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So I have this question: $\cos X = -5/13$ where $0<X<2\pi$ (less then or equal signs not just less then). Determine all possible trig ratios. I want to know how you would find the trig ratios. The answers at the back of the textbook are $$\eqalign{\sin X &= 12/13 {\rm\ or\ -12/13}\cr \tan X& = 12/5 {\rm\ or\ } -12/5\cr \sec X &= -13/5.}$$ But I don't get how they got those answers. |
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Draw a right triangle in the second quadrant that looks like this: $\ \ \ \ \lower6pt{ \llap{opp\ }|\nwarrow \rlap{\ hyp}\over adj}$ Since $\cos X=-5/13$, the hypotenuse, $hyp$, of that triangle has length 13 and the horizontal leg, $adj$, has length 5. The Pythagorean Theorem will tell you that the length of the vertical side, $opp$, of the triangle is $\sqrt{169-25}=\sqrt{144}=12$. So our triangle is $\ \ \ \ \lower6pt{ \llap{12\ }|\nwarrow \rlap{\ 13}\over 5}$ Now read the trig ratios from the triangle, attaching the appropriate sign. For example, $\tan X= {\rm opp.\over adj}=-12/5$. The negative sign is needed since $\tan$ is negative in the second quadrant. The other trig ratios I'll leave to you. You also need to do the above for a triangle in the third quadrant (since $\cos$ can be negative there). |
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You have $\sin^2 X + \cos^2 X=1$, so $\sin^2 X = 1 - \frac{25}{169} = \frac{144}{169}, \sin X = \pm \frac{12}{13}$. Then $\tan X = \frac{\sin X}{\cos X}$ and you have to check which signs are applicable. |
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