$\sin x+\sqrt{5} \cos x$ 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 $\tan$?
 A: Divide by $c$ on both sides to have $$\frac1c\sin(x)+\frac{\sqrt{5}}{c}\cos(x)=\sin(x+d)$$
You would have the angle addition rule for sine here if
$$\begin{align}
\cos(d)&=\frac1c\\
\sin(d)&=\frac{\sqrt5}{c}
\end{align}$$
Square both sides of both equations and add corresponding sides together to give yourself a quadratic equation in $c$. From there, for each of the two solutions for $c$ (which are $\pm\sqrt{6}$), use the above two equations to deduce what quadrant $d$ is in (if $c>0$, $d$ is in quadrant I, otherwise $d$ is in quadrant III), and then use the appropriate variation on $\arccos$ or $\arcsin$ to solve for $d$. (If $d$ is in quadrant I, $d=\arccos\left(\frac1c\right)+2\pi k$, and if $d$ is in quadrant III, $d=\frac\pi2-\arccos\left(\frac1c\right)+2\pi k$.)
A: 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 $\theta = \arccos(1/\sqrt{6})$.
