# 7.4 The other trigonometric functions  (Page 2/14)

 Page 2 / 14

The point $\text{\hspace{0.17em}}\left(\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right)\text{\hspace{0.17em}}$ is on the unit circle, as shown in [link] . Find $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}t,\mathrm{cos}\text{\hspace{0.17em}}t,\mathrm{tan}\text{\hspace{0.17em}}t,\mathrm{sec}\text{\hspace{0.17em}}t,\mathrm{csc}\text{\hspace{0.17em}}t,$ and $\text{\hspace{0.17em}}\mathrm{cot}\text{\hspace{0.17em}}t.$

$\mathrm{sin}t=-\frac{\sqrt{2}}{2},\mathrm{cos}t=\frac{\sqrt{2}}{2},\mathrm{tan}t=-1,sect=\sqrt{2},\mathrm{csc}t=-\sqrt{2},\mathrm{cot}t=-1$

## Finding the trigonometric functions of an angle

Find $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}t,\mathrm{cos}\text{\hspace{0.17em}}t,\mathrm{tan}\text{\hspace{0.17em}}t,\mathrm{sec}\text{\hspace{0.17em}}t,\mathrm{csc}\text{\hspace{0.17em}}t,$ and $\text{\hspace{0.17em}}\mathrm{cot}\text{\hspace{0.17em}}t.\text{\hspace{0.17em}}$ when $\text{\hspace{0.17em}}t=\frac{\pi }{6}.$

We have previously used the properties of equilateral triangles to demonstrate that $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\frac{\pi }{6}=\frac{1}{2}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\frac{\pi }{6}=\frac{\sqrt{3}}{2}.$ We can use these values and the definitions of tangent, secant, cosecant, and cotangent as functions of sine and cosine to find the remaining function values.

$\begin{array}{cccc}\hfill \text{tan}\text{\hspace{0.17em}}\frac{\pi }{6}& =\frac{\text{sin}\text{\hspace{0.17em}}\frac{\pi }{6}}{\text{cos}\text{\hspace{0.17em}}\frac{\pi }{6}}\hfill & & \\ & =\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}\hfill & =\frac{1}{\sqrt{3}}\hfill & =\frac{\sqrt{3}}{3}\hfill \\ \hfill \text{sec}\text{\hspace{0.17em}}\frac{\pi }{6}& =\frac{1}{\text{cos}\text{\hspace{0.17em}}\frac{\pi }{6}}\hfill & & \\ & =\frac{1}{\frac{\sqrt{3}}{2}}\hfill & =\frac{2}{\sqrt{3}}\hfill & =\frac{2\sqrt{3}}{3}\hfill \\ \hfill \mathrm{csc}\text{\hspace{0.17em}}\frac{\pi }{6}& =\frac{1}{\mathrm{sin}\text{\hspace{0.17em}}\frac{\pi }{6}}\hfill & =\frac{1}{\frac{1}{2}}\hfill & =2\hfill \\ \hfill \text{cot}\text{\hspace{0.17em}}\frac{\pi }{6}& =\frac{\text{cos}\text{\hspace{0.17em}}\frac{\pi }{6}}{\text{sin}\text{\hspace{0.17em}}\frac{\pi }{6}}\hfill & & \\ & =\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}\hfill & =\sqrt{3}\hfill & \end{array}$

Find $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}t,\mathrm{cos}\text{\hspace{0.17em}}t,\mathrm{tan}\text{\hspace{0.17em}}t,\mathrm{sec}\text{\hspace{0.17em}}t,\mathrm{csc}\text{\hspace{0.17em}}t,$ and $\text{\hspace{0.17em}}\mathrm{cot}\text{\hspace{0.17em}}t.\text{\hspace{0.17em}}$ when $\text{\hspace{0.17em}}t=\frac{\pi }{3}.$

$\begin{array}{c}\mathrm{sin}\frac{\pi }{3}=\frac{\sqrt{3}}{2}\hfill \\ \mathrm{cos}\frac{\pi }{3}=\frac{1}{2}\hfill \\ \mathrm{tan}\frac{\pi }{3}=\sqrt{3}\hfill \\ \mathrm{sec}\frac{\pi }{3}=2\hfill \\ \mathrm{csc}\frac{\pi }{3}=\frac{2\sqrt{3}}{3}\hfill \\ \mathrm{cot}\frac{\pi }{3}=\frac{\sqrt{3}}{3}\hfill \end{array}$

Because we know the sine and cosine values for the common first-quadrant angles, we can find the other function values for those angles as well by setting $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ equal to the cosine and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ equal to the sine and then using the definitions of tangent, secant, cosecant, and cotangent. The results are shown in [link] .

