# Using double angle theorem to solve...

I've been doing some of my AP math work, and stumbled across a question I have trouble getting the right answer for... sort of?

The question is as followed, prove that $\tan22.5^\circ = \sqrt2 -1$.

I used the identity $\tan^2 x =\frac {\sin^2 (x)}{ \cos^2 (x)}$. I later substituted the $\sin x$ and $\cos x$ by the following: $\sin^2 (x) = \frac12 -\frac12\cos2x$

$\cos^2 (x) = \frac12 + \frac12\cos2x$ .

I derived both formulas myself, and used them in the previously stated identity. My final answer for $\tan(x)$ is $\sqrt{3-2\sqrt2}$, which gives an answer of $0.414$ to $3$d.p. This answer corresponds to the answer provided by the question itself, so the calculations are right. I am confused as to why the way I did it does not correspond to the answer. In terms of my steps, I found the value of $\cos(2x)$ to be $\frac{\sqrt{2}}{2}$ and then later on I rationalised the denominator before I took the square root on $\tan^2 x$.

Would writing something like $\sqrt{3-2\sqrt2}$ = $\sqrt2 -1$ be appropriate?

• Please consider using \ and correct LaTex coding (for example, \sqrt{\sinx} instead of sqrt(sin x)), as used on this site. Otherwise, good question! Commented Feb 24, 2017 at 18:57
• $(\sqrt{2}-1)^2 = 3 - 2\sqrt{2}$, so $\sqrt{3-2\sqrt{2}}$ is a perfectly good answer; but I would definitely write a step to show that the two are equal. Commented Feb 24, 2017 at 18:59
• The half-angle formula for $\tan$ is actually very simple, $\tan \frac{\large \theta}{\large 2} = \frac{\large\sin \theta }{\large 1+\cos\theta}$ Commented Feb 24, 2017 at 19:00

Hint...To get the answer you want, you might try $$\tan 2x=\frac{2\tan x}{1-\tan^2 x}$$ using $2x=45^o$