Show that $sin^2x tan^2x = ( 3-4cos2x+cos4x ) / ( 4(1+cos2x) )$ ?

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Show that $sin^2x tan^2x = ( 3-4cos2x+cos4x ) / ( 4(1+cos2x) )$ ?

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Answer 1 · 2017-10-21T23:31:21

We will utilise the following identities:

$cos 2A -= 2cos^2A - 1 => cos^2A = 1/2(1+cos2A)$
$sin^2A + cos^2A -= 1 \ \ \ => sin^2A = 1/2(1-cos2A)$

Then:

$sin^2x tan^2x -= sin^2x sin^2x/cos^2x$

$" " = 1/2(1-cos2x) (1/2(1-cos2x)) / (1/2(1+cos2x))$

$" " = 1/2(1-cos2x) (1-cos2x) / (1+cos2x)$

$" " = ( (1-cos2x) (1-cos2x) ) / ( 2(1+cos2x) )$

$" " = ( 1-2cos2x+cos^2 2x ) / ( 2(1+cos2x) )$

$" " = ( 1-2cos2x+1/2(1+cos4x) ) / ( 2(1+cos2x) )$

$" " = ( 1/2(2-4cos2x+1+cos4x) ) / ( 2(1+cos2x) )$
$" " = ( 3-4cos2x+cos4x ) / ( 4(1+cos2x) ) \ \ \$ QED

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Show that $sin^2x tan^2x = ( 3-4cos2x+cos4x ) / ( 4(1+cos2x) )$ ?

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