First slide

Trigonometric transformations

Question

$\cos^{3} xsin 2 x = \sum_{x = 0}^{n} a_{r} \sin (rx) \forall x \in R$ then

Moderate

A

$n = 5, a_{1} = 1 / 2$
B

$n = 5, a_{1} = 1 / 4$
C

$n = 5, a_{2} = 1 / 8$
D

$n = 5, a_{2} = 1 / 4$

Solution

$\begin{array}{l} \cos^{3} xsin 2 x = \cos^{2} xcos xsin 2 x \\ = (\frac{1 + \cos 2 x}{2}) (\frac{2 \sin 2 xcos x}{2}) \\ = \frac{1}{4} (1 + \cos 2 x) (\sin 3 x + \sin x) \\ = \frac{1}{4} [\sin 3 x + \sin x + \frac{1}{2} (2 \sin 3 xcos 2 x) + \frac{1}{2} (2 \cos 2 xsin x)] \\ = \frac{1}{4} [\sin 3 x + \sin x + \frac{1}{2} (\sin 5 x + \sin x) + \frac{1}{2} (\sin 3 x - \sin x)] \\ = \frac{1}{4} [\sin x + \frac{3}{2} \sin 3 x + \frac{1}{2} \sin 5 x] \\ \Rightarrow a_{1} = 1 / 4, a_{3} = 3 / 8, n = 5 \end{array}$

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