# Trigonometric identity proof

• Jul 13th 2012, 01:57 PM
Frederick
Trigonometric identity proof
Hi all, I am stuck on this problem. Help would be appreciated.

Here's the identity I need to prove (will only take a sec to read):

View image: Untitled

Here's my attempt:

http://s13.postimage.org/jkhubi4lz/DSC03534.jpg
• Jul 13th 2012, 03:09 PM
Cbarker1
Re: Trigonometric identity proof
Change the secant and cosecant to cosine and sine. Then use Sum-Product identity then simplify the expression.
• Jul 13th 2012, 06:57 PM
Soroban
Re: Trigonometric identity proof
Hello, Frederick!

We are expected to know these two identities:

. . $\displaystyle \cos(A + B) \:=\:\cos A\cos B - \sin A\sin B$

. . $\displaystyle \cos2A \:=\:\cos^2A - \sin^2A$

Quote:

$\displaystyle \text{Prove: }\:\frac{\cos3x}{\sec x} - \frac{\sin x}{\csc3x} \:=\:\cos^22x - \sin^22x$

We have: .$\displaystyle \frac{\cos3x}{\sec x} - \frac{\sin x}{\csc3x} \;\;=\;\;\frac{\cos3x}{\frac{1}{\cos x}} - \frac{\sin x}{\frac{1}{\sin3x}} \;\;=\;\; \cos3x\cos x - \sin3x\sin x$

. . . . . . . . .$\displaystyle =\;\;\cos(3x+x) \;\;=\;\;\cos 4x \;\;=\;\;\cos^22x - \sin^22x$