Thread: Trig prob

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

Originally Posted by Sundae

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

Hi

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

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

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





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


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

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

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