First slide

Multiple and sub- multiple Angles

Question

Let $P (k) = (1 + \cos \frac{π}{4 k}) (1 + \cos \frac{(2 k - 1) π}{4 k})$ $(1 + \cos \frac{(2 k + 1) π}{4 k}) (1 + \cos \frac{(4 k - 1) π}{4 k})$ . Then

Moderate

A

$P (3) = \frac{1}{16}$
B

$P (4) = \frac{2 - \sqrt{2}}{16}$
C

$P (5) = \frac{3 - \sqrt{5}}{32}$
D

$P (6) = \frac{2 - \sqrt{3}}{16}$

Solution

$P (k) = (1 + \cos \frac{π}{4 k}) (1 + \cos (\frac{π}{2} - \frac{π}{4 k})) (1 + \cos (\frac{π}{2} + \frac{π}{4 k})) (1 + \cos (π - \frac{π}{4 k}))$
$\begin{array}{l} = (1 + \cos \frac{π}{4 k}) (1 + \sin \frac{π}{4 k}) (1 - \sin \frac{π}{4 k}) (1 - \cos \frac{π}{4 k}) \\ = (1 - \cos^{2} \frac{π}{4 k}) (1 - \sin^{2} \frac{π}{4 k}) \\ = \frac{4 \sin^{2} \frac{π}{4 k} \cdot \cos^{2} \frac{π}{4 k}}{4} \\ P (k) = \frac{1}{4} \sin^{2} (\frac{π}{2 k}) \\ \Rightarrow P (3) = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16} \end{array}$
$\begin{array}{l} \Rightarrow P (4) = \frac{π}{4} \sin^{2} \frac{π}{2 k} = \frac{1}{4} \sin^{2} \frac{π}{8} = \frac{1}{8} (1 - \cos \frac{π}{4}) = \frac{2 - \sqrt{2}}{16} \\ \Rightarrow P (5) = \frac{1}{4} \sin^{2} \frac{π}{10} = \frac{1}{8} (2 \sin^{2} \frac{π}{10}) \\ = \frac{1}{8} (1 - \cos 36^{\circ}) \\ = \frac{1}{8} (1 - \frac{\sqrt{5} + 1}{4}) = \frac{3 - \sqrt{5}}{32} \\ \Rightarrow P (6) = \frac{1}{4} \sin^{2} \frac{π}{12} = \frac{1}{8} (2 \sin^{2} \frac{π}{12}) \\ = \frac{1}{8} (1 - \cos \frac{π}{6}) \\ \Rightarrow P (6) = \frac{2 - \sqrt{3}}{16} \end{array}$

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