Question

on the following unit circle, ( \theta ) is in radians and ( \tan(\theta) approx \frac{0.08}{0.2} = 4.9 ).
image of unit circle with point ( (0.2, 0.08) )
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 tangent subtraction formula is $\tan(A - B)=\frac{\tan A - \tan B}{1 + \tan A\tan B}$. For $\tan(\pi - \theta)$, we have $A = \pi$ and $B=\theta$. We know that $\tan(\pi)=0$. So $\tan(\pi - \theta)=\frac{\tan(\pi)-\tan(\theta)}{1+\tan(\pi)\tan(\theta)}=\frac{0 - \tan(\theta)}{1 + 0\times\tan(\theta)}=-\tan(\theta)$.
We are given that $\tan(\theta)\approx4.9$, so $\tan(\pi - \theta)=- 4.9$.

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

For $\tan(2\pi - \theta)$, we use the tangent subtraction formula with $A = 2\pi$ and $B=\theta$. Since $\tan(2\pi) = 0$, then $\tan(2\pi-\theta)=\frac{\tan(2\pi)-\tan(\theta)}{1+\tan(2\pi)\tan(\theta)}=\frac{0-\tan(\theta)}{1 + 0\times\tan(\theta)}=-\tan(\theta)$? Wait, no, wait. Wait, the period of tangent is $\pi$, and also we can use the identity $\tan(2\pi-\theta)=-\tan(\theta)$? Wait, no, let's think about the unit circle. The angle $2\pi-\theta$ is in the fourth quadrant, and tangent is negative there, and $\tan(2\pi - \theta)=\tan(-\theta)$ (because $2\pi-\theta=-\theta + 2\pi$ and tangent has period $\pi$? Wait, no, tangent has period $\pi$, so $\tan(2\pi-\theta)=\tan(-\theta)$. And we know that $\tan(-\alpha)=-\tan(\alpha)$ for any $\alpha$. So $\tan(2\pi - \theta)=\tan(-\theta)=-\tan(\theta)$? Wait, no, wait, let's re - derive.

Using the formula $\tan(A - B)=\frac{\tan A-\tan B}{1 + \tan A\tan B}$, with $A = 2\pi$ and $B=\theta$. $\tan(2\pi)=0$, so $\tan(2\pi-\theta)=\frac{0-\tan(\theta)}{1+0\times\tan(\theta)}=-\tan(\theta)$. Wait, but also, we know that $\tan(\pi - \theta)=-\tan(\theta)$ and $\tan(2\pi-\theta)=-\tan(\theta)$? Wait, no, that can't be. Wait, no, let's take a simple example. Let $\theta=\frac{\pi}{4}$, $\tan(\theta) = 1$. $\tan(\pi-\frac{\pi}{4})=\tan(\frac{3\pi}{4})=- 1=-\tan(\theta)$. $\tan(2\pi-\frac{\pi}{4})=\tan(-\frac{\pi}{4})=-1=-\tan(\theta)$? Wait, no, $\tan(2\pi-\frac{\pi}{4})=\tan(\frac{7\pi}{4})=-1$, and $\tan(-\frac{\pi}{4})=-1$, and $\tan(\pi-\frac{\pi}{4})=-1$. Wait, but actually, the correct identity is $\tan(\pi - \alpha)=-\tan(\alpha)$ and $\tan(2\pi-\alpha)=-\tan(\alpha)$? Wait, no, no, $\tan(2\pi-\alpha)=\tan(-\alpha)=-\tan(\alpha)$ (because tangent is odd function: $\tan(-x)=-\tan(x)$) and $\tan(\pi - x)=-\tan(x)$ (from the subtraction formula). Wait, but in our case, we know that $\tan(\theta)=4.9$. So for $\tan(\pi - \theta)$:

From the identity $\tan(\pi - \theta)=-\tan(\theta)$, so substituting $\tan(\theta) = 4.9$, we get $\tan(\pi - \theta)=- 4.9$.

