If F(k) = (1 + sin(π)/(2k)) (1 + (k - 1)(π)/(2k))(1 + (2k + 1)(π)/(2k))(1 + (3k - 1)(π)/(2k)), then the value of F (1) + F (2) + F (3) & is equal to

Correct answer: D. (7)/(16) F(1) = (1 + sin(π)/(2))(1 + 0)(1 + sin(3π)/(2))(1 + sin π) = 2 × 1 × 0 × 1 = 0 F(2) = (1 + sin(π)/(4))(1 + sin(π)/(4))(1 + sin(5π)/(4))(1 + sin(5π)/(4)) = (1 + 1 2 ) ^ 2 (1 - 1 2 ) ^ 2 = (1 - (1)/(2)) ^ 2 = (1)/(4) F(3) = (1 + sin(π)/(6))(1 + sin(π)/(3))(1 + sin(7π)/(6))(1 + sin(4π)/(3)) = (1 + (1)/(2))(1 + 3 2 )(1 - (1)/(2))(1 - 3 2 ) = (1 - (1)/(4))(1 - (3)/(4)) = (3)/(4) × (1)/(4) = (3)/(16) ⇒ F(1) + F(2) + F(3) = 0 + (1)/(4) + (3)/(16) = (7)/(16)

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Trigonometric Ratios & Identities

NTA Abhyas 2020 · Mathematics · Trigonometric Ratios & Identities

If F(k) = (1 + sin(π)/(2k)) (1 + (k - 1)(π)/(2k))(1 + (2k +...

F\left(k\right) = \left(1 + sin\dfrac{\pi}{2k}\right) \left(1 + \sin \left(k - 1\right)\dfrac{\pi}{2k}\right)\left(1 + \sin \left(2k + 1\right)\dfrac{\pi}{2k}\right)\left(1 + \sin \left(3k - 1\right)\dfrac{\pi}{2k}\right), then

the value of F \left(1\right) + F \left(2\right)

+ F \left(3\right)

& is equal to

Options

A $\dfrac{3}{16}$
B $\dfrac{1}{4}$
C $\dfrac{5}{16}$
D $\dfrac{7}{16}$

Correct answer

D. $\dfrac{7}{16}$

Step-by-step solution

F\left(1\right) = \left(1 + sin\dfrac{\pi}{2}\right)\left(1 + \sin 0\right)\left(1 + sin\dfrac{3\pi}{2}\right)\left(1 + sin \pi\right)

= 2 \times 1 \times 0 \times 1 = 0

F\left(2\right) = \left(1 + sin\dfrac{\pi}{4}\right)\left(1 + sin\dfrac{\pi}{4}\right)\left(1 + sin\dfrac{5\pi}{4}\right)\left(1 + sin\dfrac{5\pi}{4}\right)

= {\left(1 + \dfrac{1}{\sqrt{2}}\right)}^{2}{\left(1 - \dfrac{1}{\sqrt{2}}\right)}^{2} = {\left(1 - \dfrac{1}{2}\right)}^{2} = \dfrac{1}{4}

F\left(3\right) = \left(1 + sin\dfrac{\pi}{6}\right)\left(1 + sin\dfrac{\pi}{3}\right)\left(1 + sin\dfrac{7\pi}{6}\right)\left(1 + sin\dfrac{4\pi}{3}\right)

= \left(1 + \dfrac{1}{2}\right)\left(1 + \dfrac{\sqrt{3}}{2}\right)\left(1 - \dfrac{1}{2}\right)\left(1 - \dfrac{\sqrt{3}}{2}\right)

= \left(1 - \dfrac{1}{4}\right)\left(1 - \dfrac{3}{4}\right) = \dfrac{3}{4} \times \dfrac{1}{4} = \dfrac{3}{16}

⇒ F\left(1\right) + F\left(2\right) + F\left(3\right) = 0 + \dfrac{1}{4} + \dfrac{3}{16} = \dfrac{7}{16}

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