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Welcome to NRICH.

Prove tan(4pi/11)+4sin(pi/11)=sqrt(11)

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By Tkarthik on November 16, 1998:

prove

tan(4p/11) + 4sin(p/11)=sqrt(11)

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By Alex Barnard on February 12, 1999:

I don't know how much you know so if anything below doesn't make sense then please write back.

A very effective way to prove most trig identities like the one above is to use complex numbers. To do this we use the formula

e^ix = cos(x) + isin(x)

So cos(x) = (1/2)(e^ix + e^-ix)
and sin(x) = (1/2i)(e^ix - e^-ix)

So if I set x = ip/11, then the question above is to show that:

(x⁴ - x^-4)
------- + 2 (x¹ - x^-1) = i sqrt(11)
(x⁴ + x^-4)

Now if we multiply both sides by (x⁴ + x^-4) and then square each side and simplify then we get (after a little work which I'm certainly not typing out!):

4(x¹⁰ + x⁹ + x⁸ - x⁷ + x⁶ + 2x² + 1 + x^-10 + x^-9 + x^-8 - x^-7 + x^-6 + 2x^-2) = 0

Now x¹¹ = -1 (because x¹¹ = e^ip = -1)

So -1 = x¹¹ = x⁹⁺² = x⁹ × x² => x⁹ + x² = 0

And similarly -x⁷ = x^-4

So the long equation above becomes:

(x¹⁰ + x⁸ + x⁶ + x⁴ + x² + x^-10 + x^-8 + x^-6 + x^-4 + x^-2 + 1) = 0

So if we can prove this is true then the result you want will be true.

Now, x¹¹ = -1 means that x²² = 1

So 0 = x²² - 1 = (x-1)(1 + x + x² + x³ + ... + x²¹)
= (x-1)(1+x)(1 + x² + x⁴ + x⁶ + ... + x²⁰)

Now, clearly x isn't 1 or -1, so the above means that:

1 + x² + x⁴ + ... + x²⁰ = 0

Dividing by x¹⁰ (allowed as x isn't 0) gives:

x¹⁰ + x⁸ + x⁶ + x⁴ + x² + x^-10 + x^-8 + x^-6 + x^-4 + x^-2 + 1 = 0

Which is exactly what we wanted.

So tan(4p/11) + 4sin(p/11) = sqrt(11). QED

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