Question

Find the exact values of the six trigonometric functions of the angle 0 shown in the figure

Answer 1

First solve for the hypotenuse of the given right triangle using the Pythagorean Theorem

`\begin{gathered} c^2=a^2+b^2 \\ c^2=6^2+8^2 \\ c^2=36+64 \\ c^2=100 \\ \sqrt[]{c^2}=\sqrt[]{100} \\ c=10 \end{gathered}`

The hypotenuse is 10, we can now solve for the six trigonometric ratios.

Solving for the first three trigonometric ratios.

Recall the following ratios

`\sin \theta=\frac{\text{opposite}}{\text{hypotenuse}},\cos \theta=\frac{\text{adjacent}}{\text{hypotenuse}},\tan \theta=\frac{\text{opposite}}{\text{adjacent}}`

Relative to angle Θ, we have the following sides

opposite = 8, adjacent = 6, hypotenuse = 10

Therefore, the ratio are the following

`\begin{gathered} \sin \theta=(8)/(10),\cos \theta=(6)/(10),\tan \theta=(8)/(6) \\ \sin \theta=(4)/(5),\cos \theta=(3)/(5),\tan \theta=(4)/(3)\text{ (simplified)} \end{gathered}`

Solving for the reciprocal of the trigonometric ratios

Recall their reciprocal counterparts

`\csc \theta=\frac{\text{hypotenuse}}{\text{opposite}},\sec \theta=\frac{\text{hypotenuse}}{\text{adjacent}},\cot \theta=\frac{\text{adjacent}}{\text{opposite}}`

Substitute, and we have the following

`\begin{gathered} \csc \theta=(10)/(8),\sec \theta=(10)/(6),\cot \theta=(6)/(8) \\ \csc \theta=(5)/(4),\sec \theta=(5)/(3),\cot \theta=(3)/(4)\text{ (simplified)} \end{gathered}`