Question #b8b69

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Question #b8b69

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Answer 1 · 2016-11-02T18:43:17

yes,for $x = \frac{π}{4}, {sin}^{2} x + {cos}^{2} x = 2 sin x cos x$ $x=pi/4,sin^2x+cos^2x=2sinxcosx$

Explanation:

here for $x = \frac{π}{4}, sin x = cos x = (\frac{1}{\sqrt{2}})$ $x=pi/4,sinx=cosx=(1/sqrt2)$
so , $sin x = cos x$ $sinx=cosx$
$sin x - cos x = 0$ $sinx-cosx=0$
squaring both sides we get ${(sin x - cos x)}^{2} = 0$ $(sinx-cosx)^2=0$
or , ${sin}^{2} x + {cos}^{2} x - 2 sin x cos x = 0$ $sin^2x+cos^2x-2sinxcosx=0$
or ${sin}^{2} x + {cos}^{2} x = 2 sin x cos x$ $sin^2x+cos^2x=2sinxcosx$ proved.

Answer 2 · 2016-11-05T11:30:26

While this relation is true for certain values of $x$ $x$ it is not valid in general.
I can not be proven as an identity because it is not, in general, true.

Explanation:

${sin}^{2} (x) + {cos}^{2} (x) = 1$ $sin^2(x)+cos^2(x)=1$ (this is an extension of the Pythagorean Theorem)

For x=0
$XXX 2 sin (x) \cdot cos (x) = 2 \times 0 \times 1 = 0$ $color(white)("XXX")2sin(x)⋅cos(x)=2×0×1=0$

${sin}^{2} (x) + {cos}^{2} (x) \neq 2 sin (x) \cdot cos (x)$ $sin^2(x)+cos^2(x)≠2sin(x)⋅cos(x)$

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Question #b8b69

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