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.

Printable View

- December 9th 2008, 04:13 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. - December 9th 2008, 05:47 PMskeeter
using a substitution, like u = 2x, may cut down the work some.

- December 9th 2008, 06:09 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. - December 9th 2008, 07:38 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: .

We can use two Sum-to-Product Identities:

. .

The left side is: .

The right side is: .

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