Trigonometry Examples

$\cot (x) \cos (x) + \sin (x) = \csc (x)$

Start on the left side.

$\cot (x) \cos (x) + \sin (x)$

Simplify each term.

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Rewrite $\cot (x)$ in terms of sines and cosines.

$\frac{\cos (x)}{\sin (x)} \cos (x) + \sin (x)$

Multiply $\frac{\cos (x)}{\sin (x)} \cos (x)$ .

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Combine $\frac{\cos (x)}{\sin (x)}$ and $\cos (x)$ .

$\frac{\cos (x) \cos (x)}{\sin (x)} + \sin (x)$

Raise $\cos (x)$ to the power of $1$ .

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

Raise $\cos (x)$ to the power of $1$ .

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

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Add $1$ and $1$ .

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

Apply Pythagorean identity in reverse.

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

Simplify.

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Simplify the numerator.

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

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

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

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

To write $\sin (x)$ as a fraction with a common denominator, multiply by $\frac{\sin (x)}{\sin (x)}$ .

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

$\frac{1}{\sin (x)}$

Rewrite $\frac{1}{\sin (x)}$ as $\csc (x)$ .

$\csc (x)$

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

$\cot (x) \cos (x) + \sin (x) = \csc (x)$ is an identity