Trig identity proof

1. The problem statement, all variables and given/known data

Prove the identity.

2. Relevant equations

http://postimage.org/image/vjhwki1ax/

3. The attempt at a solution

http://s13.postimage.org/jkhubi4lz/DSC03534.jpg
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 Mentor Speaking only for myself, I would be more inclined to help out if I didn't have to open one link to see the problem, and open another link to see what you did.
 I noticed that after all of your work, you got the problem to cos(3x)cos(x)-sin(x)sin(3x). Using sec(x) = 1/cos(x) and csc(x) = 1/sin(x); cos(3x)/sec(x) - sin(x)/csc(3x) cos(3x)/(1/cos(x)) - sin(x)/(1/sin(3x)) cos(3x)cos(x) - six(x)sin(3x)

Recognitions:
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Trig identity proof

 Quote by Villyer I noticed that after all of your work, you got the problem to cos(3x)cos(x)-sin(x)sin(3x).
I can't view his solution. But if he's already got it into that form, he can just use the angle sum formula for cosine to express that as $\cos kx$, where k is some positive integer (which he needs to work out). Then use the double angle formula for cosine to split it up again, yielding the required proof.

Recognitions:
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 Quote by Villyer Using sec(x) = 1/cos(x) and csc(x) = 1/sin(x); cos(3x)/sec(x) - sin(x)/csc(3x) cos(3x)/(1/cos(x)) - sin(x)/(1/sin(3x)) cos(3x)cos(x) - six(x)sin(3x)
 Quote by Curious3141 I can't view his solution.
Looking at the OP's solution, the OP went the complicated route to go from the LHS to the bolded part above. Villyer just simplified the process.
 cos(3x)cos(x) - six(x)sin(3x) cos (4x) cos^2 2x - sin^2 2x Thanks all.

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 Quote by Aaron H. cos(3x)cos(x) - six(x)sin(3x) cos (4x) cos^2 2x - sin^2 2x Thanks all.
Looks great, but you might want to put "=" signs in between the lines.