codycenters
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- Apr 12, 2011
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could anyone please help me establish the identity
sin^2(theta)*cos^2(theta)=(1/8)*(1-cos(4(theta))) step-by-step?
sin^2(theta)*cos^2(theta)=(1/8)*(1-cos(4(theta))) step-by-step?
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sin^2(theta)*cos^2(theta)=(1/8)*(1-cos(4(theta)))
- Thread starter codycenters
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codycenters said:could anyone please help me establish the identity
sin^2(theta)*cos^2(theta)=(1/8)*(1-cos(4(theta))) step-by-step?
Expand cos(4?) in terms of cos(?) and sin(?) first.
Please share your work with us, indicating exactly where you are stuck - so that we may know where to begin to help you.
Hello, codycenters1
We need these two identities:
. . \(\displaystyle \frac{1-\cos2x}{2} \:=\:\sin^2\!x \qquad\qquad \sin2x \:=\:2\sin x\cos x\)
\(\displaystyle \text{Right side: }\:\tfrac{1}{8}(1 - \cos4\theta) \;=\;\tfrac{1}{4}\left(\frac{1-\cos4\theta}{2}\right)\)
. . . . . . . . . . . . . . . . . .\(\displaystyle =\;\tfrac{1}{4}\sin^2\!2\theta\)
. . . . . . . . . . . . . . . . . .\(\displaystyle =\;\tfrac{1}{4}(2\sin\theta\cos\theta)^2\)
. . . . . . . . . . . . . . . . . .\(\displaystyle =\;\tfrac{1}{4}(4\sin^2\!\theta\cos^2\!\theta)\)
. . . . . . . . . . . . . . . . . .\(\displaystyle =\;\sin^2\!\theta\cos^2\!\theta\)
We need these two identities:
. . \(\displaystyle \frac{1-\cos2x}{2} \:=\:\sin^2\!x \qquad\qquad \sin2x \:=\:2\sin x\cos x\)
\(\displaystyle \text{Prove: }\:\sin^2\!\theta\cos^2\!\theta \:=\: \tfrac{1}{8}(1-\cos 4\theta)\)
\(\displaystyle \text{Right side: }\:\tfrac{1}{8}(1 - \cos4\theta) \;=\;\tfrac{1}{4}\left(\frac{1-\cos4\theta}{2}\right)\)
. . . . . . . . . . . . . . . . . .\(\displaystyle =\;\tfrac{1}{4}\sin^2\!2\theta\)
. . . . . . . . . . . . . . . . . .\(\displaystyle =\;\tfrac{1}{4}(2\sin\theta\cos\theta)^2\)
. . . . . . . . . . . . . . . . . .\(\displaystyle =\;\tfrac{1}{4}(4\sin^2\!\theta\cos^2\!\theta)\)
. . . . . . . . . . . . . . . . . .\(\displaystyle =\;\sin^2\!\theta\cos^2\!\theta\)