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avatar+895 

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
avatar+102417 
+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

 

 

cool cool cool

 May 12, 2019
 #2
avatar+102417 
+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)

 

 

 

cool cool cool

 May 12, 2019
 #3
avatar+895 
+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.

AdamTaurus  May 12, 2019
 #4
avatar+102417 
0

HAHAHA!!!....yeah....that can happen   !!!!

 

 

cool cool cool

CPhill  May 12, 2019

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