Question #c1e7d

Question

1007 views around the world

You can reuse this answer
Creative Commons License

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.