Trigonometry Examples

$\frac{4 \sin^{2} (x) - 1}{2 \sin (x) + 1} = 2 \sin (x) - 1$

Step 1

Start on the left side.

$\frac{4 \sin^{2} (x) - 1}{2 \sin (x) + 1}$

Step 2

Simplify.

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Step 2.1

Simplify the numerator.

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Step 2.1.1

Rewrite $4 {\sin (x)}^{2}$ as ${(2 \sin (x))}^{2}$ .

$\frac{{(2 \sin (x))}^{2} - 1}{2 \sin (x) + 1}$

Step 2.1.2

Rewrite $1$ as $1^{2}$ .

$\frac{{(2 \sin (x))}^{2} - 1^{2}}{2 \sin (x) + 1}$

Step 2.1.3

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2 \sin (x)$ and $b = 1$ .

$\frac{(2 \sin (x) + 1) (2 \sin (x) - 1)}{2 \sin (x) + 1}$

Step 2.2

Cancel the common factor of $2 \sin (x) + 1$ .

$2 \sin (x) - 1$

Step 3

Because the two sides have been shown to be equivalent, the equation is an identity.

$\frac{4 \sin^{2} (x) - 1}{2 \sin (x) + 1} = 2 \sin (x) - 1$ is an identity