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Not really :/

\(\cos 2x = 2\cos^2x-1 \)
Add 1 to both sides, what u get ?

1 + cos2x = 2 cos^2 x?

thats correct, now divide both sides by 2
what u get ?

(1+cos2x)/2 = cos^2 x?

yes, correct.
in the same way, isolate sin^2 x from \(\cos 2x = 1-2\sin^2x\)

so then cos2x-1 = -2sin^2 x?

yes, now divide both sides by -2
what u get ?

(cos2x-1)/-2 = sin^2 x?

but I thought you couldn't divide by -2 because it's with the sine?

what happened to the negative sign that was on the 2?

cos2x -1 got turned into 1-cos2x because of that negative sign.

ohh okay.
so ((1-cos^2 x)/2) ((1+cos2x)/2)) ?

how cos^2 x ? aren't both of them cos 2x ?

Darn. Yeah, you're right. I was looking at something else when I was typing

so, ((1-cos2 x)/2) ((1+cos2x)/2))
now use \((a-b)(a+b)=a^2+b^2\)
so, (1-cos2x)(1+cos2x) =... ?

1+cos2x-cos2x-2cos^2 x

check your last term.
its not -2cos^2x

errr..... -cos^2 x^2? or -cos^2 2x?

-cos^2 2x

so, you have 1-cos^2 2x
according to \(\cos 2x = 2\cos^2x-1\)
what will be cos^2 2x=... ?)

cos2x?

\(\cos 2x = 2\cos^2x-1 \\cos 2(2x) = 2\cos^2(2x)-1 \\ \cos^22x =(\cos 4x+1)/2 \)
got this ?

kind of yeah

so, whats \(\\~ \\ 1-\cos^22x =1-(\cos 4x+1)/2=..?\)

cos 2 (2x) = cos ^2 (2x) ?

ou had
\( ((1-\cos2 x)/2) (1+\cos2x)/2))= (1-\cos^2 2x)/4 \\ =(2-\cos 4x-1)=1-\cos 4x\)

@treefriendly got it now?

Sort of, but I'm a bit lost.

can you tell me where?

The last part. What happened to the 8 from the initial equation?

i dont see any 8 here

Sorry, it's getting late and I've been misreading

i see ..then i suggest you take rest
do it afresh the next day

I meant what happened to theta that was being divided from 1-cos4x?

there is no theta here

prove sin^2(x)cos^2(x) = (1-cos4x)/theta?