$\sin(x) + \cos(x) = \sqrt{2}\cdot \sin(x+\pi/4)$ but what is it as cosine?

Question

I get $-\cos(x+\pi/4)$ but it seems wrong (checking with WolframAlpha).

Answer 1

Using $\cos(-x)=\cos(x)$ and $\sin(x)=\cos(\pi/2-x)$ $$\sin{(x)}+cos{(x)}=\sqrt{2}\big(\frac{1}{\sqrt{2}}\cdot\sin{(x)}+\frac{1}{\sqrt{2}}\cos{(x)}\big)\\=\sqrt{2}\sin(x+\pi/4)=\sqrt{2}\cos{(\pi/2-(x+\pi/4))}\\ =\sqrt{2}\cos{(\pi/4-x)}=\sqrt{2}\cos(x-\pi/4)$$

Answer 2

Hint: Use $\sin(\pi/2-x) = \cos(x)$ and $\sin(x+\pi/4)=\sin(\pi/2-\pi/2+x+\pi/4)=\sin(\pi/2-(\pi/2-x-\pi/4))$.

Answer 3

Hint: Use the relation $\sin(x) = \cos(x+\pi/2)$

Answer 4

Hint: Simplify$$\sqrt{2}\cos\left(x-\frac{\pi}{4}\right)=\sqrt{2}\cos(x)\cos\left(\frac{\pi}{4}\right)+\sin(x)\sin\left(\frac{\pi}{4}\right)$$