Trig prob

**Sundae** · Apr 27th 2009, 09:34 AM

4 cosec22x + 4 cosec22x cot22x, need to simplify this and the answer is suppose 2 be 4cosec 4 2x

**running-gag** · Apr 27th 2009, 10:54 AM

Originally Posted by Sundae

4 cosec22x + 4 cosec22x cot22x, need to simplify this and the answer is suppose 2 be 4cosec 4 2x

Hi

$\frac{4}{\sin^2 2x} + \frac{4}{\sin^2 2x} \:\frac{\cos^2 2x}{\sin^2 2x} = \frac{4}{\sin^2 2x}\:\left(1 + \frac{\cos^2 2x}{\sin^2 2x}\right) = \frac{4}{\sin^2 2x}\:\frac{\sin^2 2x + \cos^2 2x}{\sin^2 2x}$

$\frac{4}{\sin^2 2x} + \frac{4}{\sin^2 2x} \:\frac{\cos^2 2x}{\sin^2 2x} = \frac{4}{\sin^4 2x}$

**Soroban** · Apr 27th 2009, 10:55 AM

Hello, Sundae!

Did you really expect us to be able to read that??

4 cosec22x + 4 cosec22x cot22x
need to simplify this and the answer is suppose 2 be 4cosec 4 2x

When u r riting math things, donut use a "2" 4 da word "to"
. . or use a "4" 4 da word "for".
Math iz hard enuf 2 read widout cutesy abbreviations n spellings 2 interpret.

Factor: . $4\csc\!22x\left(1 + \cot\!22x\right) \;=4\csc\!22x\,\csc\!22x \;=\;4\csc\!42x$ . . . Got it?

Okay, I'm kidding . . . I will take a wild guess that $22x$ is actually $^22x$

We have: . $4\csc^2\!2x + 4\csc^2\!2x\cot^2\!2x$

$\text{Factor: }\;4\csc^2\!2x\underbrace{\left[1 + \cot^2\!2x\right]}_{\text{This is }\csc^2\!2x}$

. . . . $= \;4\csc^2\!2x\cdot\csc^2\!2x \;=\;4\csc^4\!2x$

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