First slide

Multiple and sub- multiple Angles

Question

If x sin a +y sin 2a + z sin3a = sin4a, x sin b +y sin 2b + z sin3b= sin4b, x sin c+ y sin 2c +z sin 3c= sin4c, then the roots of the equation
$t^{3} - (\frac{z}{2}) t^{2} - (\frac{y + 2}{4}) t + (\frac{z - x}{8}) = 0, a, b, c \neq nπ$ are

Moderate

A

sin a, sin b, sin c
B

cos a, cos b, cos c
C

sin 2a, sin 2b, sin 2c
D

cos 2a, cos 2b cos 2c

Solution

$\begin{array}{l} xsin a + ysin 2 a + zsin 3 a = \sin 4 a \\ \Rightarrow xsin a + y \times 2 \sin acos a + z \times \sin a (3 - 4 \sin^{2} a) \\ = 2 \times 2 \sin acos acos 2 a \\ \Rightarrow x + 2 ycos a + z (3 + 4 \cos^{2} a - 4) \\ = 4 \cos a (2 \cos^{2} a - 1) [as \sin a \neq 0] \end{array}$
$\begin{array}{l} \Rightarrow 8 \cos^{3} a - 4 {zcos}^{2} a - (2 y + 4) \cos a + (z - x) = 0 \\ \Rightarrow \cos^{3} a - (\frac{z}{2}) \cos^{2} a - (\frac{y + 2}{4}) \cos a + (\frac{z - x}{8}) = 0 \end{array}$
Which shows that cos a is root of the equation
$t^{3} - (\frac{z}{2}) t^{2} - (\frac{y + 2}{4}) t + (\frac{z - x}{8}) = 0$
Similarly, from second and third equations, we can slow that cos b and cos c are the roots of the given equation.

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