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Can someone explain how this problem is done so I can understand it better? Thanks.

Let's say that, $\tan \theta = -12/5$ and $90 ^\circ < \theta < 180^\circ$:

Find the exact values of $\cos \theta/2$ and $\tan \theta/2$.

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Hint

Use the half-angle formula in a "reverse" way. If $t=\tan(\theta)$,You have $$\sin(2\theta)=\frac{2t}{1+t^2}$$ $$\cos(2\theta)=\frac{1-t^2}{1+t^2}$$ $$\tan(2\theta)=\frac{2t}{1-t^2}$$ So, from the last, you find that $$-\frac{12}{5}=\frac{2t}{1-t^2}$$ Solve the quadratic equation in $t$ and obtain the roots $t=-\frac{2}{3}$ and $t=\frac{3}{2}$.

I am sure that you can take from here.

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This solution isn't as elegant as Claude's, but here goes. Because 4 out of every 3 persons hate fractions, let's substitute $\theta = 2\phi$.

Then a restatement of the problem would be
$\tan 2\phi = -12/5$ and $45^\circ \lt \phi \lt 90^\circ$: Find the exact values of $\cos \phi$ and $\tan \phi$.

The double angle identity for $\tan$ would be $$\tan 2\phi = \frac{\sin 2\phi}{\cos 2\phi}=\frac{2 \sin \phi \cos \phi}{\cos^2\phi - \sin^2\phi}=\frac{\frac{2\sin\phi\cos\phi}{\cos^2\phi}}{\frac{\cos^2\phi - \sin^2\phi}{\cos^2\phi}}=\frac{2\tan\phi}{1-\tan^2\phi}$$

This gives us $$-12/5 = \frac{2\tan\phi}{1-\tan^2\phi}$$ $$6-6\tan^2\phi= -5\tan\phi$$ $$6\tan^2\phi-5\tan\phi-6= 0$$ $$\tan\phi = \frac{5\pm \sqrt{25-4(-36)}}{2\cdot 6}=\frac{5\pm13}{12}$$ but only $\tan\phi= \frac{3}{2}$ is consistent with $45^\circ \lt \phi \lt 90^\circ$

But because 4 out of 3 persons hate fractions I'll leave $\cos\phi$ for you to do.

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  • $\begingroup$ $\cos \phi = 2/\sqrt{13}$ $\endgroup$ – John Joy May 14 '14 at 16:01

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