# How do you find the remaining trigonometric functions of theta given costheta=24/25 and theta terminates in QIV?

Mar 30, 2017

$\sin \theta = \frac{\text{-} 7}{25}$
$\cos \theta = \frac{24}{25}$
$\tan \theta = \frac{\text{-} 7}{24}$
$\cot \theta = \frac{\text{-} 24}{7}$
$\sec \theta = \frac{1}{24}$
$\csc \theta = \frac{\text{-} 1}{7}$

#### Explanation:

Let the point $P \left(x , y\right)$ be on a circle of radius $r$ centered at the origin and $P$ is on the terminal ray of $\theta$. Let the angle $\alpha$ be the the nonreflex angle from the positive x-axis that is coterminal with $\theta$.

$\theta$ terminates in Q$\text{IV}$, so for point $P$, $x$ is positive and $y$ is negative. $\alpha$ is negative..

The line segments from $\left(0 , 0\right)$ to $\left(x , 0\right)$ and to $\left(0 , y\right)$ form a right triangle with hypotenuse $r$ and angle $\alpha$ between $x$ and $r$. $\cos \alpha = \frac{x}{r}$ and $\sin \alpha = \frac{y}{r}$.

Since $\alpha$ and $\theta$ start and end in the same place, all of their trig functions are the same, so by finding the function values for $\alpha$ we find them for $\theta$. You can say that since $\theta = \alpha + 2 \pi$, they are effectively the same angle; I will be using $\theta$ because that is what we're looking for.

$\cos \theta = \frac{x}{r}$. We are given that $\cos \theta = \frac{24}{25}$. Therefore, $x = 24$ and $r = 25$.

$x$, $y$, and $r$ are the lengths of the two legs and the hypotenuse of a right triangle, respectively. Therefore, ${x}^{2} + {y}^{2} = {r}^{2}$. Solving for $y$, $y = \pm \sqrt{{r}^{2} - {x}^{2}}$. We know the values of $x$ and $r$, so we can find that $y = \pm \sqrt{{25}^{2} - {24}^{2}} = \pm 7$. We know that $y$ is negative so $y = \text{-} 7$.

$\sin \theta = \frac{y}{r}$, so $\sin \theta = \frac{\text{-} 7}{25}$

$\tan \theta = \frac{y}{x}$, so $\tan \theta = \frac{\text{-} 7}{24}$

$\cot \theta = \frac{x}{y}$, so $\cot \theta = \frac{24}{\text{-"7)=("-} 24} / 7$

$\sec \theta = \frac{1}{x}$, so $\sec \theta = \frac{1}{24}$

$\csc \theta = \frac{1}{y}$, so $\csc \theta = \frac{1}{\text{-"7)=("-} 1} / 7$