Question #d3cc2

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Answer 1 · 2016-07-10T08:06:34

$sin^2x/(1+cosx)=1-cosx=2sin^2(x/2).$

We know that, $sin^2x=1-cos^2x=(1+cosx)(1-cosx)$

$:. sin^2x/(1+cosx)=1-cosx.$

If we use, the formula, $: 1-cos2theta=2sin^2(theta/2)$ , the above Exp. can still be expressed as $2sin^2(x/2).$

Altogether, $sin^2x/(1+cosx)=1-cosx=2sin^2(x/2).$

Answer 2 · 2016-07-10T08:06:37

$sin^2x/(1+cosx)=1-cosx=2sin^2(x/2).$

We know that, $sin^2x=1-cos^2x=(1+cosx)(1-cosx)$

$:. sin^2x/(1+cosx)=1-cosx.$

If we use, the formula, $: 1-cos2theta=2sin^2(theta/2)$ , the above Exp. can still be expressed as $2sin^2(x/2).$

Altogether, $sin^2x/(1+cosx)=1-cosx=2sin^2(x/2).$