Question

The expression ${\log }_{2}5-{\log }_2\left({\sum }_1^4\sin \left(\frac{k\pi }{5}\right)\right)$ reduces to $\frac{p}{q}$ where $p$ and $q$ are co-prime, then the value of $p^2+q^2=$

• A. $13$
• B. $15$
• C. $17$
• D. $18$

Answer 1

Answer: (C)

$17$

Let $p={\log }_{2}5-{\sum }_{k=1}^4{\log }_2\left(\sin \left(\frac{k\pi }{5}\right)\right)$

$p={\log }_{2}5-\{{\log }_2\left(\sin \left(\frac{\pi }{5}\right)\right)+{\log }_2\left(\sin \left(\frac{2\pi }{5}\right)\right)+{\log }_2\left(\sin \left(\frac{3\pi }{5}\right)\right)+{\log }_2\left(\sin \left(\frac{4\pi }{5}\right)\right)\}$

By using the product rule of logarithms, $\log a+\log b=\log ab$ we have

$p={\log }_{2}5-{\log }_2\left(\sin \frac{\pi }{5}\sin \frac{2\pi }{5}\sin \frac{3\pi }{5}\sin \frac{4\pi }{5}\right)$

$={\log }_{2}5-{\log }_2\left(\sin \frac{\pi }{5}\sin \frac{2\pi }{5}\sin (\pi -\frac{2\pi }{5})\sin (\pi -\frac{\pi }{5})\right)$

We know that $\sin (\pi -\theta )=\sin \left(\theta \right)$.

Using this, above we have

$p={\log }_{2}5-{\log }_2\left({\sin }^2\frac{\pi }{5}{\sin }^2\frac{2\pi }{5}\right)$

Using $\cos 2\theta =1-2{\sin }^2\theta$ and ${\sin }^2\theta =\frac{1-\cos 2\theta }{2}$ we get

$p={\log }_{2}5-{\log }_2\left(\frac{1-\cos \frac{2\pi }{5}}{2}\frac{1-\cos \frac{4\pi }{5}}{2}\right)$

$={\log }_{2}5-{\log }_2\left(\frac{1-\cos 72}{2}\frac{1-\cos (\pi -\frac{\pi }{5})}{2}\right)$

As $\cos (\pi -\theta )=-\cos \theta$

$={\log }_{2}5-{\log }_2\left(\frac{1-\cos 72}{2}\frac{1+\cos 36}{2}\right)$

Substituting the values of $\cos 72=\frac{\sqrt{5}-1}{4}$ and $\cos 36=\frac{\sqrt{5}+1}{4}$ we get

$p={\log }_{2}5-{\log }_2\left(\frac{1-\frac{\sqrt{5}-1}{4}}{2}\frac{1+\frac{\sqrt{5}+1}{4}}{2}\right)$

$={\log }_{2}5-{\log }_2\left(\frac{(5-\sqrt{5})(5+\sqrt{5})}{64}\right)$ ...... (On simplifying)

$={\log }_{2}5-{\log }_2\left(\frac{25-5}{64}\right)$

Using quotient law, $\log a-\log b=\log \left(\frac{a}{b}\right)$ we get

$p={\log }_2(5\times \frac{64}{20})={\log }_2\left(\frac{64}{4}\right)={\log }_2{2}^4$

As $\log a^m=m\log a$ we have

${\log }_2{2}^4=4{\log }_{2}2$

Since ${\log }_{a}a=1$ we have $p=4,q=1$

Hence, $p^2+q^2=4^2+1^2=16+1=17$