\csc\left(\frac{14\pi}{3}\right) = \frac{1}{\sin\left(\frac{14\pi}{3}\right)}$$ $$=\frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$$

Then, using quotient identities, we have $$\cot\left(\frac{14\pi}{3}\right) = \frac{\cos\left(\frac{14\pi}{3}\right)}{\sin\left(\frac{14\pi}{3}\right)}$$ $$= \frac{-1/2}{\sqrt{3}/2} = -\frac{1}{2}\cdot-\frac{2}{\sqrt{3}}$$ $$= -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}$$

Therefore, our final answers are $\csc\left(\frac{14\pi}{3}\right) = \frac{2\sqrt{3}}{3}$ and $\cot\left(\frac{14\pi}{3}\right)= -\frac{\sqrt{3}}{3}$.









Beyond the Unit Circle...Again

Once again, we can extend our definitions of these trigonometric functions beyond the unit circle using similar triangles, like we did in Section 5.1.

Circular Functions Theorem

Suppose $Q(x,y)$ is the point on the terminal side of an angle $\theta$ (plotted in standard position) which lies on the circle of radius $r$, $x^2+y^2 = r^2$. Then:

  • $\sin(\theta) = \dfrac{y}{r} = \dfrac{y}{\sqrt{x^2+y^2}}$


  • $\cos(\theta)= \dfrac{x}{r} = \dfrac{x}{\sqrt{x^2+y^2}}$


  • $\tan(\theta) = \dfrac{y}{x}$, provided $x \neq 0$.


  • $\sec(\theta) = \dfrac{r}{x} = \dfrac{\sqrt{x^2+y^2}}{x}$, provided $x \neq 0$.


  • $\csc(\theta) = \dfrac{r}{y} = \dfrac{\sqrt{x^2+y^2}}{y}$, provided $y \neq 0$.


  • $\cot(\theta) = \dfrac{x}{y}$, provided $y \neq 0$.



Suppose the terminal side of $\theta$, when plotted in standard position, contains the point $Q(3,4)$. Find the values of the six circular functions of $\theta$.


Since $x = 3$ and $y=4$, from $x^2+y^2 = r^2$, $(3)^2+(4)^2 = r^2$ so $r^2 = 25$, or $r = 5$. Our Circular Functions Theorem tells us $\sin(\theta) = \frac{4}{5}$, $\cos(\theta) = \frac{3}{5}$, $\tan(\theta) = \frac{4}{3}$, $\sec(\theta) = \frac{5}{3}$, $\csc(\theta) = \frac{5}{4}$, and $\cot(\theta) = \frac{3}{4}$.



Suppose $\theta$ is a Quadrant IV angle with $\cot(\theta) = -4$. Find the values of the five remaining circular functions of $\theta$.

In order to use Circular Functions Theorem, we need to find a point $Q(x,y)$ which lies on the terminal side of $\theta$, when $\theta$ is plotted in standard position.


We have that $\cot(\theta) = -4 = \frac{x}{y}$. Since $\theta$ is a Quadrant IV angle, we also know $x\gt 0$ and $y \lt 0$. Rewriting $-4 = \frac{4}{-1}$, we choose\footnote{We could have just as easily chosen $x=8$ and $y=-2$ - just so long as $x \gt 0$, $y\lt 0$ and $\frac{x}{y} = -4$.} $x = 4$ and $y = -1$ so that $r = \sqrt{x^2+y^2} = \sqrt{(4)^2 + (-1)^2} = \sqrt{17}$.


Applying Circular Functions Theorem, we find $\sin(\theta) =- \frac{1}{\sqrt{17}} = -\frac{\sqrt{17}}{17}$, $\cos(\theta) = \frac{4}{\sqrt{17}} = \frac{4 \sqrt{17}}{17}$, $\tan(\theta) = -\frac{1}{4}$, $\sec(\theta) = \frac{\sqrt{17}}{4}$, and $\csc(\theta) = - \sqrt{17}$.



Find $\cos\left(\theta\right)$, where $\tan(\theta) = 3$ and $\pi \lt \theta \lt \frac{3\pi}{2}$.

We are told $\tan(\theta) = 3$ and $\pi \lt \theta \lt \frac{3\pi}{2}$, so we know $\theta$ is a Quadrant III angle.


To find $\cos(\theta)$ using Circular Functions Theorem, we need to find the $x$-coordinate of a point $Q(x,y)$ on the terminal side of $\theta$, when $\theta$ is plotted in standard position, and the corresponding radius, $r$.


Since $\tan(\theta) = \frac{y}{x}$ and $\theta$ is a Quadrant III angle, we rewrite $\tan(\theta) = 3 = \frac{-3}{-1} = \frac{y}{x}$ and choose $x = -1$ and $y = -3$. From $x^2+y^2 = r^2$, we get $r = \sqrt{10}$.


