sin(180-x)-cotxsin(x-90)=cscx

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- Apr 7th 2008, 09:59 PMThelemaVerifying that the equation is an identity
sin(180-x)-cotxsin(x-90)=cscx

- Apr 8th 2008, 02:31 AMtopsquark
$\displaystyle sin(a \pm b) = sin(a)~cos(b) \pm sin(b)~cos(a)$

So:

$\displaystyle sin(180 - x) = sin(180)~cos(x) - sin(x) ~cos(180) = sin(x)$

and

$\displaystyle sin(x - 90) = sin(x)~cos(90) - sin(90)~cos(x) = -cos(x)$

So your left hand side becomes:

$\displaystyle sin(x) + cot(x)~cos(x)$

Can you take it from here?

-Dan