1. ## Verifying/proving help?

5 sin(x+arctan(4/3)) = 3 sin x + 4 cos x

(arctan is the inverse tangent)
I'm really having trouble trying to prove this. I'm mainly stuck with what to do with the inverse tangent.

2. Originally Posted by Rinnie
5 sin(x+arctan(4/3)) = 3 sin x + 4 cos x

(arctan is the inverse tangent)
I'm really having trouble trying to prove this. I'm mainly stuck with what to do with the inverse tangent.
let $\displaystyle \theta = \arctan\left(\frac{4}{3}\right)$ ... using right triangle ratios, $\displaystyle \sin{\theta} = \frac{4}{5}$ and $\displaystyle \cos{\theta} = \frac{3}{5}$

sum identity for sine ...

$\displaystyle 5 \sin(x + \theta) = 5[\sin{x}\cos{\theta} + \cos{x}\sin{\theta}]$

substitute in the values for $\displaystyle \cos{\theta}$ and $\displaystyle \sin{\theta}$ and finish it.

3. Then, of course, use $\displaystyle sin(x+ \theta)= cos(x)cos(\theta)- sin(x)sin(\theta)$.

4. Originally Posted by HallsofIvy
Then, of course, use $\displaystyle \cos(x+ \theta)= \cos(x)\cos(\theta)- \sin(x)\sin(\theta)$.
fify