Prove Trig Identity: Step-by-Step Guide

[Aaron H.]

Homework Statement

Prove the identity.

Homework Equations

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

The Attempt at a Solution

http://s13.postimage.org/jkhubi4lz/DSC03534.jpg

 

[Mark44]

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.

 

[Villyer]

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)

 

[Curious3141]

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.

 

[eumyang]

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)

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.

 

[Aaron H.]

cos(3x)cos(x) - six(x)sin(3x)

cos (4x)

cos^2 2x - sin^2 2x

Thanks all.

 

[Curious3141]

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.