Question

Prove the following identity:
(cos x + cos y)^2 + (sin x – sin y)^2 = 2 + 2cos(x + y)

Answer 1

@dtan5457 Can you help me or find someone you can?

Answer 2

@BreanneAndrews Can you help me or find someone you can?

Answer 3

Try distributing the left hand side :)

Answer 4

OKay

Answer 5

Wrote the question down wrong...the two parentheses were supposed be squared. It should be fixed now @lυἶცἶ0210

Answer 6

Yea, but distributing is still the first step :P

Reply what you get when you distbute and I will help you from there ~

Answer 7

okay, one sec

Answer 8

I'm not sure if I did this write but this is what I got:
\[\cos^2x + \cos^2y + \sin^2x - \sin^2y\]

Answer 9

Are you still there?

Answer 10

Well you got 2/3's of it, when you distribute:

\(\ (cosx+cosy)^2+(sinx-siny)^2\) =
\([(cosx*cosx)+(cosx*cosy)+(cosy*cosy)]\) +
\([(sinx*sinx)+(sinx*-siny)+(-siny*-siny) ]\)

(cut off so had to break up)

But simplifing:
\(\large = [(cos^2x+cosxcoxy +cos^2y)]+[(sin^2 x-sinxsiny +sin^2 y )\)

Make sense so far?

Answer 11

Yep

Answer 12

What's next?

Answer 13

Use trig identities and group things:

\(\large (sin^2x+cos^2x)+(sin^2y+cos^2y)+(cosxcosy-sinxsiny) \)

Do you know your trig identities?

Answer 14

Kinda

Answer 15

Don't most of the trig identities simplify things down to cos x or sin x? So wouldn't that them pointless in this situation since they're already simplified to cos x and sin x?

Answer 16

By trig identities I mean ones like the Pythagorean Identities, odd-even identities, double-angle, sum and difference, etc.

Answer 17

Oh yeah, I have a sheet next to me that has those. Do I use them now?

Answer 18

Yup, use those to try and simplify this:

\(\large (sin^2x+cos^2x)+(sin^2y+cos^2y)+(cosxcosy-sinxsiny)\)

Answer 19

kk

Answer 20

What did you end up with?

Answer 21

\[((1 - \cos2x) / 2 + ((1 + \cos2x) / 2) + ((1 - \cos2x) / 2 + ((1 + \cos2x) / 2) + 1/2 (\cos (x - y) + \cos (x +y) - 1/2 (\cos (x - y) - \cos (x + y)\]

Answer 22

Hmmm, try using the Pythagorean Identities :)

Answer 23

One sec...

Answer 24

I got 1 + 1 + (cos x cos y - sin x sin y) but I don't know what to do with the last part

Answer 25

Right :)

And the last identity isn't as common as the others, but you'll have to use the Sum and Difference Identities:
http://prntscr.com/76van2

Answer 26

So in the end it'll be 2+cos(x+y) which equals the right hand side :P

Answer 27

Well actually the right side says 2 + 2 cos (x + y) but we only have one of the twos

Answer 28

I feel silly now, I messed up in distributing in the beginning /.<

It should of been:

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

\(\large (sin^2x+cos^2x)+(sin^2y+cos^2y)+2(cosxcosy-sinxsiny) \)

Now it'll simplify to 2+2cos(x+y) :P

Answer 29

Okay, thanks! Do you mind if I ask you another question?

Answer 30

Not at all, go ahead.

Answer 31

Okay, here it is:
Derive the trigonometric addition formula for sine:
sin(a + b) = sin a cos b + cos a sin b

Answer 32

Maybe you should make another post and see if someone else can help, I'm a bit stumped, sorry :/

Answer 33

Okay, thanks for your help!