Question

Let F(K)= $\left(1+\sin \frac{\pi }{2k}\right)\left(1+\sin \left(k-1\right)\frac{\pi }{2k}\right)\left(1+\sin \left(2k+1\right)\frac{\pi }{2k}\right)\left(1+\sin \left(3k-1\right)\frac{\pi }{2k}\right)$
The value of $F\left(1\right)+F\left(2\right)+F\left(3\right)$ is equal to

• A. $\frac{3}{16}$
• B. $\frac{1}{4}$
• C. $\frac{5}{16}$
• D. $\frac{7}{16}$

Answer 1

The correct option is D $\frac{7}{16}$

$F\left(1\right)=(1+\sin \frac{\pi }{2})(1+0)(1+\sin \frac{3\pi }{2})(1+\sin \pi )=(1+1)(1+0)(1-1)(1+0)=0F\left(2\right)=(1+\sin \frac{\pi }{4})(1+\sin \frac{\pi }{4})(1+\sin \frac{5\pi }{4})(1+\sin \frac{5\pi }{4})={(1+\sin \frac{\pi }{4})}^2{(1+\sin \frac{5\pi }{4})}^2={(1+\frac{1}{\sqrt{2}})}^2{(1-\frac{1}{\sqrt{2}})}^2=\frac{1}{4}F\left(3\right)=(1+\sin \frac{\pi }{6})(1+\sin \frac{2\pi }{6})(1+\sin (\pi +\frac{\pi }{6})(1+\sin (\pi +\frac{2\pi }{6}))=(1+\sin \frac{\pi }{6}\left)\right(1-\sin \frac{\pi }{6}\left)\right(1+\sin (\frac{2\pi }{6}\left)\right(1-\sin (\frac{2\pi }{6}\left)\right)=(1-{\sin }^2\frac{\pi }{6})(1-{\sin }^2\frac{2\pi }{6})=(1-3/4)(1-1/4)=\frac{3}{16}F\left(1\right)+F\left(2\right)+F\left(3\right)=0+\frac{1}{4}+\frac{3}{16}=\frac{7}{16}$