Question

complete the identity cos(x+y)cos(x-y)

Answer 1

Might involve the use of this identity:
\[\cos(x\pm y)=\cos x\cos y\mp \sin x\sin y\]

Answer 2

How do you complete the identity?

Answer 3

By "complete the identity" do you mean "find an equivalent expression for"? Because that's what it sounds like.

Answer 4

\[\sin(A+B)=\sin A \cos B + \cos A \sin B\]\[\sin(A-B)=\sin A \cos B - \cos A \sin B\]\[\cos(A+B)=\cos A \cos B - \sin A \sin B\]\[\cos(A-B)=\cos A \cos B + \sin A \sin B\]

all of those.....

Answer 5

it basically asking you.
\[\cos(A+B)\cos(A-B)\]\[\cos(A+B) ~~~\times ~~~\cos(A-B)\]in other words, it's
\[(\cos A \cos B - \sin A \sin B)~~~ \times~~~(\cos A \cos B + \sin A \sin B)\]

Answer 6

\[(\cos A \cos B + \sin A \sin B) ~~~~~\times~~~~~ (\cos A \cos B - \sin A \sin B)\]
using \[(a+b)(a-b)=a^2-b^2\]

Answer 7

Yes. I tried multiplying it out. factoring, but i get stuck
\[A.cosa^2cosb^2+sina^2sinb^2\]\[B. \[2-\sin^2a - \sin^2b\]C. cosB^2-2\sin^2asin^2b\]\[D. \cos^b-\sin^2a\]

Answer 8

\[\cos^2A~~\cos^2B~~-~~Sin^2A~Sin^2B\]

Answer 9

so what i am doing?

Answer 10

do you want to see something cool?

Answer 11

Lets finish it off.

Answer 12

USING. \[Cos^2x=1-Sin^2x\]

Answer 13

1-sin is cos?

Answer 14

cos^2?

Answer 15

\[(1-\sin^2A)(1-Sin^2B)-Sin^2A~~Sin^2B\]

Answer 16

\[1-Sin^2A-Sin^2B+Sin^2ASin^2B~~~~~~~-Sin^2ASin^2B\]

Answer 17

\[1-Sin^2A-Sin^2B\]\[Cos^2A-Sin^2B\]is the final answer.

Answer 18

can you explain how you get to that please?

Answer 19

Are you completely lost?

Answer 20

I GOT IT
thanks

Answer 21

thank you for your help

Answer 22

Anytime, tried my best, here is your medal for willingness to learn.....