# Help Verifying Trigonometric Identity

I could really use help, hint or otherwise, in proving a trigonometric identity:

We are only allowed to work on one side of the equation.

$$\dfrac{2\sin^2(x)-5\sin(x)+2}{\sin(x)-2} = 2\sin(x)-1$$

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Is there a guide to using MathJax somewhere? I expect I'll need it in the future ... –  Daniel Ball Jun 24 '12 at 19:06

HINT: Factorize the numerator and cancel terms arguing why the terms you are canceling are not zero.

$$\dfrac{2 \sin^2(x) - 5 \sin(x) + 2}{\sin(x) - 2} = \dfrac{2 \sin^2(x) - 4 \sin(x) - \sin(x) + 2}{\sin(x) - 2}\\ = \dfrac{2 \sin(x) (\sin(x) - 2) - ( \sin(x) - 2)}{\sin(x) - 2}\\= \dfrac{(\sin(x) - 2) (2 \sin(x) - 1)}{\sin(x) - 2} = 2 \sin(x) - 1$$ We are allowed to cancel $\sin(x) - 2$ since $\sin(x) \neq 2$, $\forall x \in \mathbb{R}$.

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All of the answers were very useful, unfortunately I can only mark one and yours provides the most complete reasoning for me. Thank you very much! –  Daniel Ball Jun 24 '12 at 0:05

It may help to write $y = \sin x$. Then the equation simplifies to

$$\frac{2y^2 - 5y + 2}{y - 2} = 2y - 1.$$

To get the result, you could then try doing a polynomial long division on the left hand side.

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Thank you very much for your help, substitution and poly long division were great tips, I'll remember them for later. –  Daniel Ball Jun 24 '12 at 0:06

Factor $2\sin(x)^2-5\sin(x)+2$ to get $(2\sin(x)-1)(\sin(x)-2)$ and the result follows.

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