Question
how do you solve this trig identity? i don't get it at all!
cos(a+b)cos(a-b)=cos^2a-cos^2b-1
cos(a+b)cos(a-b)=cos^2a-cos^2b-1
Answers
Steve
cos(a+b) = cosa cosb - sina sinb
cos(a-b) = cosa cosb + sina sinb
since (x-y)(x+y) = x^2-y^2, that gives us
(cosa cosb)^2 - (sina sinb)^2
= cos^2a cos^2b - sin^2a sin^2b
= cos^2a (1-sin^2b) - (1-cos^2a) sin^2b
= cos^2a - sin^2b
= cos^2a - (1-cos^2b)
= cos^a + cos^2b - 1
I think you have a typo in your equation.
cos(a-b) = cosa cosb + sina sinb
since (x-y)(x+y) = x^2-y^2, that gives us
(cosa cosb)^2 - (sina sinb)^2
= cos^2a cos^2b - sin^2a sin^2b
= cos^2a (1-sin^2b) - (1-cos^2a) sin^2b
= cos^2a - sin^2b
= cos^2a - (1-cos^2b)
= cos^a + cos^2b - 1
I think you have a typo in your equation.