Prove these identities.

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

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

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

AdamTaurus May 13, 2019

#1**+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

CPhill May 13, 2019

#2**+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 ....

CPhill May 13, 2019

#3**+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