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\(\cos 2x = 2\cos^2x-1 = 1-2\sin^2x\) so can you find cos^2x =..... ? and sin^2x =.... from this formula ?
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?
yes, or you can say, (1-cos 2x)/2 = sin^2 x you can divide, there's no problem with that. now substitute both these results in left side of your equation sin^2x . cos^2x =... ?
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) =... ?
check your last term. its not -2cos^2x
errr..... -cos^2 x^2? or -cos^2 2x?
so, you have 1-cos^2 2x according to \(\cos 2x = 2\cos^2x-1\) what will be cos^2 2x=... ?)
\(\cos 2x = 2\cos^2x-1 \\cos 2(2x) = 2\cos^2(2x)-1 \\ \cos^22x =(\cos 4x+1)/2 \) got this ?
kind of yeah
so, whats \(\\~ \\ 1-\cos^22x =1-(\cos 4x+1)/2=..?\)
cos 2 (2x) = cos ^2 (2x) ?
ou had \( ((1-\cos2 x)/2) (1+\cos2x)/2))= (1-\cos^2 2x)/4 \\ =(2-\cos 4x-1)=1-\cos 4x\)
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?
its not theta, its 8 \(((1-\cos2 x)/2) (1+\cos2x)/2))= (1-\cos^2 2x)/4 \\\\=(1-(\cos4x+1)/2)/4\\ =(2-\cos 4x-1)/8=(1-\cos 4x)/8\)