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.
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.
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.
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.