- anonymous

Can someone help me prove sin^2(x)cos^2(x) = (1-cos4x)/theta?
I'm having trouble with trigonometry and I can't really grasp the material. Please and thank you

- schrodinger

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- hartnn

\(\cos 2x = 2\cos^2x-1 = 1-2\sin^2x\)
so can you find cos^2x =..... ? and sin^2x =.... from this formula ?

- anonymous

Not really :/

- hartnn

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

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## More answers

- anonymous

1 + cos2x = 2 cos^2 x?

- hartnn

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

- anonymous

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

- hartnn

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

- anonymous

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

- hartnn

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

- anonymous

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

- anonymous

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

- hartnn

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 =... ?

- anonymous

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

- hartnn

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

- anonymous

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

- hartnn

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

- anonymous

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

- hartnn

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

- anonymous

1+cos2x-cos2x-2cos^2 x

- hartnn

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

- anonymous

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

- hartnn

-cos^2 2x

- hartnn

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

- anonymous

cos2x?

- hartnn

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

- anonymous

kind of yeah

- hartnn

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

- anonymous

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

- hartnn

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

- AravindG

@treefriendly got it now?

- anonymous

Sort of, but I'm a bit lost.

- AravindG

can you tell me where?

- anonymous

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

- AravindG

i dont see any 8 here

- anonymous

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

- AravindG

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

- anonymous

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

- AravindG

there is no theta here

- anonymous

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

- hartnn

its not theta, its 8
\(((1-\cos2 x)/2) (1+\cos2x)/2))= (1-\cos^2 2x)/4 \\\\=(1-(\cos4x+1)/2)/4\\ =(2-\cos 4x-1)/8=(1-\cos 4x)/8\)

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