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Simplification of Expression

paulmdrdo

Active member
May 13, 2013
386
how would you simplify this

$\displaystyle \frac{1}{5}\ln|\sin5x|+\frac{1}{5}\ln|\csc5x-\cot5x|$

please explain.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
Re: simplification of expression

What have you tried?
 

paulmdrdo

Active member
May 13, 2013
386

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
Re: simplification of expression

Hello Paul! Notice you can factor the $1/5$, yielding $$\frac{1}{5} \left[ \ln |\sin 5x | + \ln |\csc 5x - \cot 5x| \right],$$ and from here use that $\ln a + \ln b = \ln (ab)$. Consequently

$$\frac{1}{5} \left[ \ln |\sin 5x | + \ln |\csc 5x - \cot 5x| \right] = \frac{1}{5} \left[ \ln |\sin (5x) \cdot (\csc (5x) - \cot (5x) )| \right] = \frac{1}{5} \left[ \ln |1 - \cos (5x)| \right].$$

Cheers. :D
 

soroban

Well-known member
Feb 2, 2012
409
Re: simplification of expression

Hello, paulmdrdo!

[tex]\text{Simplify: }\:\tfrac{1}{5}\ln|\sin5x|+\tfrac{1}{5}\ln|\csc5x-\cot5x|[/tex]

[tex]\begin{array}{ccc}\text{Factor:} & \tfrac{1}{5}\big(\ln|\sin5x| + \ln|\csc5x - \cot5x|\big) \\ \\ \text{Combine logs:} & \tfrac{1}{5}\ln|\sin5x(\csc5x - \cot5x)| \\ \\ \text{Distribute:} & \tfrac{1}{5}\ln|\sin5x\csc5x - \sin5x\cot5x| \\ \\ \text{Simplify:} & \tfrac{1}{5}\ln|1-\cos5x| \\ \\ \text{Further?} & \tfrac{1}{5}\ln\left|2\sin^2\! \tfrac{5x}{2}\right| \end{array}[/tex]
 

kaliprasad

Well-known member
Mar 31, 2013
1,322
Re: simplification of expression

Hello, paulmdrdo!


[tex]\begin{array}{ccc}\text{Factor:} & \tfrac{1}{5}\big(\ln|\sin5x| + \ln|\csc5x - \cot5x|\big) \\ \\ \text{Combine logs:} & \tfrac{1}{5}\ln|\sin5x(\csc5x - \cot5x)| \\ \\ \text{Distribute:} & \tfrac{1}{5}\ln|\sin5x\csc5x - \sin5x\cot5x| \\ \\ \text{Simplify:} & \tfrac{1}{5}\ln|1-\cos5x| \\ \\ \text{Further?} & \tfrac{1}{5}\ln\left|2\sin^2\! \tfrac{5x}{2}\right| \end{array}[/tex]
In the last step we can remove modulo sign as it positive