Math Help - need trig help

1. need trig help

10) Given that tan = -1/7 and sin > 0, use the x-y-r method to find sin and cos.

11) Sketch a 240° angle in standard position. Label a point on its terminal side, draw the right triangle in use, indicate the direction of opening, and give all six trig function values for 240°.

12) Sketch a 90° angle in standard position. Label a point on its terminal side and use the x-y-r method to compute the six trig function values for 90°.

13) In what quadrant(s) does terminate if sin > 0 and cos < 0?
Q1 Q2 Q3 Q4

14) In what quadrant(s) does terminate if tan < 0?
Q1 Q2 Q3 Q4

15) In what quadrant(s) does terminate if cos < 0?
Q1 Q2 Q3 Q4

17) Use a fundamental identity to find cos if sin = 1/5, that terminates in Q2. Show work.

2. Hello, gabby!

Here's some help . . .

10) Given that $\tan\theta = -\frac{1}{7}$ and $\sin\theta > 0$,
use the x-y-r method to find: . $\sin\theta$ and $\cos\theta$

$\tan\theta$ is negative in Q2 and Q4.
$\sin\theta$ is positive in Q1 and Q2.
. . Hence, $\theta$ is in Q2.

We have: . $\tan\theta \:=\:-\frac{1}{7} \:=\:\frac{y}{x}$

Since $\theta$ is in Q2: . $x = -7,\;y = 1$
. . Then: . $r \:=\:\sqrt{x^2+y^2} \:=\:\sqrt{(-7)^2 + 1^2} \:=\:\sqrt{50} \:=\:5\sqrt{2}$

Therefore: . $\begin{Bmatrix} \sin\theta &=&\dfrac{y}{r} &=&\dfrac{1}{5\sqrt{2}} &=& \dfrac{\sqrt{2}}{10} \\ \\[-4mm]
\cos\theta &=& \dfrac{x}{r} &=& \dfrac{-7}{5\sqrt{2}} &=& - \dfrac{7\sqrt{2}}{10} \end{Bmatrix}$

11) Sketch a 240° angle in standard position.
Label a point on its terminal side, draw the right triangle in use,
indicate the direction of opening, and give all six trig function values for 240°.
Code:
                |
|
-1  |
- - + - - * - - - -
_ |    /|
-√3 |   / |
|  /2 |
| /   |
|/    |
*     |

We have: . $x = -1,\;y = -\sqrt{3},\;r = 2$

Then: . $\begin{Bmatrix} \sin\theta &=& \frac{y}{r} &=&-\frac{\sqrt{3}}{2} \\ \\[-4mm] \cos\theta &=& \frac{x}{r} &=& -\frac{1}{2} \\ \\[-4mm] \tan\theta &=& \frac{y}{x} &=& \sqrt{3} \\ \vdots & & \vdots & & \vdots \end{Bmatrix}$

13) In what quadrant(s) does $\theta$ terminate if $\sin\theta > 0$ and $\cos\theta < 0$?

$\sin\theta$ is positive in Q1 and Q2.
$\cos\theta$ is negative in Q2 and Q3.

Therefore, $\theta$ terminates in Q2.

17) Use a fundamental identity to find $\cos\theta$
if $\sin\theta = \frac{1}{5}$, and $\theta$ terminates in Q2.

Identity: . $\sin^2\!\theta + \cos^2\!\theta \:=\:1 \quad\Rightarrow\quad \cos\theta \:=\:\pm\sqrt{1-\sin^2\!\theta}$

We have: . $\cos\theta \;=\;\pm\sqrt{1 - \left(\frac{1}{5}\right)^2} \;=\;\pm\sqrt{\frac{24}{25}} \;=\;\pm\frac{2\sqrt{6}}{5}$

Since $\theta$ is in Q2, $\cos\theta$ is negative.

Therefore: . $\cos\theta \;=\;-\frac{2\sqrt{6}}{5}$