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avatar+895 

Prove these identities.

(Gonna be honest I don't even know how to start these and my textbook has no examples. Joy.)

 

\(ln|sin\theta|=\frac{1}{2}(ln|1-cos(2\theta)|-ln2)\)

 

\(ln|cos\theta|=\frac{1}{2}(ln|1+cos(2\theta)|-ln2)\)

 May 13, 2019
 #1
avatar+105277 
+1

Note that    

 

cos2θ =  1 -2sin^2θ

 

So

 

(1/2) ( Ln l 1 - cos2θ l - ln (2) )   =

 

(1/2) ( Ln l 1 - (1 - 2sin^2θ) l - ln( 2) )  =

 

(1/2) ( Ln l 2sin^2 θl  - Ln (2) )  =

 

{By a log property  Ln (a *b)  = Ln a + Ln b )    ( and Ln l 2l    = Ln(2) )

 

(1/2) ( Ln (2) + Ln l sin^2θ l - Ln (2) ) =

 

[ Note that   l sin^2 θ l  = sin^2 θ  ]    so we have

 

(1/2) Ln (sin^2 θ)  =

 

Ln (sin^2 θ)^(1/2)  =

 

Ln √ [ sin^2 θ ]

 

And note, AT   that....... Ln  l sin θ l  is defined  as   Ln √ [ sin^2 θ ]

 

So....the right side = the left side

 

 

 

cool cool cool

 May 13, 2019
 #2
avatar+105277 
+1

The second one is similar  if you note that

 

cos 2θ   =  2cos^2θ - 1

 

See if you can do this one.....if not....I'll see if I can re-create the scene of the crime   ....

 

 

 

cool cool cool

 May 13, 2019
 #3
avatar+895 
+1

I figured out the second one thanks. It's nice on these ones because you know the answer you have to get so if you don't get it you did something wrong.

 

Also, it's always been frustrating to my OCD personality that natural log is notated ln instead of nl.

AdamTaurus  May 13, 2019
 #4
avatar+105277 
+3

LOL!!!!.....I think it comes from the New Latin term "Log Naturalis".....go figure  !!!!

 

 

 

cool cool cool

CPhill  May 13, 2019
 #5
avatar+895 
+1

Hmmm.... Cool, you learn something new every day!

AdamTaurus  May 13, 2019

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