# Math Help - Difficult Trig Proof

1. ## Difficult Trig Proof

Hey guys, I was given a hard extra credit assignment and was wondering if anyone could do this proof:
cos2x-cos10x=(tan4x)(sin2x+sin10x)

Yeah, this seems like a really tricky problem.

2. using a substitution, like u = 2x, may cut down the work some.

3. Dear C^2,

my first tipp would be that you express the equation only with the term sin(2x) and cos(2x).
e. g.
cos(10x) = cos(8x + 2x) = cos(8x)sin(2x) - sin(8x)cos(2x)
and again...and again

and finally you can substitution cos(2x) = y and sin(2x) = sqrt(1-y^2)

So you get an algebraic equation. The problem is now the too high degree of the polinom but maybe you get a special polinom which is solvable, i dunno.

4. Hello, C²!

Normally, we are not allowed to work on both sides of the identity.
But I cannot find a way to transform one side into the other.

Prove: . $\cos2x-\cos10x\:=\:\tan4x(\sin2x+\sin10x)$

We can use two Sum-to-Product Identities:

. . $\begin{array}{ccc}\cos A - \cos B &=& \text{-}2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right) \\ \\[-3mm]
\sin A + \sin B &=& 2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right) \end{array}$

The left side is: . $-2\sin(6x)\sin(-4x) \;=\;2\sin(6x)\sin(4x)$

The right side is: . $\frac{\sin(4x)}{\cos(4x)}\cdot2\sin(6x)\cos(4x) \;=\;2\sin(6x)\sin(4x)$

The two sides are equal, but it is not a satisfactory proof of an identity.