Question

prove that:
sin^2A - cos^2A.cos2B = sin^2B - cos^2B.cos2A

Answer 1

Okay start by expressing cos 2B as cos^2B-sin^2B

Sin^2A-cos^2A.(cos^2B-sin^2B)

=sin^2A-cos^2A.cos^2B+cos^2A.sin^22B

Now add and subtract cos^2B.sin^2A here

After that cos^2B as common and you get cos^2B.cos2A

The rest is easy:

=sin^2A-cos^2B.cos 2A-cos^2Bsin^2A+cos^2Asin^2B
=sin^2Asin^2B+cos^2Asin^2B-cos^2B.cos2A
=Sin^2B-cos^2B.cos 2A=RHS

Hence proved

Answer 2

Are you able to follow?

Answer 3

=sin^2A-cos^2B.cos 2A-cos^2Bsin^2A+cos^2Asin^2B
how did this come???

Answer 4

By doing the process I said. I didn't show it as I thought you can do it yourself.

Answer 5

actually, i didn't understand,, add and subtract cos^2B.sin^2A means??

Answer 6

=sin^2A-cos^2A.cos^2B+cos^2A.sin^2B+(cos^2B.sin^2A )-(cos^2B.sin^2A )

Do you get it now?

Answer 7

oh oh.. i thought of something else :p

Answer 8

lol ok Can you do the next step and get to

=sin^2A-cos^2B.cos 2A-cos^2Bsin^2A+cos^2Asin^2B ??

Answer 9

yeah.. thanks :)

Answer 10

Great! Yw :)

Answer 11

will u solve my another question too? i have already posted it!!

Answer 12

Sure but after 20 minutes maybe...Time for dinner!

Answer 13

ok...no problem..