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My cousin is working on this and showed it to me. I'm unsure how to solve it.

$x$ and $y$ represent two angles in standard position. $x$ has its terminal arm in the first quadrant and $y$ has its terminal arm in the third quadrant. If $\cos(x) = \frac{5}{13}$ and $\cos(y) = -\frac{5}{13}$ , find the value of:

$$2\sin(x) + 2\sin(y) + 2\cos(x) − 2\tan(x) + \tan(y) + 2\cos(y)$$

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    $\begingroup$ Use the identity $\sin^2 \alpha+\cos^2\alpha=1$ to get the value of $\sin x$ and $\cos x.$ $\endgroup$ – mfl Jul 9 '14 at 15:23
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When $cos (x) = \frac {5}{13}$ and x’s terminal arm is in the 1st quadrant, the x’s x-coordinate = 5 and x’s y-coordinate = …by Pythagoras theorem… = 12. This is because, in the 1st quadrant all points are in the form of (+, +).

When $cos (y) = – \frac {5}{13}$ and y’s terminal arm is in the 3rd quadrant, the y’s x-coordinate = – 5 and y’s y-coordinate = – 12. This is because, in the 3rd quadrant all points are in the form of (-, -).

$tan (z) = \frac {z’ y-coordinate}{z’ x-coordinate}$, where z can be x and can also be y.

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because $13^2-5^2=12^2$, so we have $$\sin x=a \frac{12}{13},\sin y=b \frac{12}{13}, a=\pm 1, b=\pm 1$$

$$I=2\sin(x) + 2\sin(y) + 2\cos(x) − 2\tan(x) + \tan(y) + 2\cos(y)$$

$$13I=24a + 24b + 2*5 − 2\frac{12a}{5} + \frac{12b}{-5} + 2*(-5)=\frac{96a}{5} + \frac{108b}{5}$$

$$I=\frac{12}{65}(8a+9b)$$

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