Answer:
[tex]\displaystyle \tan \left(\frac{7\pi}{8}\right) = -\left(2 + \sqrt{2}\right)[/tex].
Step-by-step explanation:
Assume that the following are known:
- The tangent half-angle identity: [tex]\displaystyle \tan\left(\frac{\theta}{2}\right) = \frac{1 - \cos(\theta)}{\sin(\theta)}[/tex].
- The sine and cosine of [tex]\displaystyle \frac{\pi}{4}[/tex]: [tex]\displaystyle \sin\left(\displaystyle \frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}[/tex], [tex]\displaystyle \cos\left(\displaystyle \frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}[/tex].
- The trigonometric identities for the sine and cosine of supplementary angles: [tex]\sin(\theta) = -\sin(2\pi - \theta)[/tex] and [tex]\cos(\theta) = \cos(2\pi - \theta)[/tex].
The angle in question [tex]\displaystyle \frac{7 \pi}{8}[/tex] isn't very well-known. However, note that:
- [tex]\displaystyle \frac{7 \pi}{8}[/tex] is equal to one-half times the angle [tex]\displaystyle \frac{7\pi}{4}[/tex];
- [tex]\displaystyle \frac{7\pi}{4}[/tex] is equal to [tex]2\pi[/tex] minus [tex]\displaystyle \frac{\pi}{4}[/tex], which is an angle with known sine and cosine values.
Since the angle [tex]\displaystyle \frac{7\pi}{4}[/tex] is equal to [tex]\displaystyle 2\pi - \frac{\pi}{4}[/tex], its sine and cosine can be found from the sine and cosine of [tex]\displaystyle \frac{\pi}{4}[/tex] using the trigonometric identities [tex]\sin(2\pi - \theta) = -\sin(\theta)[/tex] and [tex]\cos(2\pi - \theta) = \cos(\theta)[/tex]:
- [tex]\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}[/tex].
- [tex]\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}[/tex].
Apply the tangent half-angle identity to find the tangent of [tex]\displaystyle \frac{7 \pi}{8}[/tex]:
[tex]\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}[/tex].
