Q. 46 - Precalculus Enhanced with Graphing Utilities

Question

Find the exact value of $\sin t,\cos t$, and $\tan t$ if $P=(-2,5)$ is the point on the circle that corresponds to $t$.

Step-by-Step Solution

Answer

The value of $\sin t,\cos t,\tan t$ is $\frac{5\sqrt{29}}{29},-\frac{2\sqrt{29}}{29},-\frac{5}{2}$ respectively

Step 1: Given information

Given point $(-2,5)$

Step 2: Calculate the radius for circle at origin

Calculating, we get

$$\begin{gathered} r=\sqrt{x^2+y^2} \\ r=\sqrt{-2^2+5^2} \\ r=\sqrt{4+25} \\ r=\sqrt{29} \end{gathered}$$

Step 3: Calculating the sine value

Here

$$\begin{gathered} \sin t=\frac{y}{r} \\ \sin t=\frac{5}{\sqrt{29}}·\frac{\sqrt{29}}{\sqrt{29}} \\ \sin t=\frac{5\sqrt{29}}{29} \end{gathered}$$

$$\begin{gathered} \cos t=\frac{x}{r} \\ \cos t=\frac{-2}{\sqrt{29}}·\frac{\sqrt{29}}{\sqrt{29}} \\ \cos t=-\frac{2\sqrt{29}}{29} \end{gathered}$$

$$\begin{gathered} \tan t=\frac{\sin t}{\cos t} \\ \tan t=\frac{\frac{5\sqrt{29}}{29}}{-\frac{2\sqrt{29}}{29}} \\ \tan t=-\frac{5\sqrt{29}}{29}·\frac{29}{2\sqrt{29}} \\ \tan t=-\frac{5}{2} \end{gathered}$$