Solution

Before we can solve for the double angles, we first need to figure out the Sine and Cosine of the original angles. We can do that by applying our triangle drawing techniques from chapter 6. Namely, with a positive Secant and negative Sine, we are looking at a triangle in Quadrant IV:

We can then solve for the \(y\) value: \[\solve{ 12^2&=&1^1+y^2\\ 144-1&=&y^2\\ 143&=&y^2\\ -\sqrt{{143}}&=&y } \]Here I deliberately take the negative root since this is in the fourth quadrant.

Now that we have the full picture, we can determine the Sine and Cosine of \(x\): \[ \solve{ \sin(x) &=& -\dfrac{\sqrt{ 143 } }{{12}}\\ \cos(x) &=& \dfrac{{1}}{{12}} } \]With these in hand, we can find the double Sine and Cosine: \[ \solve{ \sin(2x) &=& 2\sin(x)\cos(x)\\ \sin(2x) &=& 2\left(-\dfrac{\sqrt{ 143 } }{ 12 }\right)\left(\dfrac{ 1 }{ 12 }\right)\\ \sin(2x) &=& -\dfrac{ \sqrt{ 143 } }{ 72 }\\ \\\hline\\ \cos(2x) &=& 2\cos^2(x)-1\\ \cos(2x) &=& 2\left(\dfrac{{1}}{ 12 }\right)^2-1\\ \cos(2x) &=& \dfrac{ 1 }{ 72 }-1\\ \cos(2x) &=& -\dfrac{ 71 }{ 72 }\\ \\ \hline\\ \tan(2x) &=& \dfrac{\sin(2x)}{ \cos(2x) }\\ \tan(2x) &=& \dfrac{-\frac{ \sqrt{{143}} }{ 72 } }{ -\frac{{71}}{{72}} }\\ \tan(2x) &=& -\dfrac{ \sqrt{{143}} }{ 72 }\times \left(-\dfrac{{72}}{{71}}\right)\\ \tan(2x) &=& \dfrac{ \sqrt{{143}} }{ 71 } } \]