Question

on the following unit circle, ( \theta ) is in radians and ( \tan(\theta) = \frac{0.98}{0.2} = 4.9 ). without a calculator, evaluate the following expressions to the nearest hundredth or as an exact ratio.

expression	value
( \tan(2pi - \theta) )	( square )

Explanation:

Step1: Recall the tangent subtraction formula

The formula for \(\tan(A - B)\) is \(\frac{\tan A - \tan B}{1 + \tan A\tan B}\). For \(\tan(\pi - \theta)\), let \(A=\pi\) and \(B = \theta\). We know that \(\tan(\pi)=0\), so substituting into the formula:
\[
\tan(\pi - \theta)=\frac{\tan(\pi)-\tan(\theta)}{1 + \tan(\pi)\tan(\theta)}=\frac{0 - \tan(\theta)}{1+0\times\tan(\theta)}=-\tan(\theta)
\]
We are given that \(\tan(\theta) = 4.9\), so \(\tan(\pi - \theta)=- 4.9\).

Step2: Recall the tangent subtraction formula for \(\tan(2\pi-\theta)\)

For \(\tan(2\pi - \theta)\), let \(A = 2\pi\) and \(B=\theta\). We know that \(\tan(2\pi)=0\), so substituting into the formula \(\tan(A - B)=\frac{\tan A-\tan B}{1 + \tan A\tan B}\):
\[
\tan(2\pi - \theta)=\frac{\tan(2\pi)-\tan(\theta)}{1+\tan(2\pi)\tan(\theta)}=\frac{0-\tan(\theta)}{1 + 0\times\tan(\theta)}=-\tan(\theta)
\]
Wait, no, actually, the period of tangent is \(\pi\), and the identity for \(\tan(2\pi-\theta)\) can also be derived from the unit - circle definition. The tangent function has the property \(\tan(x - y)=\frac{\tan x-\tan y}{1 + \tan x\tan y}\), and since \(\tan(2\pi)=0\), \(\tan(2\pi-\theta)=\frac{0 - \tan\theta}{1+0\times\tan\theta}=-\tan\theta\)? Wait, no, another way: \(\tan(2\pi-\theta)=\tan(-\theta)\) because the period of \(\tan\) is \(\pi\), and \(\tan(-\theta)=-\tan\theta\) (since tangent is an odd function: \(\tan(-x)=-\tan x\)). We know that \(\tan\theta = 4.9\), so \(\tan(2\pi - \theta)=-\tan\theta=- 4.9\)? Wait, no, let's check with the unit - circle. The point on the unit circle for angle \(\theta\) is \((x,y)=(0.2,0.98)\), so \(\tan\theta=\frac{y}{x}=\frac{0.98}{0.2}=4.9\). For angle \(2\pi-\theta\), the point on the unit circle is \((x,-y)=(0.2,- 0.98)\), so \(\tan(2\pi - \theta)=\frac{-y}{x}=-\frac{y}{x}=-\tan\theta=-4.9\). For angle \(\pi-\theta\), the point on the unit circle is \((-x,y)=(-0.2,0.98)\), so \(\tan(\pi - \theta)=\frac{y}{-x}=-\frac{y}{x}=-\tan\theta=-4.9\)? Wait, no, wait \(\tan(\pi - \theta)\): \(\pi-\theta\) is in the second quadrant, where tangent is negative. And \(\tan(\pi - \theta)=-\tan\theta\) (from the identity \(\tan(\pi - \alpha)=-\tan\alpha\)). And \(\tan(2\pi - \theta)=\tan(-\theta)=-\tan\theta\) (since tangent is odd and has period \(\pi\)).

Answer:

For \(\tan(\pi - \theta)\), the value is \(- 4.9\); for \(\tan(2\pi - \theta)\), the value is \(-4.9\).

So in the table:

• \(\tan(\pi - \theta)\): \(-4.9\)
• \(\tan(2\pi - \theta)\): \(-4.9\)