Question

The following identitit invovle the reciprocal, quotient, and Pythagorean relationships. Prove each one.
cos^2(x)cos^2(y) + sin^2(x)sin^2(y) +sin^2(x)cos^2(y) + sin ^2(y)cos^2

Answer 1

That's not an identity. That's an expression.

Answer 2

I copied word for word what is on the sheet, I belive i have to break this down into simplier terms

Answer 3

It doesn't equal anything?

Answer 4

Is there no way to simplify this?

Answer 5

You can simplify it, but you can't prove anything.

Answer 6

How would I simplify it?

Answer 7

The last part is incomplete.

Answer 8

that is how it is written

Answer 9

sin ^2(y)cos^2
Okay then. Good luck.

Answer 10

the last part is sin ^2(y)cos^2(x) or sin ^2(y)cos^2(y) ???

Answer 11

idk it just ended as sin^2(y) cos^2 <- it is missing x or y, possible typo on teachers part?

Answer 12

i guess the last part is sin ^2(y)cos^2(x).
if so, we can simply it be :

cos^2(x)cos^2(y) + sin^2(x)sin^2(y) +sin^2(x)cos^2(y) + sin ^2(y)cos^2(x)
= ( cos^2(x)cos^2(y) + sin ^2(y)cos^2(x) ) + ( sin^2(x)sin^2(y) +sin^2(x)cos^2(y) )
= cos^2(x) (cos^2(y) + sin^2(y)) + sin^2(x) (sin^2(y) + cos^2(y))
= cos^2(x) * 1 + sin^2(x) * 1
= cos^2(x) + sin^2(x)
= 1

Answer 13

for the 1st step, it like a+b+c+d. it would be (a+d) + (b+c)

Answer 14

you grouped them?

Answer 15

yes, then factor out which the common's and use the identity sin^2 + cos^2 = 1

Answer 16

alright I worked it out, is there any way to remember which steps to take? because after each question I have a hard time just starting :c

Answer 17

Factorize, recognize identities, simplify.