Question

cos^2 2x-cos^2 6x=sin4xsin8x

Answer 1

There's no easy way to, like, explain this one really. I think it's more "recognizable" to start with the sin4xsin8x part. It can be done going both ways, but I think starting on the right side makes the finish line visible more easily.
So we need a product to sum formula for sin. The formula is
\[\sin(x)\sin(y) = \frac{ 1 }{ 2 }[\cos(x-y)-\cos(x+y)]\]Because of how quickly we look like 2 cosine terms I chose this way. So we have sin4xsin8x, meaning our "x" must be 4x and our "y" must be 8x. SO if we plug numbers into theformula we have:
\[\frac{ 1 }{ 2 }[\cos(4x-8x) - \cos(4x+8x)]\]Now 4x - 8x will give us -4x as an angle in the first cosine, but this is perfectly okay. Cosine and secant are even functions, meaning a negative angle and a positive angle have the same value. cos4x is the same as cos-4x. So I'll rewrite it just to show:
\[\frac{ \cos(4x)-\cos(12x) }{ 2 } \]Now it does takea bit of recognition, but what we have resembles a power reducing identity. The power-reducing formula I speak of refers to cosines and is this:
\[\cos ^{2}x = \frac{ 1+\cos(2x) }{ 2 }\] SO we have cosine to thefirst power and we want to go to cosine squared like the left side of our equation is. Also notice that one of the cosine angles is 2x and we have 4x. Well, 2x is half of 4x. And same with the other angle, 6x. 6x is half of what we currently have, 12x. So it looks pretty legit that this is the power-reducing formula in reverse. Now in order to match the formula above, I need a 1 in there. So I'm going to have to add 1 somehow. I can add a 1 in the numerator as long as I also subtract a 1. So let's see what happens if we do that:
\[\frac{ 1+\cos(4x)-1-\cos(12x) }{ 2 }\]Okay, we're getting there. I'm going to factor out a negative from the right two terms to get the plus sign we need:
\[\frac{ 1+\cos(4x)-(1+\cos(12x)) }{ 2 }\]Now the last thing we need to do to match the formula in reverse is to split these two fractions. So this can be split into 2 like this:
\[\frac{ 1+\cos(4x) }{ 2 }-\frac{ 1+\cos(12x) }{ 2 }\]Now it looks like things finally match up. Starting with the cos(4x) part. In order to go in reverse back to a squared term, I cut my angle in half and eliminate everything else tagging along. So this means I can make the left fraction into:
\[\cos ^{2}(2x)\]Now just the same process with the right portion. Half the angle, raiswe the power, eliminate the extra numbers and we finally have what we need:
\[\cos ^{2}(2x)-\cos ^{2}(6x)\]
Anything more rigorous of a proof than that isn't worth the trouble for a regular question. As long as you can point out said identity and use it, that's as much as we would probably need to do to prove theidentity.