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Math - Trig - Double Angles
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Macleef
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#1
Math - Trig - Double Angles
Prove:
cos4x = 8cos^4x - 8cos^2x + 1

My Attempt:
RS:
= 4cos^2x (2cos^2x - 1) + 1
= 4 cos^2x (cos2x) + 1

LS:
= cos2(2x)
= 2cos^2(2x) - 1
= (cos^2(2)) - cos^2(2x)) - 1





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Prove:
8cos^4x = cos4x + 4cos2x + 3

My Attempt:
RS:
= cos2(2x) + 4cos(2x) + 3
= 2cos^2(2x) - 1 + 2(4)cos^2(2x) - 1 + 3

PostPosted: Sat Nov 17, 2007 5:01 pm  Back to top 
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kunny
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#2
\cos 4x=\cos 2(2x)=....
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Today's calculation of Integral Digest

PostPosted: Sat Nov 17, 2007 5:18 pm  Back to top 
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Macleef
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#3
kunny wrote:
\cos 4x = \cos 2(2x) = ....


That's not very helpful ...

PostPosted: Sat Nov 17, 2007 5:48 pm  Back to top 
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kunny
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#4
Just use \cos 2\theta =2\cos ^ 2 \theta -1 Wink
\cos 4x = \cos 2(2x) = 2\cos ^2 (2x) - 1 = 2(2\cos ^ 2 x - 1)^2-1
_________________
Today's calculation of Integral Digest

PostPosted: Sat Nov 17, 2007 6:18 pm  Back to top 
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