Hello, gabby!
Here's some help . . .
is negative in Q2 and Q4.
is positive in Q1 and Q2.
. . Hence, 
is in Q2.
We have: . 
Since is in Q2: . 
. . Then: . ^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]<br />
\cos\theta &=& \dfrac{x}{r} &=& \dfrac{-7}{5\sqrt{2}} &=& - \dfrac{7\sqrt{2}}{10} \end{Bmatrix}](http://latex.codecogs.com/png.latex?\begin{Bmatrix} \sin\theta &=&\dfrac{y}{r} &=&\dfrac{1}{5\sqrt{2}} &=& \dfrac{\sqrt{2}}{10} \\ \\[-4mm]<br />
\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: . 
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}](http://latex.codecogs.com/png.latex?\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})
is positive in Q1 and Q2.
is negative in Q2 and Q3.
Therefore, terminates in Q2.
Identity: . 
We have: . ^2} \;=\;\pm\sqrt{\frac{24}{25}} \;=\;\pm\frac{2\sqrt{6}}{5})
Since is in Q2, is negative.
Therefore: . 
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= -1/7 and sin
