More Trig Ratios

Monday 23 May 2005 at 9:42 pm | In Articles | 6 Comments

Students studying A level mathematics are expected to know exact values of a few trig ratios such as \sin\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2} (see Trig Ratios posting). But a visit to Mathworld reveals a whole world of fascinating values.

Here are just a few of them (you may wish to try proving them)

\cos\left(\dfrac{\pi}{9}\right)=2^{-\frac{4}{3}}\left(\sqrt[3]{1+i\sqrt{3}}+\sqrt[3]{1-i\sqrt{3}}\right)

\tan\left(\dfrac{3\pi}{10}\right)=\frac{1}{5}\sqrt{25+10\sqrt{5}}

\tan\left(\dfrac{\pi}{16}\right)=\sqrt{\dfrac{2-\sqrt{2+\sqrt{2}}}{2+\sqrt{2+\sqrt{2}}}}

\sin\left(\dfrac{\pi}{17}\right)=\frac{1}{8}\left[34-2\sqrt{17}-2\sqrt{2}\epsilon^*-2\sqrt{68+12\sqrt{17}+2\sqrt{2}(\sqrt{17}-1)\epsilon^*-16\sqrt{2}\epsilon}\:\right]^{\frac{1}{2}}

where \epsilon=\sqrt{17+\sqrt{17}},\ \epsilon^*=\sqrt{17-\sqrt{17}}

\sin\left(\dfrac{\pi}{18}\right)=\frac{1}{2}\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-\dots}}}} where the sequence of signs +,\;+,\;- repeats with period 3

As I said, fascinating!

Thank goodness \LaTeX can show these values easily!

6 Comments »

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  1. In re the first line: unless I’m missing some subtle joke, you want sin(45) to be sqrt(2)/2. Otherwise sin(45) is the same as sin(60), and this would make your unit circle lumpy.

    Comment by Christian Claiborn — Tuesday 24 May 2005 4:16 pm #

  2. Aaargh! How did that 4 get there? Corrected now. Thanks very much for pointing it out!

    Comment by Steve — Tuesday 24 May 2005 5:38 pm #

  3. Interesting note: you can use such results, together with the fact we know how to solve cubics/quartics explicitly, to prove that the sine of ONE degree (pi/180) is expressible as a radical!

    Comment by Mr Blobby — Thursday 26 May 2005 12:21 am #

  4. Why, I wonder, write (1/5)√(25+10√5)) rather than √(1+ 2/√5) ? … And why do my plus and minus signs vanish when I hit Preview?

    Comment by Anton Sherwood — Tuesday 14 June 2005 5:44 am #

  5. I have corrected the + sign bug, thanks for that

    Comment by steve — Tuesday 14 June 2005 5:17 pm #

  6. Well I did one exercise using complex numbers which helped me find the exact value of the cosine and sine of pi/12

    with z_a=\frac{1}{2}+i\frac{\sqrt{3}}{2}
and z_b=\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}
    now z_{a}z_{b}=
\frac{\sqrt{6}+\sqrt{2}}{4}+i\frac{\sqrt{6}-\sqrt{2}}{4}=e^{i\frac{\pi}{12}}=\cos(\frac{\pi}{12})+i\sin(\frac{\pi}{12})

    Comment by Luc — Friday 31 March 2006 5:53 pm #

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