Is My Solution for 4sinx - 3cosx Correct?

[Caldus]

I'm unsure as to whether I'm correct with this:

4sinx - 3cosx

My answer ended up being 5sin(x + sin^-1 (-3/5)).

The equations I used were:

asinx + bcosx = (a^2 + b^2)^1/2 * sin(x + angle)
angle = sin^-1 (b / (a^2 + b^2)^1/2))

Can someone clarify whether I'm correct? I would appreciate it.

 

[Hurkyl]

A good way to check your answer is to reverse the process using the angle addition identity:

sin (θ + φ) = sin θ cos φ + cos θ sin φ

And using the facts

sin arcsin (3/5) = 3/5
and
cos arcsin (3/5) = 4/5. (why?)

 

[Caldus]

I checked my answer and it equals the original equation, but have I fully solved it?

 

[Hurkyl]

If it checks (and I agree that it does), then yep!