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What is tan theta in terms of sin theta in quadrant II?
tan = sin/cos Now cos2 = 1 - sin2 so cos = +/- sqrt(1 - sin2) In the second quadrant, cos is negative, so cos = - sqrt(1 - sin2) So that tan = sin/[-sqrt(1 - sin2)] or -sin/sqrt(1 - sin2)
Find the second derivative of y equals sin sqaured x?
y = sin2(x) y' = 2sin(x)cos(x) y'' = 2 [ cos(x)cos(x) + sin(x)(-sin(x)) ] = 2 [ cos2(x) - sin2(x) ] = 2 [ 1 - sin2(x) - sin2(x) ] = 2 [ 1 - 2sin2(x) ]
What is the solution to sec plus tan equals cos over 1 plus sin?
sec + tan = cos /(1 + sin) sec and tan are defined so cos is non-zero. 1/cos + sin/cos = cos/(1 + sin) (1 + sin)/cos = cos/(1 + sin) cross-multiplying, (1 + sin)2 = cos2 (1 + sin)2 = 1 - sin2 1 + 2sin + sin2 = 1 - sin2 2sin2 + 2sin = 0 sin2 + sin = 0 sin(sin + 1) = 0 so sin = 0 or sin = -1 But sin = -1 implies that cos = 0 and cos is non-zero. Therefore sin = 0 or the solutions are k*pi radians where k is an integer.
Prove each Indentity tanx mins sinx divided by tanxsinx equals tanxsinx divided by tanx plus sinx?
(tan x - sin x)/(tan x sin x) = (tan x sin x)/(tan x + sin x)[sin x/cos x) - sin x]/[(sin x/cos x)sin x] =? [(sin x/cos x)sin x]/[sin x/cos x) + sin x][(sin x - sin x cos x)/cos x]/(sin2 x/cos x) =? (sin2 x/cos x)/[(sin x + sin x cos x)/cos x)(sin x - sin x cos x)/sin2 x =? sin2 x/(sin x + sin x cos x)[sin x(1 - cos x)]/sin2 x =? sin2 x/[sin x(1 + cos x)(1 - cos x)/sin x =? sin x/(1 + cos x)(1 - cos x)/sin x =? [(sin x)(1 - cos x)]/[(1 + cos x)(1 - cos x)](1 - cos x)/sin x =? [(sin x)(1 - cos x)]/[1 - cos2 x)(1 - cos x)/sin x =? [(sin x)(1 - cos x)]/[1 - (1 - sin2 x)](1 - cos x)/sin x =? [(sin x)(1 - cos x)]/sin2 x(1 - cos x)/sin x = (1 - cos x)/sin x True
Sec - cos equals tansin?
Prove that tan(x)sin(x) = sec(x)-cos(x) tan(x)sin(x) = [sin(x) / cos (x)] sin(x) = sin2(x) / cos(x) = [1-cos2(x)] / cos(x) = 1/cos(x) - cos2(x)/ cos(x) = sec(x)-cos(x) Q.E.D
