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How to prove this trigonometric identities ?

$$\cot{\frac{11\pi}{18}} +\cot{\frac{2\pi}{9}}=4\sin{\frac{\pi}{18}}\cot{\frac{2\pi}{9}}$$

Thank you.

kong
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  • Have you tried anything? I believe you should take $\phi = \pi/18$ for simplicity, write the problem in terms of $\phi$ and then just use simple formulas like $\cot \phi = \cos \phi/\sin \phi$, $\cos \alpha sin \beta = \sin \frac{\alpha + \beta}{2}\sin \frac{\alpha-\beta}{2}$ and so on. – Andrei Rykhalski Feb 13 '15 at 10:26

3 Answers3

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COMMENT.-In attention to the note by Community above I show the following: $$LHS=\frac{\sin(\frac{15\pi}{18})}{\sin(\frac{11\pi}{18})\sin(\frac{4\pi}{18})}\\$$ Hence one has $$\sin(\frac{15\pi}{18})=4\cos(\frac{4\pi}{18})\sin(\frac{11\pi}{18})\sin(\frac{\pi}{18})$$ Note that $\dfrac{15\pi}{18}=150^{\circ}=180^{\circ}-30^{\circ}$ so one has $$\sin(\frac{15\pi}{18})=\frac 12$$ Besides $$2\cos(\frac{4\pi}{18})\sin(\frac{11\pi}{18})=\sin(\frac{15\pi}{18})+\sin(\frac{7\pi}{18})$$ so we have $$\frac12=2\left(\frac12+\sin(\frac{7\pi}{18})\right)\sin(\frac{\pi}{18})$$ and because of $\frac{7\pi}{18}=90^{\circ}-20^{\circ}$ and $\frac{\pi}{18}=10^{\circ}$ we can reduce all to elementary calculations.

Piquito
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We have to check: $$ -\tan\frac{\pi}{9}+\cot\frac{2\pi}{9}=4\cos\frac{4\pi}{9}\cot\frac{2\pi}{9} $$ hence, by multiplying both sides by $\sin\frac{2\pi}{9}=2\sin\frac{\pi}{9}\cos\frac{\pi}{9}$: $$ -2\sin^2\frac{\pi}{9}+\cos\frac{2\pi}{9} = 4\cos\frac{4\pi}{9}\cos\frac{2\pi}{9} $$ so we just have to check that $x=\cos\frac{2\pi}{9}$ is a solution of: $$ 2x-1 = 4x(2x^2-1) $$ or a root of: $$ p(x) = 8x^3-6x+1 = 2\cdot T_3(x)-1, \tag{1}$$ where $T_3(x)=4x^3-3x$ is a Chebyshev polynomial. To check that $\cos\frac{2\pi}{9}$ is a root of the RHS of $(1)$ is trivial, since $T_3(\cos\alpha)=\cos(3\alpha)$: $$ 2\cdot T_3\left(\cos\frac{2\pi}{9}\right)-1 = 2\cdot\cos\frac{2\pi}{3}-1 = 0, $$ hence we're done.

Jack D'Aurizio
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$\cot\dfrac{11\pi}{18}=\cot\left(\pi-\dfrac{7\pi}{18}\right)=-\cot\dfrac{7\pi}{18}$

Now $\cot\dfrac{2\pi}9-\cot\dfrac{7\pi}{18}=\dfrac{\sin\left(\dfrac{7\pi}{18}-\dfrac{2\pi}9\right)}{\sin\dfrac{2\pi}9\cdot\sin\dfrac{7\pi}{18}}=\dfrac1{2\sin\dfrac{2\pi}9\cos\dfrac\pi9}$

( as $\sin\dfrac{7\pi}{18}=\cos\left(\dfrac\pi2-\dfrac{7\pi}{18}\right)=\cos\dfrac\pi9$)

which needs to be $=\sin\dfrac\pi{18}\cot\dfrac{2\pi}9$

which will be true if $8\sin\dfrac\pi{18}\cos\dfrac\pi9\cos\dfrac{2\pi}9=1$ $\iff8\cos\dfrac\pi9\cos\dfrac{2\pi}9\cos\dfrac{4\pi}9=1$

as $\dfrac{4\pi}9+\dfrac\pi{18}=\dfrac\pi2$

Now use: Upon multiplying $\cos(20^\circ)\cos(40^\circ)\cos(80^\circ)$ by the sine of a certain angle, it gets reduced. What is that angle?

Community
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lab bhattacharjee
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