Asked by Peter
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?
A. -√2
B. 0
C. 1
D. 2√2
What equation can I use to solve this problem?
Answers
Answered by
Reiny
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
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
Answered by
Peter
Oh...okay. Thanks for the help.
Answered by
Auwal rabiu dankoli
pls simplify cos2θ ÷ sinθ + cosθ
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