Question

If 3 cos theta = underoot 3 , prove that 3 sin theta - 4 sin^3 theta = 1.

Answer 1

Answer:

$3sin\theta-4sin^{3}\theta=\frac{\sqrt{2}}{3\sqrt{3}}$

Step-by-step explanation:

$Given\:3cos\theta=\sqrt{3}$

$\implies cos\theta=\frac{\sqrt{3}}{3}\\=\frac{1}{\sqrt{3}}\:---(1)$

$sin^{2}\theta=1-cos^{2}\theta\\=1-\left(\frac{1}{\sqrt{3}}\right)^{2}\\=1-\frac{1}{3}\\=\frac{3-1}{3}\\=\frac{2}{3}\:--(2)$

$\implies sin\theta = \sqrt{\frac{2}{3}}\:--(3)$

$Now, 3sin\theta-4sin^{3}\theta\\=3\times \sqrt{\frac{2}{3}}-4\times \left(\sqrt{\frac{2}{3}}\right)^{3}\\=\frac{3\sqrt{2}}{\sqrt{3}}-\frac{8\sqrt{2}}{3\sqrt{3}}\\=\frac{9\sqrt{2}-8\sqrt{2}}{3\sqrt{3}}\\=\frac{\sqrt{2}}{3\sqrt{3}}$

Therefore,

$3sin\theta-4sin^{3}\theta=\frac{\sqrt{2}}{3\sqrt{3}}$

•••♪

Answer 2

$\huge{\mathfrak{\underline{\underline{\red{Answer :-}}}}}$

$\huge{\boxed{\boxed{\pink{\frac{\sqrt{2}}{ 3\sqrt{3}}}}}}$

$\huge{\mathfrak{\blue{Explanation}}}$

✯ Given :-

$\sf{3cos{\theta} = {\sqrt3}}$

$\bf{cos{\theta} = \frac{\sqrt3}{3}}$

$\bf{cos{\theta} = \frac{1}{\sqrt3}}$

Let this be equation 2.

To Prove :-

$\sf{\large{3sin{\theta} - 4sin^{3}{\theta} = 1}}$

Let this be equation 1

✯ Solution :-

Using Formula

$\huge{\boxed{\boxed{\red{sin^{2}{\theta} = 1 - cos^{2}{\theta}}}}}$

___________[Put values]

$\sf{sin^{2}{\theta} = 1 - ({ \frac{1}{{\sqrt3}} ) }^{2}}$

$\sf{sin^{2}{\theta} = 1 - \frac{1}{3}}$

Taking LCM

$\sf{sin^{2}{\theta} = \frac{3 - 1}{3}}$

$\sf{sin^{2}{\theta} = \frac{2}{3}}$

$\bf{sin{\theta} = ({ \sqrt{ \frac{2}{3} } )}}$

Let this be equation 3

$\rule{200}{2}$

__________[Put Values in equation 1]

$\sf{3 \times \sqrt{ \frac{2}{3} } \times - 4 \times ({ \sqrt{ \frac{2}{3} } )}^{3}}$

$\sf{ \frac{ 3\sqrt{2} }{\sqrt{3} } - \frac{ 8\sqrt{2} }{3\sqrt{3}}}$

Taking LCM

$\sf{\large{ \frac{ 9\sqrt{2} - 8\sqrt{2} }{3 \sqrt{3}}}}$

$\bf{\large{ \frac{ \sqrt{2} }{ 3\sqrt{3}}}}$

$\huge{\boxed{\boxed{\red{\frac{\sqrt{2}}{ 3\sqrt{3}}}}}}$