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ylucho

  • 2 years ago

please help with verifying : sin^2x/2-2cosx=cos^2 x/2

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  1. surjithayer
    • 2 years ago
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    \[\frac{ \sin ^{2}x }{ 2-2\cos x }=\frac{ 1 }{ 2 }*\frac{\sin ^{2}x }{1-\cos x }*\frac{1+\cos x }{1+\cos x }\] \[=\frac{ 1 }{ 2 }*\frac{ \sin ^{2}x \left( 1+\cos x \right) }{ 1-\cos ^{2}x }\] you can solve and use \[1-\cos ^{2}x=\sin ^{2}x and 1+\cos x=2\cos ^{2}\frac{ x }{ 2 }\]

  2. ylucho
    • 2 years ago
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    thanks for the help!!! I never thought about splitting at the beginning into two fractions :P thanks!

  3. surjithayer
    • 2 years ago
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    yw

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