Trig Identities

**casey_k** · May 6th 2009, 04:47 PM

csc x - sin x = cot x csc x

I expanded the left side and ended up with 1-sin^2 x/sin x then i turned it into cos^2 x/ sin x and I don't know what to do next because I'm not getting the other side.

**casey_k** · May 6th 2009, 06:23 PM

bump

**Soroban** · May 6th 2009, 06:53 PM

Hello, casey_k!

There must be a typo . . . The statement is not an indentity.

Could it be: . $\csc x - \sin x \:=\: \cot x\,{\color{blue}\cos x}$

$\csc x - \sin x \;=\;\frac{1}{\sin x} - \sin x \:=\:\frac{1-\sin^2 x}{\sin x} \;=\;\frac{\cos^2\!x}{\sin x} \;=\;\frac{\cos x}{\sin x}\cdot\cos x \;=\;\cot x\cos x$

**casey_k** · May 6th 2009, 07:04 PM

I really hope it's a typo because now I can solve the problem! I'll ask my teacher about it tomorrow and thank you so much!

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