Find the Other Trig Values in Quadrant IV sin(theta)=-5/13

$\sin (θ) = - \frac{5}{13}$

Use the definition of sine to find the known sides of the unit circle right triangle. The quadrant determines the sign on each of the values.

$\sin (θ) = \frac{opposite}{hypotenuse}$

Find the adjacent side of the unit circle triangle. Since the hypotenuse and opposite sides are known, use the Pythagorean theorem to find the remaining side.

$Adjacent = \sqrt{{hypotenuse}^{2} - {opposite}^{2}}$

Replace the known values in the equation.

$Adjacent = \sqrt{{(13)}^{2} - {(- 5)}^{2}}$

Simplify inside the radical.

Raise $13$ to the power of $2$ .

Adjacent $= \sqrt{169 - {(- 5)}^{2}}$

Raise $- 5$ to the power of $2$ .

Adjacent $= \sqrt{169 - 1 \cdot 25}$

Multiply $- 1$ by $25$ .

Adjacent $= \sqrt{169 - 25}$

Subtract $25$ from $169$ .

Adjacent $= \sqrt{144}$

Rewrite $144$ as $12^{2}$ .

Adjacent $= \sqrt{12^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

Adjacent $= 12$

Find the value of cosine.

Use the definition of cosine to find the value of $\cos (θ)$ .

$\cos (θ) = \frac{adj}{hyp}$

Substitute in the known values.

$\cos (θ) = \frac{12}{13}$

Find the value of tangent.

Use the definition of tangent to find the value of $\tan (θ)$ .

$\tan (θ) = \frac{opp}{adj}$

Substitute in the known values.

$\tan (θ) = \frac{- 5}{12}$

Move the negative in front of the fraction.

$\tan (θ) = - \frac{5}{12}$

Find the value of cotangent.

Use the definition of cotangent to find the value of $\cot (θ)$ .

$\cot (θ) = \frac{adj}{opp}$

Substitute in the known values.

$\cot (θ) = \frac{12}{- 5}$

Move the negative in front of the fraction.

$\cot (θ) = - \frac{12}{5}$

Find the value of secant.

Use the definition of secant to find the value of $\sec (θ)$ .

$\sec (θ) = \frac{hyp}{adj}$

Substitute in the known values.

$\sec (θ) = \frac{13}{12}$

Find the value of cosecant.

Use the definition of cosecant to find the value of $\csc (θ)$ .

$\csc (θ) = \frac{hyp}{opp}$

Substitute in the known values.

$\csc (θ) = \frac{13}{- 5}$

Move the negative in front of the fraction.

$\csc (θ) = - \frac{13}{5}$

This is the solution to each trig value.

$\sin (θ) = - \frac{5}{13}$

$\cos (θ) = \frac{12}{13}$

$\tan (θ) = - \frac{5}{12}$

$\cot (θ) = - \frac{12}{5}$

$\sec (θ) = \frac{13}{12}$

$\csc (θ) = - \frac{13}{5}$

Find the Other Trig Values in Quadrant IV sin(theta)=-5/13

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