Verify the following trigonometric identity.

Question

Verify the following trig identity: $$\sin(3\theta)-\sin\theta = 2\cos(2\theta)\sin\theta$$

Here is my work so far.

$\sin(3\theta)-\sin\theta = 2\cos(2\theta)\sin\theta$

LHS:$$\sin(\theta+2\theta)-\sin\theta$$ $$\sin\theta \cos(2\theta)+\sin(2\theta)\cos\theta-\sin\theta$$ $$\sin\theta \cos(2\theta)+(2\sin\theta \cos\theta)\cos\theta-\sin\theta$$

Where do I go from here? I think I should leave the first term in the line above as is, and try and manipulate the second two terms to equal $\sin\theta \cos(2\theta)$, then the LHS will add together to equal $2\cos(2\theta)\sin\theta$, and the identity will be verified. How do you suggest I get there?

Any hints or advice would be appreciated.

EDIT:

I ended up verifying this identity using the identity frank000 mentioned in comments. Thanks to everyone for the input, it was all very helpful.

Do you know how to prove $\sin(x)-\sin(y)=2\cos(\frac{x+y}{2})\sin(\frac{x-y}{2})$ in general? — , Oct 12 '15 at 02:03
Just for this question $sin(3x)=3sin(x)-4sin^3(x)$, $cos(2x)=1-2sin^2(x)$ — , Oct 12 '15 at 02:04
@frank000 I'm familiar with that identity, but don't exactly know how to prove it... — McB, Oct 12 '15 at 02:04
You could also use complex numbers for a very short and easy prove that both sides are identical. — Stefan Gruenwald, Oct 12 '15 at 04:07

score 4 · Answer 1 · answered Oct 12 '15 at 02:09

4

You're almost done! In the second and third term, factor the $\sin \theta$ and you'll get $2 \cos^2 \theta -1$ which you might recognize as $\cos 2\theta$.

answered Oct 12 '15 at 02:09

Faraz Masroor

1,210

score 2 · Accepted Answer · answered Oct 12 '15 at 02:05

2

hint: also use $\sin(\theta) = \sin(2\theta - \theta) = \sin(2\theta)\cos(\theta)-\sin(\theta)\cos(2\theta)$

answered Oct 12 '15 at 02:05

DeepSea

77,651

score 2 · Answer 3 · answered Oct 12 '15 at 02:09

2

$\sin(3θ)−\sinθ=2\cos(2θ)\sinθ$

$\sinθ\cos(2θ)+\sin(2θ)\cosθ−sinθ=2(\cos^2θ-\sin^2θ)\sinθ$

$\sinθ(\cos^2θ-\sin^2θ)+(2\sinθ\cosθ)\cosθ−\sinθ=2(\cos^2θ-\sin^2θ)\sinθ$

$\cos^2θ-\sin^2θ−1=-2\sin^2θ$

$-\sin^2θ-\sin^2θ=-2\sin^2θ$

$-2\sin^2θ=-2\sin^2θ$

answered Oct 12 '15 at 02:09

jimbo

2,156

Fourth and subsequen lines should be augmented by "or $\sin \theta = 0$" since division by zero is undefined. – Eric Towers Oct 12 '15 at 02:54

Verify the following trigonometric identity.

EDIT:

3 Answers3