# Help needed in verifying a trigonometric identity

I have the following identity: $$32\sin^{2}\left(\theta\right)\cos^{4}\left(\theta\right) =2 + \cos\left(2\theta\right) - 2\cos\left(4\theta\right) -\cos\left(6\theta\right)$$

I've tried all sorts of combination of things, which i'd love to write out however given the nature my approach the working out becomes very nested and convoluted.

However here are a couple of things that I tried that I thought were promising:

1. Using the LHS and noticing that it is in a form similar to the double angle identity for sine and trying to taken advantage of this, i.e. $$32{\sin ^2}\theta {\cos ^4}\theta = 8{\cos ^2}\theta {(2\sin\theta \cos\theta )^2} = 8{\cos ^2}\theta {(\sin 2\theta )^2}$$

2. Using the RHS and trying to put everything in terms of $\cos(2\theta)$, once this was done I attempted to simplify, however things got very messy.

3. Using the RHS I saw that there were two expressions that could be written using the sum to product identity. If memory serves me right after doing this I had the following expression:

$2-(\cos(6\theta)-\cos(2\theta))-2\cos(4\theta)$

$= 2-(-2\sin(4\theta)\sin(2\theta))-2\cos(4\theta)$

$= 2+2\sin(4\theta)\sin(2\theta)-2\cos(4\theta)$

$=2(1+\sin (4\theta)\sin (2\theta)-\cos(4\theta))$

However after expanded this all out using the double angle identities I end up with expressions that I feel I cant really manipulate. Any help or direction would be appreciated, thanks.

• Are you allowed to use complex numbers? – Avitus Aug 2 '14 at 17:28
• No, I wouldnt think so as that isnt covered yet in the textbook – seeker Aug 2 '14 at 17:29
• already tried with $\cos(4\theta)=cos(2\theta+2\theta)=\cos^2(2\theta)-\sin^2(2\theta)$ and similarly for $\cos(6\theta)=\cos(2\theta+4\theta)=...$? – Avitus Aug 2 '14 at 17:33
• I dont understand you? – seeker Aug 2 '14 at 17:34
• I would try to reduce the RHS of you equation to an expression in $\cos\theta$, $\sin\theta$ and powers. To do so I would use the formulae $\cos(\alpha+\beta)=\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)$ on $\cos(2\theta)$, $\cos(4\theta)$ and $\cos(6\theta)$ starting with $\cos(2\theta)$ and using $\alpha=\theta$, $\beta=\theta$...and then go on with $\cos(4\theta)$ and $\cos(6\theta)$ with $\alpha=\beta=2\theta$ in the first case and $\alpha=2\theta$, $\beta=4\theta$ in the second – Avitus Aug 2 '14 at 17:37

Using Double angle formula, $$2-2\cos4x=2(1-\cos4x)=2(2\sin^22x)$$
Using Prosthaphaeresis Formula, $$\cos2x-\cos6x=2\sin4x\sin2x=2(2\sin2x\cos2x)\sin2x=4\sin^22x\cos2x$$
Add them and use $\displaystyle\sin2x=2\sin x\cos x$ and $\displaystyle1+\cos2x=2\cos^2x$