Question

If 2 sin theta-1=0, then prove that sin 3 theta =3 sin theta-4sin^3 theta

Answer 1

$\textbf{Given:}$

$2\;sin\theta-1=0$

$\textbf{To prove:}$

$sin3\theta=3\,sin\theta-4\,sin^3\theta$

$\text{Consider,}$

$2\;sin\theta-1=0$

$\implies\,sin\theta=\frac{1}{2}$

$\implies\bf\theta=30^{\circ}$

$\text{Now,}$

$\bf\,sin3\theta$

$=sin3(30^{\circ})$

$=sin\,90^{\circ}$

$=1$.......(1)

$\bf\,3\,sin\theta-4\,sin^3\theta$

$=3\,sin(30^{\circ})-4\,sin^3(30^{\circ})$

$=3(\dfrac{1}{2})-4(\dfrac{1}{2})^3$

$=\dfrac{3}{2}-4(\dfrac{1}{8})$

$=\dfrac{3}{2}-\dfrac{1}{2}$

$=1$.......(2)

$\text{From (1) and (2), we get}$

$\boxed{\bf\,sin3\theta=3\,sin\theta-4\,sin^3\theta}$

$\textbf{Hence proved}$

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Answer 2

Answer:

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Step-by-step explanation:

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