Question

prove the identity (last one)

Answer 1

@Vocaloid I would give a try but I haven't touched precal in a week (cramming for APs) lol.

Answer 2

\[\cos^2x+2\cos(x)\cos(y)+\cos^2(y)\]

Answer 3

\[\sin^2x-2\sin(x)\sin(y)+\sin^2(y)\]

Answer 4

\[\cos 2 x+2\cos(x)\cos(y)+ \cos 2 (y) \sin 2 x−2\sin(x)\sin(y)+ \sin 2 (y)\]

Answer 5

let's back up a bit, your first two steps are good, not sure about your third step

think about how you can use the identity cos^2 + sin^2 = 1 to simplify the sum

Answer 6

2cos(x)cos(y)

Answer 7

wouldnt it cancel out

Answer 8

\[\cos^2x+2\cos(x)\cos(y)+\cos^2(y)+\sin^2x-2\sin(x)\sin(y)+\sin^2(y)\]

notice how you have cos^2(x) + sin^2(x) and cos^2(y) + sin^2(y)

what do these add up to, using the identity cos^2(theta)+sin^2(theta) = 1

Answer 9

cos^2(x) + sin^2(x) = 1
as does
cos^2(y) + sin^2(y) = 1

therefore our new sum is

2 + 2cos(x)cos(y) - 2sin(x)2sin(y)

Answer 10

if you look at the third identity, cos(x+y) = cos(x)cos(y) - sin(x)sin(y)

therefore you can re-write the equation

2 + 2cos(x)cos(y) - 2sin(x)2sin(y)

as

2 + 2cos(x+y)

thus completing your poof