Hence, $\cos(\theta) = \frac{x}{r} = \frac{-1}{\sqrt{10}} = -\frac{\sqrt{10}}{10}$.



Practice Problems

In the following exercises, find the exact value of the cosine and sine of the given angle.

1. $\tan \left( \dfrac{\pi}{4} \right)$ {$\csc \left( \dfrac{5\pi}{6} \right)$}

2. $\sec \left( \dfrac{\pi}{6} \right)$ {$\csc \left( \dfrac{5\pi}{6} \right)$}

3. $\csc \left( \dfrac{5\pi}{6} \right)$

4. $\cot \left( \dfrac{4\pi}{3} \right)$

5. $\tan \left( -\dfrac{11\pi}{6} \right)$

6. $\sec \left( -\dfrac{3\pi}{2} \right)$

7. $\csc \left( -\dfrac{\pi}{3} \right)$

8. $\cot \left( \dfrac{13\pi}{2} \right)$

9. $\tan \left( 117\pi \right)$

10. $\sec \left( -\dfrac{5\pi}{3} \right)$

11. $\csc \left( 3\pi \right)$

12. $\cot \left( -5\pi \right)$

13. $\tan \left( \dfrac{31\pi}{2} \right)$

14. $\sec \left( \dfrac{\pi}{4} \right)$

15. $\csc \left( -\dfrac{7\pi}{4} \right)$

16. $\cot \left( \dfrac{7\pi}{6} \right)$

17. $\tan \left( \dfrac{2\pi}{3} \right)$

18. $\sec \left( -7\pi \right)$

19. $\csc \left( \dfrac{\pi}{2} \right)$

20. $\cot \left( \dfrac{3\pi}{4} \right)$

1.$\tan \left( \dfrac{\pi}{4} \right) = 1$

2. $\sec \left( \dfrac{\pi}{6} \right) = \dfrac{2\sqrt{3}}{3}$

3. $\csc \left( \dfrac{5\pi}{6} \right) = 2$

4. $\cot \left( \dfrac{4\pi}{3} \right) = \dfrac{\sqrt{3}}{3}$

5. $\tan \left( -\dfrac{11\pi}{6} \right) = \dfrac{\sqrt{3}}{3}$

6. $\sec \left( -\dfrac{3\pi}{2} \right)$ is undefined

7. $\csc \left( -\dfrac{\pi}{3} \right) = -\dfrac{2\sqrt{3}}{3}$

8. $\cot \left( \dfrac{13\pi}{2} \right) = 0$

9. $\tan \left( 117\pi \right) = 0$

10. $\sec \left( -\dfrac{5\pi}{3} \right) = 2$

11.$\csc \left( 3\pi \right)$ is undefined

12. $\cot \left( -5\pi \right)$ is undefined

13. $\tan \left( \dfrac{31\pi}{2} \right)$ is undefined

14. $\sec \left( \dfrac{\pi}{4} \right) = \sqrt{2}$

15. $\csc \left( -\dfrac{7\pi}{4} \right) = \sqrt{2}$

16. $\cot \left( \dfrac{7\pi}{6} \right) = \sqrt{3}$

17. $\tan \left( \dfrac{2\pi}{3} \right) = -\sqrt{3}$

18. $\sec \left( -7\pi \right) = -1$

19. $\csc \left( \dfrac{\pi}{2} \right) = 1$

20. $\cot \left( \dfrac{3\pi}{4} \right) = -1$



In the following exercises, use the given the information to find the exact values of the circular functions of $\theta$.

21. $\sin(\theta) = \dfrac{3}{5}$ with $\theta$ in Quadrant II

22. $\tan(\theta) = \dfrac{12}{5}$ with $\theta$ in Quadrant III

23. $\csc(\theta) = \dfrac{25}{24}$ with $\theta$ in Quadrant I

24. $\sec(\theta) = 7$ with $\theta$ in Quadrant IV

25. $\csc(\theta) = -\dfrac{10\sqrt{91}}{91}$ with $\theta$ in Quadrant III

26. $\cot(\theta) = -23$ with $\theta$ in Quadrant II

27. $\tan(\theta) = -2$ with $\theta$ in Quadrant IV.

28. $\sec(\theta) = -4$ with $\theta$ in Quadrant II.

29. $\cot(\theta) = \sqrt{5}$ with $\theta$ in Quadrant III.

30. $\cos(\theta) = \dfrac{1}{3}$ with $\theta$ in Quadrant I.

31. $\cot(\theta) = 2$ with $0 \lt \theta \lt \dfrac{\pi}{2}$.

32. $\csc(\theta) = 5$ with $\dfrac{\pi}{2} \lt \theta \lt \pi$.

33. $\tan(\theta) = \sqrt{10}$ with $\pi \lt \theta \lt \dfrac{3\pi}{2}$.

34. $\sec(\theta) = 2\sqrt{5}$ with $\dfrac{3\pi}{2} \lt \theta \lt 2\pi$.

