Question

Question 3 (Essay Worth 4 points)

Answer 1

Answer:

$\displaystyle \tan \left(\frac{7\pi}{8}\right) = -\left(2 + \sqrt{2}\right)$.

Step-by-step explanation:

Assume that the following are known:

• The tangent half-angle identity: $\displaystyle \tan\left(\frac{\theta}{2}\right) = \frac{1 - \cos(\theta)}{\sin(\theta)}$.
• The sine and cosine of $\displaystyle \frac{\pi}{4}$: $\displaystyle \sin\left(\displaystyle \frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$, $\displaystyle \cos\left(\displaystyle \frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$.
• The trigonometric identities for the sine and cosine of supplementary angles: $\sin(\theta) = -\sin(2\pi - \theta)$ and $\cos(\theta) = \cos(2\pi - \theta)$.

The angle in question $\displaystyle \frac{7 \pi}{8}$ isn't very well-known. However, note that:

• $\displaystyle \frac{7 \pi}{8}$ is equal to one-half times the angle $\displaystyle \frac{7\pi}{4}$;
• $\displaystyle \frac{7\pi}{4}$ is equal to $2\pi$ minus $\displaystyle \frac{\pi}{4}$, which is an angle with known sine and cosine values.

Since the angle $\displaystyle \frac{7\pi}{4}$ is equal to $\displaystyle 2\pi - \frac{\pi}{4}$, its sine and cosine can be found from the sine and cosine of $\displaystyle \frac{\pi}{4}$ using the trigonometric identities $\sin(2\pi - \theta) = -\sin(\theta)$ and $\cos(2\pi - \theta) = \cos(\theta)$:

• $\displaystyle \sin\left(\frac{7\pi}{4}\right) = -\sin\left(2\pi - \frac{7\pi}{4}\right) = - \sin\left(\frac{\pi}{4}\right) = - \frac{\sqrt{2}}{2}$.
• $\displaystyle \cos\left(\frac{7\pi}{4}\right) = \cos\left(2\pi - \frac{7\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$.

Apply the tangent half-angle identity to find the tangent of $\displaystyle \frac{7 \pi}{8}$:

$\begin{aligned}\tan\left(\frac{7\pi}{8}\right) &= \tan\left(\frac{7\pi/4}{2}\right) = \frac{1 - \cos(7\pi/4)}{\sin(7\pi/4)} \\ &= \frac{1 - \left(\sqrt{2}/2\right)}{\sqrt{2} / 2} = \frac{2 - \sqrt{2}}{\sqrt{2}} = \frac{\left(\sqrt{2}\right)^2 - \sqrt{2}}{\sqrt{2}} = \sqrt{2} - 1 \end{aligned}$.