# Trig Identity

• May 3rd 2010, 11:43 PM
Deimos
Trig Identity
Hello all. I am having trouble with a trig identity, and was wondering if you could help. This is a problem from a take home quiz, and unfortunately I can't find anything in my text about how to deal with the $\displaystyle \cos{4\theta}$.

Here is the identity.

$\displaystyle \sin^2\theta\cos^2\theta=\frac{1}{8}[1-\cos{4\theta}]$
• May 3rd 2010, 11:58 PM
harish21
Quote:

Originally Posted by Deimos
Hello all. I am having trouble with a trig identity, and was wondering if you could help. This is a problem from a take home quiz, and unfortunately I can't find anything in my text about how to deal with the $\displaystyle \cos{4\theta}$.

Here is the identity.

$\displaystyle \sin^2\theta\cos^2\theta=\frac{1}{8}[1-\cos{4\theta}]$

You can write the given expression as:
$\displaystyle 8 \sin^2\theta\cos^2\theta=[1-\cos{4\theta}]$

Firstly, I assume you know that:

$\displaystyle 1 - cos2x = 2sin^{2}x$

$\displaystyle \therefore RHS = 1 - cos(4\theta) = 1 - cos2(2\theta) = 2 sin^{2}(2\theta) = 2 \times [sin(2\theta)]^2$$\displaystyle = 2 \times [ 2 \sin(\theta) \cos(\theta)]^2 = 2 \times 4 \sin^{2}(\theta) \cos^{2}(\theta) = 8 \sin^{2}(\theta) \cos^{2}(\theta) = LHS$
• May 4th 2010, 12:03 AM
Deimos
Thank you very much!