Question

Let ${\sec }^2\left(\frac{\pi }{9}\right)+{\sec }^2\left(\frac{2\pi }{9}\right)+{\sec }^2\left(\frac{4\pi }{9}\right)=S$ and ${\sum }_{k=1}^{89}{\cos }^6\left(k^{\circ }\right)=\frac{a}{b}$, where $a,b$ are coprime, then which of the following is/are correct?

• A. Number of positive divisors of $a+b+S$ is $4$.
• B. Number of positive divisors of $a+b+S$ is $8$
• C. $S$ is a perfect square
• D. $a+b$ is a prime number.

Answer 1

Answer: (A), (C), (D)

$a+b$ is a prime number.

$S={\sec }^2\left(\frac{\pi }{9}\right)+{\sec }^2\left(\frac{2\pi }{9}\right)+{\sec }^2\left(\frac{4\pi }{9}\right)$

$=\frac{{\cos }^2\frac{2\pi }{9}{\cos }^2\frac{4\pi }{9}+{\cos }^2\frac{4\pi }{9}{\cos }^2\frac{\pi }{9}+{\cos }^2\frac{\pi }{9}{\cos }^2\frac{2\pi }{9}}{{\left(\cos \frac{\pi }{9}\cos \frac{2\pi }{9}\cos \frac{4\pi }{9}\right)}^2}$

$=\frac{1}{4}\left[\frac{{(\cos \frac{2\pi }{3}+\cos \frac{2\pi }{9})}^2+{(\cos \frac{5\pi }{9}+\cos \frac{\pi }{3})}^2+{(\cos \frac{\pi }{3}+\cos \frac{\pi }{9})}^2}{\left(\frac{1}{64}\right)}\right]$

$=16[\frac{3}{4}+{\cos }^2\frac{\pi }{9}+{\cos }^2\frac{2\pi }{9}+{\cos }^2\frac{4\pi }{9}+\underset{\left(0\right)}{\underset{}{\cos \frac{\pi }{9}-\cos \frac{2\pi }{9}-\cos \frac{4\pi }{9}}}]$

$=16(\frac{3}{4}+\frac{3}{2})=36$

$\sum _{k=1}^{89}{\cos }^6\left(k^{\circ }\right)=\sum _{k=1}^{89}{\sin }^6\left(k^{\circ }\right)=\frac{1}{2}\left(\sum _{k=1}^{89}\right({\sin }^6\left(k^{\circ }\right)+{\cos }^6\left(k^{\circ }\right)\left)\right)$

$=\frac{1}{2}\left(\sum _{k=1}^{89}(1-\frac{3}{4}{\sin }^2(2k^{\circ }\left)\right)\right)$

$=\frac{89}{2}-\frac{3}{8}\sum _{k=1}^{89}{\sin }^2\left(2k^{\circ }\right)$

$=\frac{89}{2}-\frac{3}{8}\times 45=\frac{221}{8}(\because \sum _{k=1}^{89}{\sin }^2(2k^{\circ })=\sum _{k=1}^{90}{\sin }^2(2k^{\circ })=\sum _{k=1}^{90}{\cos }^2(2k^{\circ })=\frac{1}{2}\sum _{k=1}^{90}({\sin }^2\left(2k^{\circ }\right)+{\cos }^2\left(2k^{\circ }\right))=45$

$a=221,b=8$