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  • If   <img alt="\\x \sin a+y \sin 2 a+z \sin 3 a=\sin 4 a\\x \sin b+y \sin 2 b+z \sin 3 b=\sin 4 b, \\x \sin c+y \sin 2 c+z \sin 3 c=\sin 4 c" src="https://entrancecorner.oncodec

If  

\\x \sin a+y \sin 2 a+z \sin 3 a=\sin 4 a\\x \sin b+y \sin 2 b+z \sin 3 b=\sin 4 b, \\x \sin c+y \sin 2 c+z \sin 3 c=\sin 4 c

Then the roots of the equation 

t^3-\left(\frac{z}{2}\right) t^2-\left(\frac{y+2}{4}\right) t+\left(\frac{z-x}{8}\right)=0, a, b, c \neq n \pi

Option: 1

\\\sin a, \sin b, \sin c\\


Option: 2

\\\cos a, \cos b, \cos c\\


Option: 3

\\\sin 2 a, \sin 2 b, \sin 2 c\\


Option: 4

\\\cos 2 a, \cos 2 b, \cos 2 c\\


Answers (1)

best_answer

Equation first can be written as 

\begin{gathered} x \sin a+y \times 2 \sin a \cos a+z \times \sin a\left(3-4 \sin ^2 a\right) \\ \\=2 \times 2 \sin a \cos a \cos 2 a \\ \\\Rightarrow x+2 y \cos a+z\left(3+4 \cos ^2 a-4\right) \\ \\=4 \cos a\left(2 \cos ^2 a-1\right) \text { as } \sin a \neq 0 \end{gathered}

\begin{aligned} & \Rightarrow 8 \cos ^3 a-4 z \cos ^2 a-(2 y+4) \cos a+(z-x)=0 \\ \\& \Rightarrow \cos ^3 a-\left(\frac{z}{2}\right) \cos ^2 a-\left(\frac{y+2}{4}\right) \cos a+\left(\frac{z-x}{8}\right)=0 \end{aligned}

Which is shows that  cos a is root of the equation

           t^3-\left(\frac{z}{2}\right) t^2-\left(\frac{y+2}{4}\right) t+\left(\frac{z-x}{8}\right)=0

Similarly, from second and third equation we can verify \cos \: b\: \text{and} \cos\: c are the roots of the given equation.

Posted by

Gaurav

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