LinkBack URL
About LinkBacksHello 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.
Here is the identity.
![\sin^2\theta\cos^2\theta=\frac{1}{8}[1-\cos{4\theta}]](http://latex.codecogs.com/png.latex?\sin^2\theta\cos^2\theta=\frac{1}{8}[1-\cos{4\theta}])
You can write the given expression as: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
Here is the identity.
Firstly, I assume you know that:
![\therefore RHS = 1 - cos(4\theta) = 1 - cos2(2\theta) = 2 sin^{2}(2\theta) = 2 \times [sin(2\theta)]^2](http://latex.codecogs.com/png.latex?\therefore RHS = 1 - cos(4\theta) = 1 - cos2(2\theta) = 2 sin^{2}(2\theta) = 2 \times [sin(2\theta)]^2 )
![= 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](http://latex.codecogs.com/png.latex?= 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 )
  |