Angle $0$
Cosine 1 $\frac{\sqrt{3}}{2}$ $\frac{\sqrt{2}}{2}$ $\frac{1}{2}$ 0
Sine 0 $\frac{1}{2}$ $\frac{\sqrt{2}}{2}$ $\frac{\sqrt{3}}{2}$ 1
Tangent 0 $\frac{\sqrt{3}}{3}$ 1 $\sqrt{3}$ Undefined
Secant 1 $\frac{2\sqrt{3}}{3}$ $\sqrt{2}$ 2 Undefined
Cosecant Undefined 2 $\sqrt{2}$ $\frac{2\sqrt{3}}{3}$ 1
Cotangent Undefined $\sqrt{3}$ 1 $\frac{\sqrt{3}}{3}$ 0

## Using reference angles to evaluate tangent, secant, cosecant, and cotangent

We can evaluate trigonometric functions of angles outside the first quadrant using reference angles as we have already done with the sine and cosine functions. The procedure is the same: Find the reference angle    formed by the terminal side of the given angle with the horizontal axis. The trigonometric function values for the original angle will be the same as those for the reference angle, except for the positive or negative sign, which is determined by x - and y -values in the original quadrant. [link] shows which functions are positive in which quadrant.

To help remember which of the six trigonometric functions are positive in each quadrant, we can use the mnemonic phrase “A Smart Trig Class.” Each of the four words in the phrase corresponds to one of the four quadrants, starting with quadrant I and rotating counterclockwise. In quadrant I, which is “ A ,” a ll of the six trigonometric functions are positive. In quadrant II, “ S mart,” only s ine and its reciprocal function, cosecant, are positive. In quadrant III, “ T rig,” only t angent and its reciprocal function, cotangent, are positive. Finally, in quadrant IV, “ C lass,” only c osine and its reciprocal function, secant, are positive.

Given an angle not in the first quadrant, use reference angles to find all six trigonometric functions.

1. Measure the angle formed by the terminal side of the given angle and the horizontal axis. This is the reference angle.
2. Evaluate the function at the reference angle.
3. Observe the quadrant where the terminal side of the original angle is located. Based on the quadrant, determine whether the output is positive or negative.

## Using reference angles to find trigonometric functions

Use reference angles to find all six trigonometric functions of $\text{\hspace{0.17em}}-\frac{5\pi }{6}.$

The angle between this angle’s terminal side and the x -axis is $\text{\hspace{0.17em}}\frac{\pi }{6},$ so that is the reference angle. Since $\text{\hspace{0.17em}}-\frac{5\pi }{6}\text{\hspace{0.17em}}$ is in the third quadrant, where both $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ are negative, cosine, sine, secant, and cosecant will be negative, while tangent and cotangent will be positive.

$\begin{array}{cccc}\hfill \text{cos}\left(-\frac{5\pi }{6}\right)& =-\frac{\sqrt{3}}{2},\text{sin}\left(-\frac{5\pi }{6}\right)\hfill & =-\frac{1}{2},\text{tan}\left(\frac{5\pi }{6}\right)\hfill & =\frac{\sqrt{3}}{3},\hfill \\ \hfill \text{sec}\left(-\frac{5\pi }{6}\right)& =-\frac{2\sqrt{3}}{3},\text{csc}\left(-\frac{5\pi }{6}\right)\hfill & =-2,\text{cot}\left(-\frac{5\pi }{6}\right)\hfill & =\sqrt{3}\hfill \end{array}$

Find that number sum and product of all the divisors of 360
what is subgroup
Prove that: (2cos&+1)(2cos&-1)(2cos2&-1)=2cos4&+1
e power cos hyperbolic (x+iy)
10y
Michael
tan hyperbolic inverse (x+iy)=alpha +i bita
prove that cos(π/6-a)*cos(π/3+b)-sin(π/6-a)*sin(π/3+b)=sin(a-b)
why {2kπ} union {kπ}={kπ}?
why is {2kπ} union {kπ}={kπ}? when k belong to integer
Huy
if 9 sin theta + 40 cos theta = 41,prove that:41 cos theta = 41
what is complex numbers
give me treganamentry question
Solve 2cos x + 3sin x = 0.5
madras university algebra questions papers first year B. SC. maths
Hey
Rightspect
hi
chesky
Give me algebra questions
Rightspect
how to send you
Vandna
What does this mean
cos(x+iy)=cos alpha+isinalpha prove that: sin⁴x=sin²alpha
cos(x+iy)=cos aplha+i sinalpha prove that: sinh⁴y=sin²alpha
rajan
cos(x+iy)=cos aplha+i sinalpha prove that: sinh⁴y=sin²alpha
rajan
is there any case that you can have a polynomials with a degree of four?
victor
***sscc.edu/home/jdavidso/math/catalog/polynomials/fourth/fourth.html
Oliver
can you solve it step b step
give me some important question in tregnamentry
Anshuman