Question #a0e09

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Question #a0e09

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Answer 1 · 2016-10-10T04:40:24

Use that sine is an odd function, cosine is an even function, and the identity ${cos}^{2} (θ) + {sin}^{2} (θ) = 1$ $cos^2(theta)+sin^2(theta) = 1$

Explanation:

We will use the following:

${cos}^{2} (θ) + {sin}^{2} (θ) = 1$ $cos^2(theta)+sin^2(theta) = 1$
$cos (- θ) = cos (θ)$ $cos(-theta) = cos(theta)$
$sin (- θ) = - sin (θ)$ $sin(-theta) = -sin(theta)$

With those,

$cos (θ) cos (- θ) - sin (θ) sin (- θ)$ $cos(theta)cos(-theta)-sin(theta)sin(-theta)$

$= cos (θ) (cos (θ)) - sin (θ) (- sin (θ))$ $=cos(theta)(cos(theta))-sin(theta)(-sin(theta))$

$= cos (θ) cos (θ) + sin (θ) sin (θ)$ $=cos(theta)cos(theta)+sin(theta)sin(theta)$

$= {cos}^{2} (θ) + {sin}^{2} (θ)$ $=cos^2(theta)+sin^2(theta)$

$= 1$ $=1$

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Question #a0e09

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