Rewriting expression to use at most one trigonometric function I have the expression: 
$$41\sqrt{2}\cos(v) + 41\sqrt{2}\sin(v)$$
And I want to rewrite it like an expression in v ∈ R that contains at most one trigonometric function.
What I have tried to do is:
$$41\sqrt{2}(\cos(v) + \sin(v))$$
But now I don't know where to go from here. Is there a formula that I somehow can use to get further or can I move in the constants into the trigonometric functions sin and cos? 
 A: This is a known trick.
If you compare 
$$a\cos\theta+b\sin\theta$$
to 
$$c\cos(\theta-\phi)=c\cos\theta\cos\phi+c\sin\theta\sin\phi$$
you see that there can be a match when
$$\begin{cases}a=c\cos\phi,\\b=c\sin\phi.\end{cases}$$
Furthermore, by eliminating one unknown or the other,
$$\begin{cases}c=\sqrt{a^2+b^2},\\\tan\phi=\dfrac ba.\end{cases}$$

$82\cos\left(\theta-\dfrac\pi4\right).$

A: It can be written like this as Donald Splutterwit mentioned,
$$82\sin(v + \frac{\pi}{4} )$$
Factor out $41\sqrt{2}$,
$$41\sqrt{2}(\cos(v) + \sin(v))$$
Then apply this formula (the key step your looking for),
$$\cos(v)+\sin(v) = \sqrt{2}\sin(v+\frac{\pi}{4})$$
This gives us,
$$41\sqrt{2}\sqrt{2}\sin(v+\frac{\pi}{4})$$
Because $\sqrt{2}\sqrt{2} = 2$ we have,
$$82\sin(v + \frac{\pi}{4} )$$
In addition it can be expressed without using any trigonometric terms if we decide to allow complex numbers,
$$\frac{(41 + 41 i)e^{-iv}}{\sqrt{2}}+\frac{(41 - 41 i)e^{iv}}{\sqrt{2}}$$
A: Note that $\sin( \pi/4) = \cos( \pi/4) = 1/\sqrt{2}$ and using the $ \sin$ addition formula 
\begin{eqnarray*}
\sin(v+w) = \sin(v) \cos(w) +  \cos(v) \sin(w). 
\end{eqnarray*}
Gives
\begin{eqnarray*}
41 \sqrt{2} \cos(v) + 41 \sqrt{2} \sin(v) = 82 \sin(v + \pi/4 ).
\end{eqnarray*}
