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#### paulmdrdo

##### Active member

- May 13, 2013

- 386

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

please explain.

- Thread starter paulmdrdo
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- Thread starter
- #1

- May 13, 2013

- 386

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

please explain.

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- #2

- Jan 26, 2012

- 4,202

What have you tried?

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- May 13, 2013

- 386

i just factored out the 1/5 and i'm stucked.What have you tried?

- Feb 29, 2012

- 342

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.

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]

- Mar 31, 2013

- 1,356

In the last step we can remove modulo sign as it positiveHello, 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]