+895 Prove these identities.
(Gonna be honest I don't even know how to start these and my textbook has no examples. Joy.)
\(ln|sin\theta|=\frac{1}{2}(ln|1-cos(2\theta)|-ln2)\)
\(ln|cos\theta|=\frac{1}{2}(ln|1+cos(2\theta)|-ln2)\)
+129852 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
+129852 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 ....
+895 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.
