Question

Use the addition formulae to find an expression for \[\cos^2(A+B) + \sin^2(A+B). Verify~that~your ~expression~ reduces \to 1.\]

Answer 1

wow, according to the instructions, this could get long....

Answer 2

Yes, it can... I got up to a step where cosA^2AcosB^2+sinA^2SinB^2+sinA^2cosB^2+cosA^2SinB^2.... But I have no clue if I'm correct...!

Answer 3

cos^2A(sin^2B + cos^2B) + sin^2B(cos^2a + sin^2a)...
I am sure you will get something like this after taking some terms common.

#I would have left this question, the answer is obvious and the solution according to the instructions is pretty boring,

Answer 4

I'd still like a run through to see if someone gets that answer, and then how to get 1 from there...

Answer 5

ok so lets see if you're on the right track...
cos(A+B) = cosAcosB - sinAsinB
sin(A+B) = 2sinAcosB

so cos^2(A+B) + sin^2(A+B) = (cosAcosB - sinAsinB)^2 + (2sinAsinB)^2
still with me?

Answer 6

Yes, I am. Sorry, my internet is going on and off!

Answer 7

pay no attention to my previous post. there was an error in the last statement.

using sum rule for cosine and sine we get...

(cosAcosB - sinAsinB)^2 + (2sinAcosB)^2

so FOILing this out we get:
(cosAcosB)^2 - 2*(cosAcosB)(sinAsinB) + (sinAsinB)^2 +4sin^2Acos^2B

Answer 8

(cosAcosB)^2 - 2*(cosAcosB)(sinAsinB) + (sinAsinB)^2 +4sin^2Acos^2B
cos^2A*cos^2B - 2*cosA*cosB*sinA*sinB + sin^2A*sin^2B +4*sin^2A*cos^2B

Answer 9

cos^2A*cos^2B - 2*cosA*cosB*sinA*sinB + sin^2A*sin^2B +4*sin^2A*cos^2B
cos^2A*cos^2B - 2*cosA*cosB*sinA*sinB + (sin^2A*sin^2B +4*sin^2A*cos^2B)
cos^2A*cos^2B - 2*cosA*cosB*sinA*sinB + sin^2A(sin^2B +4*cos^2B)

this is where I am so far.
gimme a sec.... lemme do it on paper...

Answer 10

Ok. Sure. No problem.

Answer 11

ok, no wonder I wasn't getting anywhere....
sin(A+B) = sinAcosB + cosAsinB

sorry 'bout that.
continuing...

Answer 12

Sure :)

Answer 13

so,
\[\cos ^{2}(A+B)=(cosAcosB - sinAsinB)^{2}\]
and
\[\sin ^{2}(A+B) = (sinAcosB + cosAsinB)^{2}\]

Answer 14

FOILing out the right side of both of these and adding them together will give you
\[(cosAcosB)^{2}+(sinAcosB)^{2}+(cosAsinB)^{2}\]

Answer 15

the middle terms cancelled each other out when you added the two right sides together.

Answer 16

Yes, that's where I got :) I just don't know how to carry on from there.

Answer 17

Wait, I think you're missing one
What of (sinAsinB)^2?

Answer 18

group the firs and last terms and factor out cos^2A:
(cosAcosB)^2 + (sinAsinB)^2 + (sinAcosB)^2 + (cosAsinB)^2
yes... you're right. almost done though....
are you up to this point? (double check for me)

Answer 19

Yes, I'm up to this point. I double checked, and you only left out the (sinAsinB)^2. Thanks!! I hope you can get to the 1 and show me :)

Answer 20

piece of cake from here....

Answer 21

cosAcosB)^2 + (sinAsinB)^2 + (sinAcosB)^2 + (cosAsinB)^2
= [cosAcosB)^2 + (cosAsinB)^2] + [(sinAsinB)^2 + (sinAcosB)^2]
= [cos^2A*cos^2B + cos^2A*sin^2B] + [sin^2A*sin^2B + sin^2A*cos^2B]
= cos^2A*[cos^2B + sin^2B] + sin^2A*[sin^2B + cos^2B]
= cos^2A*[1] + sin^2A*[1]
= cos^2A + sin^2A
= 1 . wheWWWWWWW

Answer 22

i guess i gotta review my trig formulas!!!

Answer 23

Thanks for all your help!

Answer 24

nw

Answer 25

Wait, how did you get rid of cos&2B + sin^2B?