Need help with one trig identity problem.

[joec]

Hello, I am new here and looking for some assistance.

I have two problems that I have been working on, and I can't seem to find out the right path to take. Any help would be appreciated.


2. Verify the identity, cos^2(5x)-cos^2(x) = -sin4x sin6x
cos^2(5x)-cos^2(x) = cos^2(5x)-cos^2(x) <-- common sense.
cos^2(5x)-1-sin^2(x)
-sin^2(x)+sin^2(5x)
(-sinx)^2 + (sin5x)^2

I am not sure if I went the right direction with number two. Any help would be appreciated, I have these last two problems in some homework and I need them done soon

 

[joec]

I figured out number 1, still stuck on number 2. Any thoughts?

 

[soroban]

Hello, joec!

Welcome aboard!

2. Verify the identity: \(\displaystyle \,\cos^2(5x)\,-\,\cos^2(x) \:=\: -\sin(4x)\sin(6x)\)

There are some sum-to-product identites that make this problem quite easy.

. . \(\displaystyle \cos A\,+\,\cos B\;=\;2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)\)

. . \(\displaystyle \cos A\,-\,\cos B\;=\;-2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)\)


We have the difference of two squares: \(\displaystyle \:\cos^2(5x)\,-\,\cos^2(x)\)

. . \(\displaystyle =\;\left[\cos(5x)\,+\,\cos(x)\right]\,\left[\cos(5x)\,-\,\cos(x)\right]\)

. . \(\displaystyle =\;\left[2\cos(3x)\cos(2x)\right]\,\left[-2\sin(3x)\sin(2x)\right]\)

. . \(\displaystyle =\;-\left[2\sin(2x)\cos(2x)\right]\,\left[2\sin(3x)\cos(3x)\right]\)
. . . . . . . . . . . \(\displaystyle \searrow\) . . . . . . . . \(\displaystyle \swarrow\)
. . \(\displaystyle =\;\;\;\;\;\;\;\;\;-\sin(4x)\;\sin(6x)\)

 

[joec]

Thanks a lot!