sin(A+B)sin(A-B) = sin^2A-sin^2B is what I have to prove

So

LHS= (sinAcosB+cosAsinB)(sinAcosB-cosAsinB)

This is obviously a quadratic. I just don't know how it would multiply

e.g sinAcosB x sinAcosB = ???

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- May 6th 2009, 08:20 AMgamboocompound-angle formulae with trig identity question
sin(A+B)sin(A-B) = sin^2A-sin^2B is what I have to prove

So

LHS= (sinAcosB+cosAsinB)(sinAcosB-cosAsinB)

This is obviously a quadratic. I just don't know how it would multiply

e.g sinAcosB x sinAcosB = ??? - May 6th 2009, 08:27 AMbebrave
==>sin^2A.cos^2B-sin^2Bcos^2A

===> sin^2A(1-sin^2B)-sin^2B(1-sin^2A)

===> sin^2A-sin^2Asin^2 B-sin^2B+sin^2Asin^2B

===> sin^2A-sin^2B. - May 6th 2009, 08:43 AMbebrave
sin(A+B)=sinAcosB+cosAsinB

sin(x) = cos(90- x) and cos(x) = sin(90 - x)

let x = A + B,

we have,

sin(A+B)=cos (90-(A+B))

then i can write this

cos90-A-B)as cos((90-A)-B)

cos(A-B) =cos(A)cos(B) + sin(A)sin(B)

sin(A+B)=cos(90-A)cosB +sin(90-A)sin(B)

sin(A+B)=sinAcosB+cosAsinB

i think these prooves can help you... - May 6th 2009, 10:03 AMgamboo
Thankyou

- May 6th 2009, 10:06 AMbebrave
you are welcome ;)