# Verify a trig identity

• Jun 20th 2013, 08:43 AM
Ragnarok
Verify a trig identity
Hello, I'm doing an induction problem and in the basis step I have to verify this identity (assuming $\sin{x/2}\neq 0$):

$\sin{x}=\frac{\sin{(2x)}}{4\sin^2{\left(\frac{x}{2 } \right) }}-\frac{2\cos{\left( \frac{3x}{2} \right) }}{2\sin{\left( \frac{x}{2} \right) }}$

Getting a common denominator, I get the RHS to be

$\frac{\sin{(2x)}-4\sin{\left( \frac{x}{2} \right) }\cos{\left( \frac{3x}{2} \right) }}{4\sin^2{\left( \frac{x}{2} \right) }}$

But where do I go from here? I'm afraid I have almost no background in trigonometry, and so don't know the strategy for these types of problems.
• Jun 20th 2013, 09:27 AM
ebaines
Re: Verify a trig identity
There may be an easier way, but if you convert the right hand side to be all in terms of (x/2) it works out pretty easily. Start with:

$\sin (2x) = \sin (\frac x 2 + \frac {3x} 2) = \sin (\frac x 2 ) \cos (\frac {3x} 2) + \cos (\frac x 2 ) \sin (\frac {3x} 2 )$

then apply the identities:

$\cos (3a) = \cos^3(a) - 3 \sin^2 (a) \cos(a)$

$\sin(3a) = 3 \cos^2(a) \sin (a) - \sin^3(a)$

using a = x/2.

You should end up with the right hand side being: $2 \sin (\frac x 2) \cos ( \frac x 2)$, which is equivalent to $\sin (2 \times \frac x 2) = \sin(x)$
• Jun 21st 2013, 03:50 PM
Ragnarok
Re: Verify a trig identity
Thanks a ton!
• Jun 21st 2013, 08:31 PM
ibdutt
Re: Verify a trig identity
You can also consider the attached solution.
Attachment 28658