Question

Find the exact value by using a half-angle identity. tan 7pi/8

Answer 1

`\tan^2x=(\sin^2x)/(\cos^2x)`

`\tan^2x=\frac{\frac{1-\cos2x}2}{\frac{1+\cos2x}2}`

`\tan^2x=(1-\cos2x)/(1+\cos2x)`

`\tan x=\pm\sqrt{(1-\cos2x)/(1+\cos2x)}`

When
`\frac\pi2<x<\pi`, you have
`\tan x<0`, so for
`x=\frac{7\pi}8` you would take the negative root. Now,


`\tan\frac{7\pi}8=-\sqrt{\frac{1-\cos\frac{7\pi}4}{1+\cos\frac{7\pi}4}}`

`\tan\frac{7\pi}8=-\sqrt{\frac{1-\frac1{\sqrt2}}{1+\frac1{\sqrt2}}}`

`\tan\frac{7\pi}8=-√(3-2\sqrt2)`