Trigonometry Examples

$\sec (x) = \frac{13}{12}$ , $\csc (x) = \frac{13}{5}$

To find the value of $\sin (x)$ , use the fact that $\frac{1}{\csc (x)}$ then substitute in the known values.

$\sin (x) = \frac{1}{\csc (x)} = \frac{1}{\frac{13}{5}}$

Simplify the result.

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Multiply the numerator by the reciprocal of the denominator.

$\sin (x) = \frac{1}{\csc (x)} = 1 (\frac{5}{13})$

Multiply $\frac{5}{13}$ by $1$ .

$\sin (x) = \frac{1}{\csc (x)} = \frac{5}{13}$

To find the value of $\cos (x)$ , use the fact that $\frac{1}{\sec (x)}$ then substitute in the known values.

$\cos (x) = \frac{1}{\sec (x)} = \frac{1}{\frac{13}{12}}$

Simplify the result.

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Multiply the numerator by the reciprocal of the denominator.

$\cos (x) = \frac{1}{\sec (x)} = 1 (\frac{12}{13})$

Multiply $\frac{12}{13}$ by $1$ .

$\cos (x) = \frac{1}{\sec (x)} = \frac{12}{13}$

To find the value of $\tan (x)$ , use the fact that $\tan (x) = \frac{\sin (x)}{\cos (x)}$ then substitute in the known values.

$\tan (x) = \frac{\sin (x)}{\cos (x)} = \frac{\frac{5}{13}}{\frac{12}{13}}$

Simplify the result.

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Multiply the numerator by the reciprocal of the denominator.

$\tan (x) = \frac{\sin (x)}{\cos (x)} = \frac{5}{13} \cdot \frac{13}{12}$

Cancel the common factor of $13$ .

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Factor out the greatest common factor $13$ .

$\tan (x) = \frac{\sin (x)}{\cos (x)} = \frac{5}{13 \cdot 1} \cdot \frac{13 \cdot 1}{12}$

Cancel the common factor.

$\tan (x) = \frac{\sin (x)}{\cos (x)} = \frac{5}{13 \cdot 1} \cdot \frac{13 \cdot 1}{12}$

Rewrite the expression.

$\tan (x) = \frac{\sin (x)}{\cos (x)} = \frac{5}{1} \cdot \frac{1}{12}$

Multiply $\frac{5}{1}$ and $\frac{1}{12}$ .

$\tan (x) = \frac{\sin (x)}{\cos (x)} = \frac{5}{12}$

To find the value of $\cot (x)$ , use the fact that $\frac{1}{\tan (x)}$ then substitute in the known values.

$\cot (x) = \frac{1}{\tan (x)} = \frac{1}{\frac{5}{12}}$

Simplify the result.

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Multiply the numerator by the reciprocal of the denominator.

$\cot (x) = \frac{1}{\tan (x)} = 1 (\frac{12}{5})$

Multiply $\frac{12}{5}$ by $1$ .

$\cot (x) = \frac{1}{\tan (x)} = \frac{12}{5}$

The trig functions found are as follows:

$\sin (x) = \frac{5}{13}$

$\cos (x) = \frac{12}{13}$

$\tan (x) = \frac{5}{12}$

$\cot (x) = \frac{12}{5}$

$\sec (x) = \frac{13}{12}$

$\csc (x) = \frac{13}{5}$

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