# Trigonometry proof- finding an angle

Suppose $\alpha$ is some angle less than $45^\circ$. If $a=\cos^2\alpha - \sin^2\alpha$ and $b = 2\sin\alpha\cos\alpha$, show that there is an angle $\theta$ such that $a = \cos\theta$ and $b = \sin\theta$.

Thanks.

• $2\sin\theta\cos\theta=\sin 2\theta.$ – Bumblebee Oct 14 '14 at 11:15
• You are using $a$ in two different ways in the equation $a = \cos^2a - \sin^2a$ since the left side is a number while the $a$ on the right side is a degree measure. – N. F. Taussig Oct 14 '14 at 11:20
• Thanks. I have fixed this. Any tips on how to approach the problem? – user183974 Oct 14 '14 at 11:28
• Yes that would be wonderful. Thanks. – user183974 Oct 14 '14 at 11:43
• For future reference, if you want to write $45^\circ$, write 45^\circ between dollar signs. – N. F. Taussig Oct 14 '14 at 12:04

there is a lemma on Trigenometry book by I.M gelfand that states:

if $a$ and $b$ are some pair of positive numbers such that $a^2+b^2=1$, then there exits an angle $\theta$ such that $a=cos \theta$ and $b=sin \theta$.

so for your question just square $a$ and $b$ and you can see $a^2+b^2=1$. so then there exits an angle $\theta$ such that $a=cos \theta$ and $b=sin \theta$.

As Nilan pointed out, $2\sin x \cos x=\sin 2x$. Also, $\cos^2 x + \sin^2 x =1$, so $\sin^2 x =1-\cos^2 x$, from which it follows that $a=2\cos^2 x-1=\cos 2x$

The double angle formulas for cosine and sine are given below

\begin{align*} \cos(2\varphi) & = \cos^2\varphi - \sin^2\varphi\\ & = 2\cos^2\varphi - 1\\ & = 1 - 2\sin^2\varphi\\ \sin(2\varphi) & = 2\sin\varphi\cos\varphi \end{align*}

Hence, $a = \cos^2\alpha - \sin^2\alpha = \cos(2\alpha)$ and $b = 2\sin\alpha\cos\alpha = \sin(2\alpha)$. Hence, $\theta = 2\alpha$.