Verifying Trigonometric Identities Help! sin^2a - sin^2b = sin(a+b)*sin(a-b)?

Question

So Far

(sinacosb + cosasinb)(sinacosb - cosasinb)
sin^2acos^2b - cos^2asin^2b
?

Answer 1

sin^2(a) - sin^2(b) = sin(a + b) * sin(a - b)

You were correct in choosing the more complex side, the right hand side (RHS).

sin(a + b) * sin(a - b)

By the sine addition identity, sin(x + y) = sinxcosy + sinycosx.
Similarly, sin(x - y) = sinxcosy - sinycosx. So we have

[sinacosb + sinbcosa] [sinacosb - sinbcosa]

Which is a difference of squares, like you've already stated.

sin^2(a) cos^2(b) - sin^2(b)cos^2(a)

HERE is the major trick, that involves mathematical ingenuity. What you're going to do is add "ZERO" at this point. That is,
you're going to add sin^2(a)sin^2(b) - sin^2(a)sin^2(b) and write it smack in the middle between the two.

sin^2(a) cos^2(b) + sin^2(a)sin^2(b) - sin^2(a)sin^2(b) - sin^2(b)cos^2(a)

Factor the first two terms and the last two terms.

sin^2(a) [cos^2(b) + sin^2(b)] - sin^2(b) [sin^2(a) + cos^2(a)]

Note the popular identities in the brackets; they're both equal to 1. That is, sin^2(x) + cos^2(x) = 1.

sin^2(a) [1] - sin^2(b) [1]
sin^2(a) - sin^2(b) = LHS

As a side note, this "addition of 0" to solve proofs is a recurring theme, and is used in deriving the quadratic formula (completing the square) and the product rule of derivatives in Calculus. It's also used in mathematical induction. It just happened to come in handy here, as well.

Answer 2

that's no longer actual. attempt A = B = C = pi/3. Then the LHS is sin(2pi/3) = sqrt(3) /2 and the RHS is two*sin^2(pi/3)*cos(pi/3) = 2*(sqrt(3) /2)^2 *(a million/2) = 3/4. i ended up with sin(2A) - sin(2B) + sin(2C) = 4*cos(A)*sin(B)*cos(C), which does paintings with the above occasion information. i'm going to attend to hearken to from you appropriate to the accuracy of the given assertion in the past spending the time writing out a data.

Answer 3

sin^2a - sin^2b = (sina + sinb)(sina - sinb)
= 4sin[(a+b)/2] * cos[(a-b)/2]*sin[(a-b)/2] * cos[(a+b)/2]
= sin(a+b) * sin(a-b)

we have (a + b)(a - b) = a^2 - b^2
thus
(sinacosb + cosasinb)(sinacosb - cosasinb) = sin^2acos^2b - cos^2asin^2b

Answer 4

Puggy's answer is excellent. I can't beat him of this one.
=)

Ana