How do you prove $1 + sin 2 x = {(sin x + cos x)}^{2}$ $1 + sin 2x = (sin x + cos x)^2$ ?

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How do you prove $1 + sin 2 x = {(sin x + cos x)}^{2}$ $1 + sin 2x = (sin x + cos x)^2$ ?

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Answer 1 · 2016-02-19T01:30:35

Please refer to explanation below

Explanation:

Remember : ${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x + cos^2x = 1$

$2 sin x cos x = sin 2 x$ $2sinx cosx = sin2x$

Step 1: Rewrite the problem as it is

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

Step 2: Pick a side you want to work on - (right hand side is more complicated)

$1 + sin (2 x) = (sin x + cos x) (sin x + cos x)$ $1+ sin (2x) = (sin x+ cos x)(sin x+cosx)$

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

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

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

$= 1 + 2 sin x cos x$ $= 1 + 2sinx cos x$

= $1 + sin 2 x$ $1 + sin 2x$

Q.E.D

Noted: the left hand side is equal to right hand side, this meant this expression is correct. We can conclude the proof by add QED (in Latin meant quod erat demonstrandum, or "which is what had to be proven")

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How do you prove $1 + sin 2 x = {(sin x + cos x)}^{2}$ $1 + sin 2x = (sin x + cos x)^2$ ?

1 Answer

Explanation:

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