Posted by **Peter** on Tuesday, July 31, 2012 at 8:50pm.

If θ represents an angle such that sin2θ = tanθ - cos2θ, then sin θ - cosθ =

A. -√2

B. 0

C. 1

D. 2√2

What equation can I use to solve this problem?

- Pre-Calculus -
**Reiny**, Tuesday, July 31, 2012 at 10:07pm
sin2A + cos2A - sinA/cosA = 0

2sinAcosA + 1 - 2sin^2 A - sinA/cosA = 0

2sinAcos^2 A + cosA - 2sin^2 A cosA - sinA = 0

2sinAcosA(cosA - coA) + (cosA - sinA) =

(cosA - sinA)(2sinAcosA + 1) = 0

cosA - sinA = 0 or ......

got it! That's all they asked for.

cosA - sinA = 0

- Pre-Calculus -
**Peter**, Tuesday, July 31, 2012 at 10:32pm
Oh...okay. Thanks for the help.

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