Question

(sinA+sinB)^2+(cosA+cosB)^2-2 write as a single trig function

Answer 1

expand those squares with (x+y)^2 = x^2+2xy+y^2

Answer 2

then use sin^2 θ + cos^2 θ =1

Answer 3

most of the terms will cancel out

Answer 4

finally, use

cos u cos v + sin u sin v = sin (u + v)

Answer 5

@UnkleRhaukus Uh, that last identity doesn't look right to me. Try it with \( u = v = \pi/6\)

I think \[\cos u \cos v + \sin u \sin v = \cos (u-v)\]is what you want there...

Answer 6

your right @whpalmer4

Answer 7

im not sure what my deal is still not getting it the whole thing about trig id's

Answer 8

Can you show us what you've done so far?

Answer 9

2-(sinA-cosB)^2 -(cosA+sinB)^2 i expand the sq of both then i combine like terms and now im left with:2-sin^2A+cos^2B - cos^2A+2cosAsinB+Sin^2B

Answer 10

It looks like you're starting in the wrong place:
\[(\sin A+ \sin B)^2+(\cos A+\cos B)^2-2\]is what the problem statement reads.

Answer 11

thats the problem out my book

Answer 12

Well, that's not what you put up top!

Answer 13

But no matter. You work this problem in much the same way. Expand the squares of both and you should get \[-2 \cos (x) \sin (y)+2 \sin (x) \cos (y)-\sin ^2(x)-\cos ^2(x)-\sin ^2(y)-\cos ^2(y)+2\]
Use \(\sin^2(x) + \cos^2(x)= 1\) to simplify that (also do it for \(y\). When you are done, you'll be left with the \(\sin * \cos\) product terms, and you should be able to find them on your trig identity sheet.

Answer 14

where did you get that from?

Answer 15

oops the problem up top is a diff. i dont see how -sin^2A-cos^2A=-1 i kno sin^2a+cos^a=1

Answer 16

\[-\sin^2x -\cos^2x = -1\]Factor out a -1
\[-1(\sin^2x+\cos^2x) = -1\]\[-1(1) = -1\]