Question

show steps to identify: (cos x + cos y)^2 + (sin x – sin y)^2 = 2 + 2cos(x + y)

Answer 1

Use
\[(a\pm b)^2 = a^2 \pm 2 a b + b^2
\]

Answer 2

Then use
\[
\cos ( x + y) =\cos (x) \cos(y) - \sin(x) \sin(y)
\]

Answer 3

could you please explain this a little more? I don't really understand how to use them

Answer 4

Some moderators will issue warning if we try to help too much.!!!!

Answer 5

\[(cos x + cos y)^2 = \cos^2(x) + 2 \cos(x) \sin(x) + \sin^2(x)= 1 + 2\cos(x) \sin(x)

\]

Answer 6

Do the same for the next term and expand it.

Answer 7

Eh...
\[(cos x + cos y)^2=cos^2x + 2cosxcosy +cos^2y\]\[(sin x – sin y)^2 = sin^2x - 2sinxsiny +sin^2y\]
So,
\[(cos x + cos y)^2 + (sin x – sin y)^2 = (cos^2x + 2cosxcosy +cos^2y)+(sin^2x - 2sinxsiny +sin^2y)\]\[=(cos^2x +sin^2x) + (cos^2y+sin^2y) + 2(cosxcosy-sinxsiny)\]\[=1+1+ 2 cos(x+y)\]\[=can \ you \ continue?\]

Answer 8

For the fifth line:
\[(cosx+cosy)^2 + (sinx-siny)^2\]\[=(cos^2x+2cosxcosy+cos^2y)+(sin^2x-2sinxsiny+sin^2y)\]