Why to simplify $\sin x - 1 =\cos x$ you have to multiply both side by $(\frac{\sqrt 2}{2}) $? Why do you have to multiply both side of $$\sin x - 1 = \cos x$$
by:
$$(\frac{\sqrt 2}{2}) $$ to simplify it?
I am doing Shaum's pre-calculus and the final answer to this question is: π but I have no clue where the idea of multiplying by $(\frac{\sqrt 2}{2})$ came from. Like what is the taught process behind it?
 A: It is because $\dfrac{\sqrt{2}}{2}=\sin\dfrac{\pi}{4}=\cos\dfrac{\pi}{4}$. Then we get
$$\dfrac{\sqrt{2}}{2}\sin x-\dfrac{\sqrt{2}}{2}\cos x=\cos\dfrac{\pi}{4}\sin x-\sin\dfrac{\pi}{4}\cos x=\sin(x-\dfrac{\pi}{4}).$$
Generally, for expressions like $$a\cos x+b\sin x=c$$
we do $$\frac{a}{\sqrt{a^2+b^2}}\cos x+\frac{b}{\sqrt{a^2+b^2}}\sin x=\frac{c}{\sqrt{a^2+b^2}}$$
Now considering $$\alpha=\sin^{-1}\frac{a}{\sqrt{a^2+b^2}}=\cos^{-1}\frac{b}{\sqrt{a^2+b^2}}$$
we reach to
$$\sin\alpha\cos x+\cos\alpha \sin x =\frac{c}{\sqrt{a^2+b^2}}$$
or $$\sin(\alpha+x)=\frac{c}{\sqrt{a^2+b^2}}$$
which can be solved easily.
A: Another solution you can consider is $$\sin x -\cos x = 1$$ then square both sides to get $$\sin^2x - 2\sin x \cos x + \cos^2x = 1$$ so $$-2\sin x \cos x  = 0$$
This is equivalent to $\sin(2x) = 0$, so $x$ is any $\frac{n\pi}{4}$ , $n \in \mathbb{Z}$ 
However, a danger with this approach is that you may introduce extra solutions; as when $$\sin^2x - 2\sin x \cos x + \cos^2x = 1$$ this also introduces $$\sin x -\cos x = -1$$ as possible roots.
A: When you do that we get $\sin x 1/\sqrt{2}-1/\sqrt{2}\cos x=1/\sqrt{2}$ which is nothing but $\sin x  \cos (\pi/4)- \sin (\pi/4) \cos x=   \sin (\pi/4), $ 
which gives $\sin(x-\pi/4)=\sin (\pi/4), $ by which you can get the solutions for $x.$
