Question

Find the exact value of tan 5pi / 4. No decimals are allowed, no calculators are allowed. Show your work.
Part 1: Sketch the angle. Identify the reference angle on your sketch.
Part 2: Identify the terminal point associated with the angle, then find the tangent of the angle.

Answer 1

Answer:

Step-by-step explanation:

Most people are more familiar with degrees than with radians. So perhaps what is giving a bad time is the pi/4

pi/4 = 45 degrees.

Tangent of 45 degrees is 1. So part A is  = 5 * 45 degrees = 225 degrees which is in the 3 rd quadrant. (Remember that the 4 goes into defining the 45 degree angle from pi/4 radians.)

The tangent is positive in quad 3. So you draw a line 1/2 way between the minus x and minus y axis to get 5/4 pi.

The tangent of the angle is 1. The terminal point is (-1 , - 1)

Answer 2

Answer:

The exact value of $\tan (\frac{5\pi}{4})$ is 1.

The reference angle is $\frac{\pi}{4}$

The terminal point is $(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}})$.

Step-by-step explanation:

The given expression is

$\tan (\frac{5\pi}{4})$

Since $\pi<\frac{5\pi}{4}<\frac{3\pi}{2}$, so it lies in third quadrant.

We need to find the exact value of given expression.

$\tan (\frac{5\pi}{4})=\tan (pi+\frac{\pi}{4})$

In third quadrant, tangent is positive.

$\tan (\frac{5\pi}{4})=\tan (pi+\frac{\pi}{4})$

$\tan (\frac{5\pi}{4})=\tan (\frac{\pi}{4})$       $[\because \tan (\pi+\theta)=\tan \theta]$

$\tan (\frac{5\pi}{4})=1$       $[\because \tan (\frac{\pi}{4})=1]$

Part 1: Reference angle for third quadrant is

$\text{Reference angle}=\theta -\pi$

$\text{Reference angle}=\frac{5\pi}{4} -\pi$

$\text{Reference angle}=\frac{\pi}{4}$

Therefore the reference angle is $\frac{\pi}{4}$.

Part 2: Terminal point associated with the angle be (x,y).

For a unit circle hypotenuse is 1, perpendicular is y and base is x.

$\sin (\theta)=\frac{perpendicular}{hypotenuse}$

$\sin (\frac{\pi}{4})=\frac{y}{1}$

$\frac{1}{\sqrt{2}}=y$

$\cos (\theta)=\frac{base}{hypotenuse}$

$\cos (\frac{\pi}{4})=\frac{x}{1}$

$\frac{1}{\sqrt{2}}=x$

In third quadrant sine and cosine both are negative. So the terminal point is $(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}})$.