For $\tan(2\pi - \theta)$:

We know that $\tan(2\pi - \theta)=\tan(-\theta)$ (since the period of tangent is $\pi$, $2\pi-\theta=-\theta + 2\pi$, and $\tan(x + 2\pi)=\tan(x)$ for all $x$ in domain). And since $\tan(-x)=-\tan(x)$ (tangent is an odd function), so $\tan(2\pi - \theta)=\tan(-\theta)=-\tan(\theta)$. Substituting $\tan(\theta)=4.9$, we get $\tan(2\pi - \theta)=-4.9$? Wait, no, wait, that's a mistake. Wait, no, let's check with the unit circle. The point on the unit circle is $(x,y)=(0.2,0.08)$. So $\tan(\theta)=\frac{y}{x}=\frac{0.08}{0.2}=0.4$? Wait, wait, the original problem says $\tan(\theta)\approx\frac{0.08}{0.2}=4.9$? Wait, no, $\frac{0.08}{0.2}=0.4$, not 4.9. Oh, there's a typo in the original problem? Wait, 0.08 divided by 0.2 is 0.4. Maybe it's a typo, 0.98 instead of 0.08? Let's assume that maybe it's a typo, but according to the point $(0.2,0.08)$, $x = 0.2$, $y = 0.08$, so $\tan(\theta)=\frac{y}{x}=\frac{0.08}{0.2}=0.4$.

Let's correct that. So $\tan(\theta)=\frac{0.08}{0.2}=0.4$.

Now, for $\tan(\pi - \theta)$:

Using the identity $\tan(\pi - \theta)=-\tan(\theta)$, so $\tan(\pi - \theta)=- 0.4$.

For $\tan(2\pi - \theta)$:

Using the identity $\tan(2\pi - \theta)=\tan(-\theta)=-\tan(\theta)$ (since $\tan$ is odd), so $\tan(2\pi - \theta)=- 0.4$? Wait, no, wait, $\tan(2\pi - \theta)$: the angle $2\pi-\theta$ is in the fourth quadrant, where tangent is negative, and $\tan(2\pi - \theta)=\frac{\sin(2\pi - \theta)}{\cos(2\pi - \theta)}$. We know that $\sin(2\pi - \theta)=-\sin(\theta)$ and $\cos(2\pi - \theta)=\cos(\theta)$. So $\tan(2\pi - \theta)=\frac{-\sin(\theta)}{\cos(\theta)}=-\tan(\theta)$. And $\tan(\pi - \theta)=\frac{\sin(\pi - \theta)}{\cos(\pi - \theta)}=\frac{\sin(\theta)}{-\cos(\theta)}=-\tan(\theta)$. So both $\tan(\pi - \theta)$ and $\tan(2\pi - \theta)$ are equal to $-\tan(\theta)$.

Since $\tan(\theta)=\frac{y}{x}=\frac{0.08}{0.2}=0.4$, then:

$\tan(\pi - \theta)=- 0.4$

$\tan(2\pi - \theta)=- 0.4$

Wait, but the original problem had a typo in the calculation of $\tan(\theta)$. Let's go back. The point is $(0.2,0.08)$, so $x = 0.2$, $y = 0.08$. Then $\tan(\theta)=\frac{y}{x}=\frac{0.08}{0.2}=0.4$, not 4.9. So probably a typo in the problem statement (0.98 instead of 0.08 would give 4.9). But based on the point $(0.2,0.08)$, we use $y = 0.08$ and $x = 0.2$.

So, step - by - step:

For $\tan(\pi - \theta)$:

1. Recall the identity $\tan(\pi - \alpha)=-\tan(\alpha)$.
2. Substitute $\alpha=\theta$, so $\tan(\pi - \theta)=-\tan(\theta)$.
3. Calculate $\tan(\theta)=\frac{y}{x}=\frac{0.08}{0.2}=0.4$.
4. Then $\tan(\pi - \theta)=- 0.4$.

For $\tan(2\pi - \theta)$:

1. Recall the identity $\tan(2\pi - \alpha)=-\tan(\alpha)$ (or use $\tan(2\pi - \alpha)=\tan(-\alpha)=-\tan(\alpha)$ since $\tan$ is odd and has period $2\pi$).
2. Substitute $\alpha = \theta$, so $\tan(2\pi - \theta)=-\tan(\theta)$.
3. Since $\tan(\theta)=0.4$, then $\tan(2\pi - \theta)=- 0.4$.

Answer:

For $\tan(\pi - \theta)$: $\boldsymbol{-0.4}$

For $\tan(2\pi - \theta)$: $\boldsymbol{-0.4}$