How do you prove the identity $cos 2 x - sin x = 0$ $Cos2x-sinx=0$ ?

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How do you prove the identity $cos 2 x - sin x = 0$ $Cos2x-sinx=0$ ?

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Answer 1 · 2015-10-20T18:54:06

cos2x = ${cos}^{2} x - {sin}^{2} x$ $cos^2x -sin^2x$
SO,
${cos}^{2} x - {sin}^{2} x$ $cos^2x -sin^2x$ - sinx =0
1- ${sin}^{2} x$ $sin^2x$ - ${sin}^{2} x$ $sin^2x$ -sinx=0
1-2 ${sin}^{2} x$ $sin^2x$ -sinx=0
2 ${sin}^{2} x$ $sin^2x$ +sinx -1 =0
2 ${sin}^{2} x$ $sin^2x$ +(2-1) sinx -1=0
2 ${sin}^{2} x$ $sin^2x$ +2sinx - sinx -1 =0
2sinx(sinx +1) -1(sinx +1)=0
(sinx+1) ( 2sinx -1) =0

either or

sinx=-1 sinx=1/2

So you can solve for these values of sinx.

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How do you prove the identity $cos 2 x - sin x = 0$ $Cos2x-sinx=0$ ?

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How do you prove the identity cos2x−sinx=0 Cos2x-sinx=0?

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How do you prove the identity $cos 2 x - sin x = 0$ $Cos2x-sinx=0$ ?