# trig identities

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• Sep 5th 2010, 02:56 AM
sa-ri-ga-ma
In post#11 Archie Meade reached the step

sinA + sinB = 2sin(A+B)/2*cos(A-B)/2

sinA - sinB = 2cos(A+B)/2*sin(A-B)/2.

From that step

(sinA + sinB)(sinA - sinB) = [2sin(A+B)/2*cos(A+B)/2][2sin(A-B)/2*cos(A-B)/2]

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

When you try to solve the equation, it is better to take A>B. So I sugested to write the equation as

sin^(2x) - sin^2(x-30) = 0
• Sep 5th 2010, 03:35 AM
jayshizwiz
Quote:

(sinA + sinB)(sinA - sinB) = [2sin(A+B)/2*cos(A+B)/2][2sin(A-B)/2*cos(A-B)/2]
I believe (sinA + sinB)(sinA - sinB) = [2sin(A+B)/2*cos(A-B)/2][2sin(A-B)/2*cos(A+B)/2]

and further... how do we get from [2sin(A+B)/2*cos(A+B)/2][2sin(A-B)/2*cos(A-B)/2] to sin(A+B)sin(A-B)
• Sep 5th 2010, 04:39 AM
jayshizwiz
What I'm trying to say is that it's easy to go from sin(A+B)sin(A-B) to sin^2A - sin^2B
but I can't figure out how to go from sin^2A - sin^2B to sin(A+B)sin(A-B). Can someone please show me the proof??
• Sep 5th 2010, 05:54 AM
sa-ri-ga-ma
Quote:

Originally Posted by jayshizwiz
I believe (sinA + sinB)(sinA - sinB) = [2sin(A+B)/2*cos(A-B)/2][2sin(A-B)/2*cos(A+B)/2]

and further... how do we get from [2sin(A+B)/2*cos(A+B)/2][2sin(A-B)/2*cos(A-B)/2] to sin(A+B)sin(A-B)

I am using sin(2θ) = 2sinθcos(θ). So [2sin(A+B)/2*cos(A+B)/2] = sin(A+B)
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