# 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.

• Use your formula, and get, say, $a=2.6$. Then you know that $R\cos(x-2.6)$ is what you're looking for. Rewrite to $R\cos(x+(-2.6))$, and you have it on the form you were asked for. May 21, 2015 at 11:06

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 $0.64$