# Evaluate $\dfrac{-\sin{x}+\sin{2x}}{1-\cos{x}+\cos{2x}}$

A rational trigonometric expression is given in this problem to evaluate its value. In both numerator and denominator, trigonometric functions sine and cosine appear with double angles. So, the given trigonometric expression in rational form can be simplified by using the double angle trigonometric identities.

$\dfrac{-\sin{x}+\sin{2x}}{1-\cos{x}+\cos{2x}}$

### Simplify the expression in the numerator

Let’s concentrate on the expression in the numerator. The negative sign of sine of angle $x$ is added to the sine of double angle $2x$. For simplifying the expression in the numerator, the sine of angle $2x$ can be expanded by the sine double angle identity.

$=\,\,\,$ $\dfrac{-\sin{x}+2\sin{x}\cos{x}}{1-\cos{x}+\cos{2x}}$

In this trigonometric expression, there is a $\sin{x}$ factor common in both terms. So, it can be taken out common from the both terms for simplifying this expression further.

$=\,\,\,$ $\dfrac{(-1) \times \sin{x}+2 \times \sin{x} \times \cos{x}}{1-\cos{x}+\cos{2x}}$

$=\,\,\,$ $\dfrac{(-1) \times \sin{x}+\sin{x} \times 2 \times \cos{x}}{1-\cos{x}+\cos{2x}}$

$=\,\,\,$ $\dfrac{(-1) \times \sin{x}+\sin{x} \times 2\cos{x}}{1-\cos{x}+\cos{2x}}$

$=\,\,\,$ $\dfrac{\sin{x} \times \Big((-1)+2\cos{x}\Big)}{1-\cos{x}+\cos{2x}}$

$=\,\,\,$ $\dfrac{\sin{x} \times (-1+2\cos{x})}{1-\cos{x}+\cos{2x}}$

### Simplify the expression in the denominator

It is time to focus on simplifying the trigonometric expression in the denominator. There is a cosine term with double angle. It can be expanded to express it in terms of cosine of angle as per cos of double angle formula.

$=\,\,\,$ $\dfrac{\sin{x} \times (-1+2\cos{x})}{1-\cos{x}+2\cos^2{x}-1}$

Now, let us try to simplify the trigonometric expression in the denominator.

$=\,\,\,$ $\dfrac{\sin{x} \times (-1+2\cos{x})}{\cancel{1}-\cos{x}+2\cos^2{x}-\cancel{1}}$

$=\,\,\,$ $\dfrac{\sin{x} \times (-1+2\cos{x})}{-\cos{x}+2\cos^2{x}}$

You can easily understand that there is a common factor in terms of cosine in the denominator. So, it can be taken common, from the two terms of the expression.

$=\,\,\,$ $\dfrac{\sin{x} \times (-1+2\cos{x})}{-\cos{x}+2\cos{x} \times \cos{x}}$

$=\,\,\,$ $\dfrac{\sin{x} \times (-1+2\cos{x})}{(-1) \times \cos{x}+2\cos{x} \times \cos{x}}$

$=\,\,\,$ $\dfrac{\sin{x} \times (-1+2\cos{x})}{\cos{x} \times \Big((-1)+2\cos{x}\Big)}$

$=\,\,\,$ $\dfrac{\sin{x} \times (-1+2\cos{x})}{\cos{x} \times (-1+2\cos{x})}$

### Simplify the whole expression

The trigonometric expressions in both numerator and denominator are simplified to their maximum level. So, it is time to simplify the whole trigonometric function for evaluating its value.

$=\,\,\,$ $\dfrac{\sin{x}}{\cos{x}}$ $\times$ $\dfrac{-1+2\cos{x}}{-1+2\cos{x}}$

$=\,\,\,$ $\dfrac{\sin{x}}{\cos{x}}$ $\times$ $\dfrac{\cancel{-1+2\cos{x}}}{\cancel{-1+2\cos{x}}}$

$=\,\,\,$ $\dfrac{\sin{x}}{\cos{x}} \times 1$

$=\,\,\,$ $\dfrac{\sin{x}}{\cos{x}}$

According to the quotient identity of sine and cosine functions, the quotient of sine of angle $x$ by cosine of angle $x$ is equal to the tan of angle $x$.

$=\,\,\,$ $\tan{x}$

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