How do you prove $cos (2 x) = {[cos (x)]}^{2} - {[sin (x)]}^{2}$ $cos(2x) = [cos(x)]^2 - [sin(x)]^2$ and $sin (2 x) = 2 sin (x) cos (x)$ $sin(2x) = 2sin(x)cos(x)$ using Euler's Formula: $e^{i x} = cos (x) + i sin (x)$ $e^(ix) = cos(x) + isin(x)$ ?

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How do you prove $cos (2 x) = {[cos (x)]}^{2} - {[sin (x)]}^{2}$ $cos(2x) = [cos(x)]^2 - [sin(x)]^2$ and $sin (2 x) = 2 sin (x) cos (x)$ $sin(2x) = 2sin(x)cos(x)$ using Euler's Formula: $e^{i x} = cos (x) + i sin (x)$ $e^(ix) = cos(x) + isin(x)$ ?

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Answer 1 · 2015-08-25T06:47:14

Use Euler's Formula to evaluate $cos (2 x) + i sin (2 x)$ $cos(2x)+i sin(2x)$ , then equate real and imaginary parts to get the double angle formulae.

Explanation:

$cos (2 x) + i sin (2 x)$ $cos(2x)+i sin(2x)$

$= e^{2 i x} = {(e^{i x})}^{2}$ $= e^(2ix) = (e^(ix))^2$

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

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

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

Equate real and imaginary parts to get:

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

$sin (2 x) = 2 sin (x) cos (x)$ $sin(2x) = 2 sin(x)cos(x)$

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How do you prove $cos (2 x) = {[cos (x)]}^{2} - {[sin (x)]}^{2}$ $cos(2x) = [cos(x)]^2 - [sin(x)]^2$ and $sin (2 x) = 2 sin (x) cos (x)$ $sin(2x) = 2sin(x)cos(x)$ using Euler's Formula: $e^{i x} = cos (x) + i sin (x)$ $e^(ix) = cos(x) + isin(x)$ ?

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Explanation:

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How do you prove cos(2x)=[cos(x)]2−[sin(x)]2cos(2x) = [cos(x)]^2 - [sin(x)]^2 and sin(2x)=2sin(x)cos(x)sin(2x) = 2sin(x)cos(x) using Euler's Formula: eix=cos(x)+isin(x)e^(ix) = cos(x) + isin(x)?

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Explanation:

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How do you prove $cos (2 x) = {[cos (x)]}^{2} - {[sin (x)]}^{2}$ $cos(2x) = [cos(x)]^2 - [sin(x)]^2$ and $sin (2 x) = 2 sin (x) cos (x)$ $sin(2x) = 2sin(x)cos(x)$ using Euler's Formula: $e^{i x} = cos (x) + i sin (x)$ $e^(ix) = cos(x) + isin(x)$ ?