Help Verifying Trigonometric Identity

Question

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.

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

Answer 1

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 $\sin(x) \neq 2$, $\forall x \in \mathbb{R}$.

Answer 2

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.

Answer 3

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