# How to prove $\cos \theta + \sin \theta =\sqrt{2} \cos\theta$.

I'm learning Trigonometry right now with myself. I'm stuck in a problem from sometimes (If $\cos \theta - \sin \theta =\sqrt{2} \sin\theta$, proof that $\cos \theta + \sin \theta =\sqrt{2} \cos\theta$) . I don't know what to do next. Please have a look at the pictures of my solution.  I don't know how to continue the proof from the last line. Also have not yet read about product to sum or, sum to product formula and have read only some conversion(such as $\sin$ to $\cos$, $1 + \tan^2=sec^2\theta$ etc) and about Periods of the trigonometric function and some other few topics. The question is given in an exercise in my book S.L. Loney.

$\cos \theta + \sin \theta =\sqrt{2} \cos\theta$

There is no need of computing $\theta$ in a explicit way.

$\cos\theta-\sin\theta=\sqrt{2}\sin\theta$ is equivalent to $\cos\theta = (1+\sqrt{2})\sin\theta$, that gives $\cot\theta=1+\sqrt{2}$.

So we have: $$\tan\theta = \frac{1}{\sqrt{2}+1} = \sqrt{2}-1$$ and by multiplying both sides by $\cos\theta$, then rearranging, $$\sin\theta+\cos\theta = \sqrt{2}\cos\theta$$ follows.

\begin{align} \cos{\theta}-\sin{\theta} &=\sqrt{2}\sin{\theta} \\ \cos{\theta} &= (1+\sqrt{2})\sin{\theta} \\ (1-\sqrt{2})\cos{\theta} &=(1-\sqrt{2})(1+\sqrt{2})\sin{\theta}=(1-2)\sin{\theta} \\ \cos{\theta}+\sin{\theta} &= \sqrt{2}\cos{\theta} \end{align}

An easier approach would be as follows:

$$\cos\theta-\sin\theta=\sqrt{2}\sin\theta \implies \cos\theta=\sin\theta(1+\sqrt{2}) \implies$$ $$\cos\theta+\sin\theta=\sin\theta+\sin\theta(1+\sqrt{2}) \implies \cos\theta+\sin\theta=\sqrt{2}\sin\theta(1+\sqrt{2}) \implies$$ $$\cos\theta+\sin\theta=\sqrt{2}\cos\theta$$

$$\cos \theta - \sin \theta = \sqrt{2}\sin \theta$$ $$\cos \theta = (\sqrt{2} + 1)\sin \theta$$ $$\cot \theta = \sqrt{2} + 1$$ $$\tan \theta = \frac{1}{\sqrt{2} + 1} = \sqrt{2} - 1$$ $$\sin \theta = (\sqrt{2} - 1)\cos \theta$$ $$\cos \theta + \sin \theta = \sqrt{2}\cos \theta$$