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How to find $\tan(-\frac{5\pi}{16})$ with half-angle formulas?

I tried the $\pm \sqrt{\frac{1-\cos{A}}{1+\cos{A}}}$ and $\frac{\sin{A}}{1+\cos{A}}$ but got stuck once there were square roots on top and bottom like $\frac{\sqrt{...}}{1-\sqrt{...}}.$

Using the cosine over cosine in square root I got up to

$$=-\sqrt{ \frac{ 1+\sqrt{\frac{1+\cos(5\pi/3)}{2}}}{1-\sqrt{\frac{1+\cos(5\pi/3)}{2}}} }$$

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    This angle is half of $-5\pi/8$; unless you happen to know its sine/cosine, you'll have to half-angle this one as well. Either way, could you try to type up what you have, or at least what you got, and why you're not happy with it? Some things with radicals can't really be simplified; maybe you're just at that spot. – 2017-01-29
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    @pjs36 I added where I got to. :o – 2017-01-29

5 Answers 5

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I assume you know (or can figure out) $$\cos(-5\pi/4) = -\frac{\sqrt{2}}2,\ \sin(-5\pi/4) = +\frac{\sqrt{2}}2 $$ Applying the half-angle formulas to that, and noting that $-\pi < -5\pi/8 < -\pi/2$ $$ \eqalign{\cos(-5\pi/8) &= - \sqrt{\frac{1+\cos(-5\pi/4)}{2}} = -\frac{\sqrt{2-\sqrt{2}}}{2}\cr \sin(-5\pi/8) &= - \sqrt{\frac{1-\cos(-5\pi/4)}{2}} = - \frac{\sqrt{2+\sqrt{2}}}{2}}$$ Similarly, since $-\pi/2 < -5\pi/16 < 0$, $$ \eqalign{\cos(-5\pi/16) &= +\sqrt{\frac{1+\cos(-5\pi/8)}{2}} = \frac{\sqrt{2-\sqrt{2-\sqrt{2}}}}{2}\cr \sin(-5\pi/16) &= - \sqrt{\frac{1-\cos(-5\pi/8)}{2}} = - \frac{\sqrt{2+\sqrt{2-\sqrt{2}}}}{2}}$$ so that $$ \tan(-5\pi/16) = \frac{\sin(-5\pi/16)}{\cos(-5\pi/16)} = - \frac{\sqrt{2+\sqrt{2-\sqrt{2}}}}{\sqrt{2-\sqrt{2-\sqrt{2}}}}$$

It turns out (but this is somewhat harder) that you can do some simplification here: you can write it as

$$ \tan(-5\pi/16) = 1 - \sqrt{2} - \sqrt{4-2\sqrt{2}} $$

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    Also possible if one just guessed it: Use $\tan{A}+\tan{B} \equiv (1-\tan{A}\tan{B})\tan{(A+B)}$ with $A =\frac{\pi}{12}, B = -\frac{5\pi}{16}$ and the nested radical is the only thing that remains upon simplification. – 2017-01-30
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Fill in details:

$$r:=\tan\left(-\frac{5\pi}{16}\right)=\tan\frac{11\pi}{16}\implies$$

$$s:=\tan\frac{11\pi}8=\tan\left(2\cdot\frac{11\pi}{16}\right)=\frac{2\tan\frac{11\pi}{16}}{1-\tan^2\frac{11\pi}{16}}=\frac{2r}{1-r^2}$$

and

$$-1=\tan\frac{11\pi}4=\tan\left(2\cdot\frac{11\pi}8\right)=\frac{2s}{1-s^2}\implies$$

$$s^2-2s-1=0\implies s_{1,2}=1\pm\sqrt2$$

and now

$$sr^2+2r-s=0\implies r_{1,2}=\frac{-2\pm\sqrt{4+4s^2}}{2s}=\frac{-1\pm\sqrt{1+s^2}}{s}$$

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    Of course $s^2-1=(1+s^2)-2=2s$ so $\sqrt{1+s^2}=\sqrt{2(s+1)}$. – 2017-01-29
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    @ThomasAndrews It's late here and I'm tired: are you trying to make a point? – 2017-01-29
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    Just an additional simplification to that RHS. Not meant to be a bug or anything. – 2017-01-29
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    @ThomasAndrews Oh, I see. Thanks for that. +1\ – 2017-01-29
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$$\tan{\frac{A}{2}} \equiv \frac{\sin{A}}{1+\cos{A}}$$

$$\tan{\left(-\frac{5\pi}{16}\right)} \equiv \frac{-\sin{\frac{5\pi}{8}}}{1+\cos{\frac{5\pi}{8}}}$$

Solving for the larger angle ratios should be relatively straightforward, however, it will not be as direct due to square roots.

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Apart from a couple of threes that might be fours ? You are fine upto here \begin{eqnarray*} \tan(-\frac{5\pi}{16})=-\sqrt{ \frac{ 1+\sqrt{\frac{1+\cos(5\pi/4)}{2}}}{1-\sqrt{\frac{1+\cos(5\pi/4)}{2}}}}. \end{eqnarray*} Now $\cos(5\pi/4)=\frac{ 1}{\sqrt{2}}$ & after a little bit algebra ... \begin{eqnarray*} \tan(-\frac{5\pi}{16})=-\sqrt{ \frac{ 8^{1/4}+\sqrt{\sqrt{2}-1}}{ 8^{1/4}-\sqrt{\sqrt{2}-1}}}. \end{eqnarray*} My Casio fx-83MS gives both sides equal to $-1.4966 \cdots $.

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The problem is straightforward, it's just a form which is more trouble to rationalize that it's worth. And there are several equivalent expressions depending upon which identities are used.

$$ \tan\left(-\frac{5\pi}{16}\right)=-\frac{\sin\left(\frac{5\pi}{8}\right)}{1+\cos\left(\frac{5\pi}{8}\right)}$$

and

$$ \sin\left(\frac{5\pi}{8}\right)=\sqrt{\frac{1-\cos(5\pi/4)}{2}}=\frac{1}{2}\sqrt{2+\sqrt{2}}$$

whereas

$$ \cos\left(\frac{5\pi}{8}\right)=-\sqrt{\frac{1+\cos(5\pi/4)}{2}}=-\frac{1}{2}\sqrt{2-\sqrt{2}}$$

Therefore

$$\tan\left(-\frac{5\pi}{16}\right)=\frac{\sqrt{2+\sqrt{2}}}{\sqrt{2-\sqrt{2}}-2} $$

It takes many steps to rationalize this denominator so one may as well leave it in this form.