Question

1. a. sin 1°

find sin θ and cos θ.
17.
18.
19.
20.

Explanation:

Step1: Recall the definitions of sine and cosine in a right - triangle in the coordinate plane

For a point $(x,y)$ on the terminal side of an angle $\theta$ in standard position, the distance from the origin $r=\sqrt{x^{2}+y^{2}}$, $\sin\theta=\frac{y}{r}$ and $\cos\theta=\frac{x}{r}$.

Step2: Solve for problem 17

Given the point $(x = 3,y = 4)$, first find $r$:
$r=\sqrt{3^{2}+4^{2}}=\sqrt{9 + 16}=\sqrt{25}=5$.
Then $\sin\theta=\frac{y}{r}=\frac{4}{5}$ and $\cos\theta=\frac{x}{r}=\frac{3}{5}$.

Step3: Solve for problem 18

Given the point $(x=-3,y = - 1)$, find $r$:
$r=\sqrt{(-3)^{2}+(-1)^{2}}=\sqrt{9 + 1}=\sqrt{10}$.
Then $\sin\theta=\frac{y}{r}=\frac{-1}{\sqrt{10}}=-\frac{\sqrt{10}}{10}$ and $\cos\theta=\frac{x}{r}=\frac{-3}{\sqrt{10}}=-\frac{3\sqrt{10}}{10}$.

Step4: Solve for problem 19

In the right - triangle, if the opposite side to $\theta$ is $y = 12$ and the hypotenuse $r = 13$, using the Pythagorean theorem to find the adjacent side $x$.
$x=\sqrt{r^{2}-y^{2}}=\sqrt{13^{2}-12^{2}}=\sqrt{169 - 144}=\sqrt{25}=5$.
So $\sin\theta=\frac{y}{r}=\frac{12}{13}$ and $\cos\theta=\frac{x}{r}=\frac{5}{13}$.

Answer:

1. $\sin\theta=\frac{4}{5},\cos\theta=\frac{3}{5}$
2. $\sin\theta=-\frac{\sqrt{10}}{10},\cos\theta=-\frac{3\sqrt{10}}{10}$
3. $\sin\theta=\frac{12}{13},\cos\theta=\frac{5}{13}$