Trigonometry express $4\cos x+3\sin x$ in the form $R \cos (x+a)$. I have been asked to express $4\cos x+3\sin x$ in the form $R \cos (x+a)$.
I know that the formula to express it in that form is $a \cos x+b\sin x=R \cos (x-a)$.
But as the question is asking me to express it in where it is $(x+a)$ instead of $(x-a)$ I am unsure of what to do.
Any help is much appreciated. 

I have never seen this kind of question but here is my attempt for a solution.
Assume, a $\triangle ABC$ right angled at B, opposite angle $\angle CAB$ as $a$, and hypotenuse as AC.
Now, presume BC is 4 units and BA is 3. So, AC becomes 5 units.
That implies, $\cos a=\frac{4}{5}$, $\sin a=\frac{3}{5}$
Coming back to your original question.
$4\cos x+3\sin x=k$
Divide by 5 both sides,
$\frac{4}{5}\cos x+\frac{3}{5} \sin x=\frac{k}{5}$
which is the form of
$\cos a\times\cos x+\sin a\times\sin x=\frac{k}{5}$
$\implies \cos(a-x)=\cos(x-a)=\frac{k}{5}$
Or,
$k=5\cos(x-a)=5\cos(x+(-a))$
$a$ is approximately $0.64$
