Question #c42b1

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Question #c42b1

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Answer 1 · 2016-06-12T23:30:24

We will use the following:

$cos (α + β) = cos (α) cos (β) - sin (α) sin (β)$ $cos(alpha+beta) = cos(alpha)cos(beta)-sin(alpha)sin(beta)$
$cos (α - β) = cos (α) cos (β) + sin (α) sin (β)$ $cos(alpha - beta) = cos(alpha)cos(beta) + sin(alpha)sin(beta)$
$cos (45^{\circ}) = sin (45^{\circ}) = \frac{\sqrt{2}}{2}$ $cos(45^@) = sin(45^@) = sqrt(2)/2$

With those, we have

$cos (θ + 45^{\circ}) = cos (θ) cos (45^{\circ}) - sin (θ) sin (45^{\circ})$ $cos(theta+45^@) = cos(theta)cos(45^@)-sin(theta)sin(45^@)$

$= \frac{\sqrt{2}}{2} cos (θ) - \frac{\sqrt{2}}{2} sin (θ)$ $=sqrt(2)/2cos(theta)-sqrt(2)/2sin(theta)$

$cos (θ - 45^{\circ}) = cos (θ) cos (45^{\circ}) + sin (θ) sin (45^{\circ})$ $cos(theta-45^@) = cos(theta)cos(45^@)+sin(theta)sin(45^@)$

$= \frac{\sqrt{2}}{2} cos (θ) + \frac{\sqrt{2}}{2} sin (θ)$ $=sqrt(2)/2cos(theta)+sqrt(2)/2sin(theta)$

Adding the two, we get

$cos (θ + 45^{\circ}) + cos (θ - 45^{\circ})$ $cos(theta+45^@)+cos(theta-45^@)$

$= \frac{\sqrt{2}}{2} cos (θ) - \frac{\sqrt{2}}{2} sin (θ) + \frac{\sqrt{2}}{2} cos (θ) + \frac{\sqrt{2}}{2} sin (θ)$ $=sqrt(2)/2cos(theta)-sqrt(2)/2sin(theta)+sqrt(2)/2cos(theta)+sqrt(2)/2sin(theta)$

$= 2 (\frac{\sqrt{2}}{2} cos (θ))$ $=2(sqrt(2)/2cos(theta))$

$= \sqrt{2} cos (θ)$ $=sqrt(2)cos(theta)$

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Question #c42b1

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