How Do You Prove:- ( cos x + sin x) ( cos x - sin x ) = 2 cos^ x-1 ??

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How Do You Prove:- ( cos x + sin x) ( cos x - sin x ) = 2 cos^ x-1 ??

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Answer 1 · 2018-04-18T15:24:10

See method below

Explanation:

$(cos x + sin x) (cos x - sin x) = {cos}^{2} x - {sin}^{2} x$ $(cosx + sinx)(cosx - sinx) = cos^2x- sin^2x$

since ${cos}^{2} x + {sin}^{2} x = 1 \Rightarrow {cos}^{2} x + {sin}^{2} x - 1 = 0$ $cos^2x + sin^2x = 1 => cos^2x + sin^2x - 1 = 0$

Adding zero to the expression wont change the value and since ${cos}^{2} x + {sin}^{2} x - 1 = 0$ $cos^2x + sin^2x - 1=0$ we can simply add that without changing the value of the expression.

$(cos x + sin x) (cos x - sin x) = {cos}^{2} x - {sin}^{2} x + {cos}^{2} x + {sin}^{2} x - 1$ $(cosx + sinx)(cosx - sinx) = cos^2x- sin^2x + cos^2x + sin^2x - 1$

${sin}^{2} x$ $sin^2x$ cancels out

$(cos x + sin x) (cos x - sin x) = {cos}^{2} x + {cos}^{2} x - 1$ $(cosx + sinx)(cosx - sinx) = cos^2x + cos^2x - 1$
$(cos x + sin x) (cos x - sin x) = 2 {cos}^{2} x - 1$ $(cosx + sinx)(cosx - sinx) = 2cos^2x - 1$

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How Do You Prove:- ( cos x + sin x) ( cos x - sin x ) = 2 cos^ x-1 ??

1 Answer

Explanation:

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