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Pre-Calculus

posted by on .

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 - ,

    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 - ,

    Oh...okay. Thanks for the help.

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