Question

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

Answer 1

Multiply to expand it out

Answer 2

so would it be (cos x^2 + cos y^2) + (sin x^2 -sin y^2)

Answer 3

No, you can't just pull the exponent in

Answer 4

Okay, I just started on this unit, and I am having a bit of trouble. Do you mind helping me through the steps?

Answer 5

\[(a+b)^2 \text{ means } (a+b)(a+b) \text{ which also means } a^2+2ab+b^2 \]

Answer 6

\[
(\cos x+ \cos y)^2 =(\cos x+ \cos y)(\cos x+ \cos y)
\]This is like foil

Answer 7

would you do the same for (sin x - sin y)^2 or just (cos x+ cos y)^2?

Answer 8

like this \[(\cos x + \cos y)(\cos x + \cos y)+(sinx + \sin y)(\sin x+ \sin y)\] ?

Answer 9

Im not really sure what I do next.

Answer 10

I think you had (sin(x)-sin(y))^2 not (sin(x)+sin(y))^2

Answer 11

but as @wio said you need to multiply (cos(x)+cos(y)) and (cos(x)+cos(y))
and also do the same for the other thing you have squared

Answer 12

\[(\text{ kitty } - \text{ doggy})^2 \\ = (\text{ kitty } - \text{ doggy})(\text{ kitty } - \text{ doggy}) \\ =\text{ kitty } \cdot \text{ kitty} - \text{ kitty} \cdot \text{ doggy}-\text{ kitty} \cdot \text{ doggy}+ \text{ doggy} \cdot \text{ doggy} \\ =(\text{ kitty} )^2-2 \text{ kitty} \cdot \text{ doggy}+ (\text{ doggy})^2\]

Answer 13

Okay Ill try that

Answer 14

so \[(cosx)^2-2\cos x \cos y +(\cos y)^2\] right?

Answer 15

except you had (cos(x)+cos(y))^2

so you should have (cos(x))^2+2cos(x)cos(y)+(cos(y))^2

Answer 16

now expand also the (sin(x)-sin(y))^2

Answer 17

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

Answer 18

there is a minus sign in between sin(x) and sin(y) not a +

Answer 19

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

Answer 20

now use some identities

Answer 21

like the Pythagorean identity
and the sum/difference identity for cosine