Question

Prove that sin teta=sin3 teta/1+2 cos teta
and hence deduce the value of sin 15 degree​

Answer 1

Answer:

Correct Question:-

Prove that $\sin\theta = \dfrac{sin3\theta}{1+2cos2\theta}$ and deduce the value of $sin15\degree$.

Solution :-

Taking R.H.S :-

$\sf\dfrac{sin3\theta}{1+2cos2\theta}$

$\sf sin3\theta = 3sin\theta- 4sin^{3}\theta$

$\sf cos2\theta = 1- 2sin^{2}\theta$.

Substituting values:-

$\sf\dfrac{3sin\theta- 4sin^{3}\theta}{1 + 2(1- 2sin^{2}\theta}$

= $\sf\dfrac{3sin\theta- 4sin^{3}\theta}{1 + 2 - 4sin^{2}\theta}$

= $\sf\dfrac{sin\theta(3 - 4sin^{2}\theta}{3 - sin^{2}\theta}$

= $\sf sin\theta$

Hence L.H.S = R.H.S

Value of $\sf sin15\degree$

$\sf sin15\degree = \dfrac{sin3(15\degree)}{1 + 2cos2(15\degree)}$

$\sf sin15\degree = \dfrac{sin45\degree}{1 + 2cos30\degree}$

$\sf sin45\degree = \dfrac{1}{\sqrt{2}}$

$\sf cos30\degree = \dfrac{\sqrt{3}}{2}$

Substituting values:-

$\sf sin15\degree = \dfrac{\dfrac{1}{\sqrt{2}}}{1 + 2(\dfrac{\sqrt{3}}{2})}$

= $\sf\dfrac{1}{\sqrt{2}(1+\sqrt{3})}$

= $\sf\dfrac{1}{\sqrt{2} + \sqrt{6}}$

Rationalising Denominater :-

$\frac{1}{ \sqrt{6} + \sqrt{2} } \times \frac{ \sqrt{6} - \sqrt{2} }{ \sqrt{6} - \sqrt{2} }$

$\frac{ \sqrt{6} - \sqrt{2} }{4}$

$\sqrt{2} (\frac{ \sqrt{3} - 1 }{4})$

$\frac{ \sqrt{3} - 1 }{2 \sqrt{2} }$

$\sf\therefore sin15\degree = \dfrac{\sqrt{3}-1}{2\sqrt{2}}$