Question

If $\sin \theta =\frac{5}{13}$, theta is in quadrant II, how do you find the exact value of each of the remaining trigonometric functions of the theta?

Answer 1

Find the exact value of each of the remaining trigonometric functions of theta in quadrant II:

Given that, $\sin \theta =\frac{5}{13}$

Use the definition of the sine formula to find the known sides of the given expression:

$\begin{array}{rcl}s\mathrm{in}\left(\mathrm{\theta }\right)& =& \frac{\mathrm{opposite}}{\mathrm{hypotenuse}}\\ & =& \frac{5}{13}\end{array}$

Since the hypotenuse and opposite sides are known, use the Pythagorean theorem to find the remaining side.

Step-1: Find the adjacent side.

Use the Pythagorean theorem formula:

$\begin{array}{rcl}{a}^2+b^2& =& c^2\\ a^2& =& c^2-b^2\\ a& =& \sqrt{c^2-b^2}\end{array}$

Here, the theta value of sin in quadrant II, so the value will be negative,

$\begin{array}{rcl}& \Rightarrow & \mathrm{Adjacent}=-\sqrt{\mathrm{hypotenus}{\mathrm{e}}^2-{\mathrm{opposite}}^2}\\ & =& -\sqrt{{13}^2-5^2}\\ & =& -\sqrt{169-24}\\ & =& -\sqrt{144}\\ & =& -12\end{array}$

Step-2: Find the value of cosine.

Use the cosine formula:

$\begin{array}{rcl}& \Rightarrow & \cos \left(\mathrm{\theta }\right)=\frac{{\displaystyle \mathrm{adjacent}}}{\mathrm{hypotenuse}}\\ & =& \frac{-12}{13}\end{array}$

Step-3: Find the value of the tangent.
Use the tangent formula:

$\begin{array}{rcl}& \Rightarrow & \tan \left(\mathrm{\theta }\right)=\frac{{\displaystyle \mathrm{opposite}}}{\mathrm{adjacent}}\\ & =& \frac{5}{-12}\end{array}$

Step-4: Find the value of the cotangent.

Use the cotangent formula:

$\begin{array}{rcl}& \Rightarrow & \cot \left(\mathrm{\theta }\right)=\frac{{\displaystyle \mathrm{adjacent}}}{\mathrm{opposite}}\\ & =& \frac{-12}{5}\end{array}$

Step-5: Find the value of the secant.

Use the secant formula:

$\begin{array}{rcl}& \Rightarrow & \sec \left(\mathrm{\theta }\right)=\frac{{\displaystyle \mathrm{hypotenuse}}}{\mathrm{adjacent}}\\ & =& \frac{13}{-12}\end{array}$

Step-6: Find the value of the cosecant.

Use the cosecant formula:

$\begin{array}{rcl}& \Rightarrow & \mathrm{cosec}\left(\mathrm{\theta }\right)=\frac{{\displaystyle \mathrm{hypotenuse}}}{\mathrm{opposite}}\\ & =& \frac{13}{5}\end{array}$

Hence, the other trigonometric functions of theta in quadrant II are,

$\begin{array}{rcl}\mathbf{cos}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}& \mathbf{=}& \frac{\mathbf{-}\mathbf{12}}{\mathbf{13}}\\ \mathbf{tan}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}& \mathbf{=}& \frac{\mathbf{}\mathbf{}\mathbf{}\mathbf{5}}{\mathbf{-}\mathbf{12}}\\ \mathbf{cot}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}& \mathbf{=}& \frac{\mathbf{-}\mathbf{12}}{\mathbf{5}}\\ \mathbf{sec}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}& \mathbf{=}& \frac{\mathbf{}\mathbf{}\mathbf{13}}{\mathbf{-}\mathbf{12}}\\ \mathbf{cosec}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}& \mathbf{=}& \frac{\mathbf{13}}{\mathbf{5}}\end{array}$