Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

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

share|cite|improve this question
Is there a guide to using MathJax somewhere? I expect I'll need it in the future ... – Daniel B. Jun 24 '12 at 19:06
up vote 7 down vote accepted

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

Move your mouse over the gray area for the answer.

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

share|cite|improve this answer
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 B. 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.

share|cite|improve this answer
Thank you very much for your help, substitution and poly long division were great tips, I'll remember them for later. – Daniel B. 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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.