$\sin x+\sqrt{5} \cos x$ in a form $c\cdot \sin (x+d)$ How can I rewrite $\sin x + \sqrt{5}\cdot \cos x$ in a form $c \cdot \sin (x+d)$??? How can I find the values for $c$ and $d$? I have no idea how to solve that algebraically.
Is there also a possibility to rewrite it in termes of $\cos$ instead of $\sin$? Or maybe even with $\tan$?

In general, if we have: $a\sin x + b\cos x$, you can write that as: $$\sqrt{a^2 + b^2}\left(\frac{a}{\sqrt{a^2+b^2}}\sin x + \frac{b}{\sqrt{a^2 + b^2}}\cos x \right)$$
and call $y = \arccos\left(\frac{a}{\sqrt{a^2 + b^2}}\right)$, to finally use the formula for $\sin(x+y)$ on the other direction. Applying that to your specific problem, we have: $$\sin x + \sqrt{5}\cos x = \sqrt{6}\left(\frac{1}{\sqrt{6}}\sin x + \frac{\sqrt{5}}{\sqrt{6}}\cos x\right) = \sqrt{6}\sin(x + \theta)$$
where $\theta = \arccos(1/\sqrt{6})$.
