Expert: Sherry Wallin - 3/22/2010

Question
Can you plaese help me verify the following identity?

(3sinQ-2cos^2Q)/2+sinQ = 2sinQ-1

Answer
First Jim you need to write the problem more clearly...is it (3sinQ-2cos^2Q)/2 and then add sinQ or is it (3sinQ-2cos^2Q)/(2+sinQ)? If it is the latter (which I suspect it is) then you need to put parentheses around the first part, i.e.: (3sinQ-2cos^2Q)/(2 + sinQ).

To verify the identity I would replace cos^2Q with 1-sin^2Q giving me the new expression:
[3sinQ-2(1-sin^2Q)]/(sinQ+2) noting that 2+sinQ is the same as sinQ+2
=[3sinQ-2+2sin^2Q]/(sinQ+2)
=(2sinQ-1)(sinQ+2)/sinQ+2) note that 3sinQ-2+2sin^2Q = 2sin^2Q+3sinQ-2 and let sinQ = x so that what you have to factor is 2x^2+3x-2 = (2x-1)(x+2) and now substitute back in for x = sinQ and cancel the sinQ+2 in the numerator and the denominator:

(2sinQ-1)(sinQ+2)/sinQ+2) = 2sinQ-1 as desired

Math Prof