21. $\sin(\theta) = \frac{3}{5}, \cos(\theta) = -\frac{4}{5}, \tan(\theta) = -\frac{3}{4}, \csc(\theta) = \frac{5}{3}, \sec(\theta) = -\frac{5}{4}, \cot(\theta) = -\frac{4}{3}$

22. $\sin(\theta) = -\frac{12}{13}, \cos(\theta) = -\frac{5}{13}, \tan(\theta) = \frac{12}{5}, \csc(\theta) = -\frac{13}{12}, \sec(\theta) = -\frac{13}{5}, \cot(\theta) = \frac{5}{12}$

23. $\sin(\theta) = \frac{24}{25}, \cos(\theta) = \frac{7}{25}, \tan(\theta) = \frac{24}{7}, \csc(\theta) = \frac{25}{24}, \sec(\theta) = \frac{25}{7}, \cot(\theta) = \frac{7}{24}$

24. $\sin(\theta) = \frac{-4\sqrt{3}}{7}, \cos(\theta) = \frac{1}{7}, \tan(\theta) = -4\sqrt{3}, \csc(\theta) = -\frac{7\sqrt{3}}{12}, \sec(\theta) = 7, \cot(\theta) = -\frac{\sqrt{3}}{12}$

25. $\sin(\theta) = -\frac{\sqrt{91}}{10}, \cos(\theta) = -\frac{3}{10}, \tan(\theta) = \frac{\sqrt{91}}{3}, \csc(\theta) = -\frac{10\sqrt{91}}{91}, \sec(\theta) = -\frac{10}{3}, \cot(\theta) = \frac{3\sqrt{91}}{91}$

26. $\sin(\theta) = \frac{\sqrt{530}}{530}, \cos(\theta) = -\frac{23\sqrt{530}}{530}, \tan(\theta) = -\frac{1}{23}, \csc(\theta) = \sqrt{530}, \sec(\theta) = -\frac{\sqrt{530}}{23}, \cot(\theta) = -23$

27. $\sin(\theta) = -\frac{2\sqrt{5}}{5}, \cos(\theta) = \frac{\sqrt{5}}{5}, \tan(\theta) = -2, \csc(\theta) = -\frac{\sqrt{5}}{2}, \sec(\theta) = \sqrt{5}, \cot(\theta) = -\frac{1}{2}$

28. $\sin(\theta) = \frac{\sqrt{15}}{4}, \cos(\theta) = -\frac{1}{4}, \tan(\theta) = -\sqrt{15}, \csc(\theta) = \frac{4\sqrt{15}}{15}, \sec(\theta) = -4, \cot(\theta) = -\frac{\sqrt{15}}{15}$

29. $\sin(\theta) = -\frac{\sqrt{6}}{6}, \cos(\theta) = -\frac{\sqrt{30}}{6}, \tan(\theta) = \frac{\sqrt{5}}{5}, \csc(\theta) = -\sqrt{6}, \sec(\theta) = -\frac{\sqrt{30}}{5}, \cot(\theta) = \sqrt{5}$

30. $\sin(\theta) = \frac{2\sqrt{2}}{3}, \cos(\theta) = \frac{1}{3}, \tan(\theta) = 2\sqrt{2}, \csc(\theta) = \frac{3\sqrt{2}}{4}, \sec(\theta) = 3, \cot(\theta) = \frac{\sqrt{2}}{4}$

31. $\sin(\theta) = \frac{\sqrt{5}}{5}, \cos(\theta) = \frac{2\sqrt{5}}{5}, \tan(\theta) = \frac{1}{2}, \csc(\theta) = \sqrt{5}, \sec(\theta) = \frac{\sqrt{5}}{2}, \cot(\theta) = 2$

32. $\sin(\theta) = \frac{1}{5}, \cos(\theta) = -\frac{2\sqrt{6}}{5}, \tan(\theta) = -\frac{\sqrt{6}}{12}, \csc(\theta) = 5, \sec(\theta) = -\frac{5\sqrt{6}}{12}, \cot(\theta) = -2\sqrt{6}$

33. $\sin(\theta) = -\frac{\sqrt{110}}{11}, \cos(\theta) = -\frac{\sqrt{11}}{11}, \tan(\theta) = \sqrt{10}, \csc(\theta) = -\frac{\sqrt{110}}{10}, \sec(\theta) = -\sqrt{11}, \cot(\theta) = \frac{\sqrt{10}}{10}$

34. $\sin(\theta) = -\frac{\sqrt{95}}{10}, \cos(\theta) = \frac{\sqrt{5}}{10}, \tan(\theta) = -\sqrt{19}, \csc(\theta) = -\frac{2\sqrt{95}}{19}, \sec(\theta) = 2\sqrt{5}, \cot(\theta) = -\frac{\sqrt{19}}{19}$