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.

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- Dec 9th 2008, 03:13 PMC^2Difficult 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. - Dec 9th 2008, 04:47 PMskeeter
using a substitution, like u = 2x, may cut down the work some.

- Dec 9th 2008, 05:09 PMSkalkaz
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. - Dec 9th 2008, 06:38 PMSoroban
Hello, Cē!

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.

Quote:

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

We can use two Sum-to-Product Identities:

. . $\displaystyle \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: .$\displaystyle -2\sin(6x)\sin(-4x) \;=\;2\sin(6x)\sin(4x)$

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

The two sides are equal, but it is__not__a satisfactory proof of an identity.