2.sinX - sin2X+ cotg2X/2 = 10/3

2.sinX + sin2X

I have an exam tomorrow.... please help

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- May 6th 2009, 05:58 AM #1

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- May 6th 2009, 07:10 AM #2

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I will guess that "cotg" means "cotangent", that decimal points mean multiplication, and that 2s after function names indicate squaring of the function, so the equation is meant to be as follows:

. . . . .$\displaystyle \frac{\sin(x)\, -\, \sin^2(x)}{2\sin(x)\, +\, \sin^2(x)}\, +\, \cot^2\left(\frac{x}{2}\right)\, =\, \frac{10}{3}$

What were the instructions for this exercise? What have you tried so far? Where are you stuck?

Please be complete. Thank you!

- May 6th 2009, 08:09 AM #3

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- May 6th 2009, 08:43 AM #4

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