Question #c42b1

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Answer 1 · 2016-06-12T23:30:24

We will use the following:

With those, we have

$cos(theta+45^@) = cos(theta)cos(45^@)-sin(theta)sin(45^@)$

$=sqrt(2)/2cos(theta)-sqrt(2)/2sin(theta)$

$cos(theta-45^@) = cos(theta)cos(45^@)+sin(theta)sin(45^@)$

$=sqrt(2)/2cos(theta)+sqrt(2)/2sin(theta)$

Adding the two, we get

$cos(theta+45^@)+cos(theta-45^@)$

$=sqrt(2)/2cos(theta)-sqrt(2)/2sin(theta)+sqrt(2)/2cos(theta)+sqrt(2)/2sin(theta)$

$=2(sqrt(2)/2cos(theta))$

$=sqrt(2)cos(theta)$