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:


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

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

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

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.

  • $\begingroup$ Are you allowed to use complex numbers? $\endgroup$ – Avitus Aug 2 '14 at 17:28
  • $\begingroup$ No, I wouldnt think so as that isnt covered yet in the textbook $\endgroup$ – seeker Aug 2 '14 at 17:29
  • $\begingroup$ already tried with $\cos(4\theta)=cos(2\theta+2\theta)=\cos^2(2\theta)-\sin^2(2\theta)$ and similarly for $\cos(6\theta)=\cos(2\theta+4\theta)=...$? $\endgroup$ – Avitus Aug 2 '14 at 17:33
  • $\begingroup$ I dont understand you? $\endgroup$ – seeker Aug 2 '14 at 17:34
  • $\begingroup$ 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 $\endgroup$ – 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 $\displaystyle1+\cos2x=2\cos^2x$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.