Show angle exists by making use of trigonometric ratios

Question

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

Exercise 1: Suppose $\alpha$ is some angle less than $45°$. 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 $b = \sin \theta$.

(from: Trigonometry: I.M. Gelfand, Mark Saul)

It was ok for me to follow that book, but that chapter in which the above lemma is introduced, along with the exercise threw me off a bit.

There is also a proof for the lemma, but I couldn't get it together with the exercise.

How would I go about finding a solution to the above exercise, assuming the book wants us to make use of the given lemma.

Answer 1

As $a = \cos(2\alpha)$ and $b = \sin(2\alpha)$ you can proof the excercise. Moreover, you can compute $a^2 + b^2$ and using the lemma: $$a^2+b^2 =\cos^4\alpha + \sin^4\alpha - 2\cos^2\alpha\sin^2\alpha + 4\cos^2\alpha\sin^2\alpha = \cos^4\alpha + \sin^4\alpha + 2\cos^2\alpha\sin^2\alpha = (\cos^2\alpha + \sin^2\alpha)^2 = 1$$

Answer 2

If you were wondering about doing it without using the lemma, you can show it by using the trig identities:

$ \sin(2\alpha) = 2\sin(\alpha)\cos(\alpha) $

and

$ \cos(2\alpha)=\cos^2(\alpha)-\sin^2(\alpha) $

Thus, $a=\cos(2\alpha)$ and $b=\sin(2\alpha)$, meaning that if you let $\theta=2\alpha$, you have shown that you can rewrite $a=\cos(\theta)$ and $b=\sin(\theta)$