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# Prove

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Prove these identities.

$$tan(\frac{v}{2})=csc(v)-cot(v)$$

$$1-\frac{1}{2}sin(2\theta)=\frac{sin^3\theta+cos^3\theta}{sin\theta+cos\theta}$$

May 12, 2019
edited by AdamTaurus  May 12, 2019

#1
+1

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   May 12, 2019
#2
+1

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)   May 12, 2019
#3
+1

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.   