+895 Prove these identities.
tan(v2)=csc(v)−cot(v)
1−12sin(2θ)=sin3θ+cos3θsinθ+cosθ
+130474 tan (v/2) =
sin(v/2) √ [ (1 - cosv) / 2 ] √ [ 1 - cos v]
_______ = _______________ = _____________ multiply top/bottom by √ [1 - cos v ] and we have
cos(v/2) √ [ (1 + cos v)/ 2 ] √ [ 1 + cos v ]
√[1 - cosv ] √ [ 1 - cos v ] 1 - cos v 1 - cos v 1 - cos v
____________________ = ____________ = ___________ = _________ =
√ [ 1 + cos v]√ [ 1 - cos v] √ [ 1 - cos^2v ] √ sin^2v sin v
1 - cos v
____ _____ =
sin v sin v
csc v - cot v
+130474 Note that sin^3 θ + cos^3 θ factors as a sum of cubes = ( sin θ + cos θ) ( sin^2 θ - sin θ cos θ + cos^2 θ)
So....the first factor cancels with the factor in the denominator on the right and we are left with
(sin^2 θ - sin θ cos θ + cos^2 θ) =
( sin^2 θ + cos ^2 θ) - sin θcos θ =
1 - (1/2)sin (2 θ)
+895 Thanks, CPhill. Some of these tick me off, because looking at how you did it, it was the same way I did, I just took one tiny wrong step and got WAYY off base.
