Question #c1e7d

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Answer 1 · 2016-12-13T03:41:21

Please see the explanation

Verify: $(sin(theta) + cos(theta))^2 = 1 + 2sin(theta)cos(theta)$

I will only make changes to the left side.

Expand the square:

$sin^2(theta) + 2sin(theta)cos(theta) + cos^2(theta) = 1 + 2sin(theta)cos(theta)$

Move the swap the second and third terms:

$sin^2(theta) + cos^2(theta) + 2sin(theta)cos(theta) = 1 + 2sin(theta)cos(theta)$

Use the identity $1 = sin^2(theta) + cos^2(theta)$ :

$1 + 2sin(theta)cos(theta) = 1 + 2sin(theta)cos(theta)$

Verified.