2

How to compute

$$S=\csc^2\frac{\pi}{14}+\csc^2\frac{3\pi}{14}+\csc^2\frac{5\pi}{14}$$

I tried to rewrite it in terms of $\sin$

$$ \csc^2\frac{\pi}{14}+\csc^2\frac{3\pi}{14}+\csc^2\frac{5\pi}{14}= \frac{1}{\sin^2\frac{\pi}{14}}+\frac{1}{\sin^2\frac{3\pi}{14}}+\frac{1}{\sin^2\frac{5\pi}{14}}=\\ \\ \frac{\sin^2\frac{3\pi}{14}\sin^2\frac{5\pi}{14}+\sin^2\frac{\pi}{14}\sin^2\frac{5\pi}{14}+\sin^2\frac{\pi}{14}\sin^2\frac{3\pi}{14}}{\sin^2\frac{\pi}{14}\sin^2\frac{3\pi}{14}\sin^2\frac{5\pi}{14}}$$

then i used

$$2\cos x\cos y=\cos(x-y)+\cos(x+y)\\ 2\sin x\sin y=\cos(x-y)-\cos(x+y)$$

then i found \begin{align} S &= 2\frac{\left(\cos\frac{\pi}{7}-\cos\frac{4\pi}{7}\right)^2+\left(\cos\frac{2\pi}{7}-\cos\frac{3\pi}{7}\right)^2+\left(\cos\frac{\pi}{7}-\cos\frac{2\pi}{7}\right)^2}{\left(\cos\frac{\pi}{7}-\cos\frac{4\pi}{7}\right)\left(\cos\frac{2\pi}{7}-\cos\frac{3\pi}{7}\right)\left(\cos\frac{\pi}{7}-\cos\frac{2\pi}{7}\right)} \end{align}

i made a simplification and used again the transformation of product on sum and arrived at \begin{align} S &= 4\frac{6-5\cos\frac{\pi}{7}+2\cos\frac{2\pi}{7}-4\cos\frac{3\pi}{7}+2\cos\frac{4\pi}{7}-4\cos\frac{5\pi}{7}+\cos\frac{6\pi}{7}}{5-8\cos\frac{\pi}{7}+6\cos\frac{2\pi}{7}-5\cos\frac{3\pi}{7}+4\cos\frac{4\pi}{7}-3\cos\frac{5\pi}{7}+\cos\frac{6\pi}{7}} \\ & \cos\frac{6\pi}{7}=-\cos\frac{\pi}{7} \hspace{5mm} \cos\frac{5\pi}{7}=-\cos\frac{2\pi}{7} \hspace{5mm} \cos\frac{4\pi}{7}=-\cos\frac{3\pi}{7} \\ S &= 4\frac{6+6\left(-\cos\frac{\pi}{7}+\cos\frac{2\pi}{7}-\cos\frac{3\pi}{7}\right)}{5+9\left(-\cos\frac{\pi}{7}+\cos\frac{2\pi}{7}-\cos\frac{3\pi}{7}\right)} \end{align}

$$-\cos\frac{\pi}{7}+\cos\frac{2\pi}{7}-\cos\frac{3\pi}{7}=-\frac{1}{2}$$

$$S=4\frac{6-3}{5-\frac{9}{2}}=4\frac{3}{\frac{1}{2}}=4\cdot3\cdot2=24$$

Leucippus
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cand
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1 Answers1

5

$$S=\csc^2\frac{\pi}{14}+\csc^2\frac{3\pi}{14}+\csc^2\frac{5\pi}{14}$$

Let $s:=\sin,c:=\cos$ and the subscript as in $s_k$ denote $\sin\left(\frac{k\pi}{14}\right)$ and similiarly for other trigonometric ratios. $$\csc^2\frac{\pi}{14}+\csc^2\frac{3\pi}{14}+\csc^2\frac{5\pi}{14}=\frac{(s_3s_5)^2+(s_5s_1)^2+(s_1s_3)^2}{(s_1s_3s_5)^2}\\ =\frac{(c_2-c_8)^2+(c_4-c_6)^2+(c_2-c_4)^2}{4s_1^2s_3^2s_5^2} $$ Now, numerator is: $$\begin{align} (c_2-c_8)^2+(c_4-c_6)^2+(c_2-c_4)^2 &=2c_2^2+2c_4^2+c_6^2+c_8^2-2(c_2c_4+c_4c_6+c_2c_8)\\ &=1+c_4+1+c_8+1+\frac12(c_{12}+c_{16})-2(c_6+c_2+c_{10})\\ &=3+c_4+c_8-c_2-2(c_2+c_6+c_{10})\\ &=3-c_2+c_4-c_6-2(c_2-c_4+c_6)\\ &=3(1-c_2+c_4-c_6)\\ &=3(c_0+c_4+c_8+c_{12})\\ &=\frac3{2s_2}2s_2(c_0+c_4+c_8+c_{12})\\ &=\frac3{2s_2}(2s_2c_0+2s_2c_4+2s_2c_8+2s_2c_{12})\\ &=\frac3{2s_2}(s_2+s_2+s_6-s_2+s_{10}-s_6+s_{14}-s_{10})\\ &=\frac3{2s_2}(s_2)\\ &=\frac32 \end{align}$$ One thing to note is $\displaystyle c_0+c_4+c_8+c_{12}=\frac12$.

Now denominator is (do similarly for this):

$$\begin{align} 4s_1^2s_3^2s_5^2 &=4\frac1{16}(2(2s_1s_3)s_5)^2\\ &=\frac1{4}(2(c_2-c_4)s_5)^2\\ &=\frac1{4}(s_7+s_3-s_9-s_1)^2\\ &=\frac1{4}(1-s_1+s_3-s_5)^2\\ &=\frac1{4}(c_0+c_8+c_4+c_{12})^2\\ &=\frac1{4}\left(\frac12\right)^2\\ &=\frac{1}{16} \end{align}$$

Jean-Claude Arbaut
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RE60K
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