Question

What is the exact value of tan(pi/24) in fractional form

Answer 1

Recall the half angle identities for sine and cosine:


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

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

So you have


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

This is useful because we have


`\tan^2\frac\pi{24}=\frac{1-\cos\frac\pi{12}}{1+\cos\frac\pi{12}}`

Unfortunately, we need to know the value of
`\cos\frac\pi{12}` to continue. But we can use either of the half-angle identities to figure this out.


`\cos^2\frac\pi{12}=\frac{1+\cos\frac\pi6}2`

`\cos^2\frac\pi{12}=\frac{1+\frac{\sqrt3}2}2`

`\cos^2\frac\pi{12}=\frac{2+\sqrt3}4`

`\cos\frac\pi{12}=\frac{√(2+\sqrt3)}2`

where you take the positive root since
`\cos x` is positive when
`0\le x<\frac\pi2`.

So, you have


`\tan^2\frac\pi{24}=\frac{1-\frac{√(2+\sqrt3)}2}{1+\frac{√(2+\sqrt3)}2}`

`\tan^2\frac\pi{24}=(2-√(2+\sqrt3))/(2+√(2+\sqrt3))`

`\tan\frac\pi{24}=\sqrt{(2-√(2+\sqrt3))/(2+√(2+\sqrt3))}`

again taking the positive root because
`\tan x` is positive when
`0<x<\frac\pi2`.