Question

$if \: cos \frac{ \alpha }{3} = \frac{1}{2} \: find \: the \: value \: of \: sin \alpha .$

Answer 1

$\sf {\underline {Solution }: }$

$\mathcal {\underline{Here,}}$

$\sf{ Cos \frac{ \alpha }{3} = \frac{1}{2} }$

$\mathcal {\underline{Then,}}$

$\sf Sin \frac{ \alpha }{3} = \sqrt{1 - {Cos}^{2} \frac{ \alpha }{3} } = \sqrt{1 - ( { \frac{1}{2} })^{2} }$

$\sf \implies \sqrt{ \frac{4 - 1}{4} } = \frac{ \sqrt{3} }{2}$

$\mathcal {\underline{Now,}}$

$\sf Sin \: \alpha = 3 \: Sin \frac{ \alpha }{3} - 4 \: {Sin}^{3} \frac{ \alpha }{3}$

$\sf{ \implies 3 \: . \: \frac{ \sqrt{3} }{2} } - 4 (\frac{ \sqrt{3} }{2} ) ^{3}$

$\sf \implies \frac{3 \sqrt{3} }{2} - 4 \: . \: \frac{3 \sqrt{3} }{8}$

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

$\sf \implies \: 0$

$\red {\boxed { \green{ \mathcal{Hence, The \: value \: of \: Sin \: \alpha \: i s \: 0.}}}}$

Answer 2

Answer:

So ,welcome to the concept of trigonometric formulas :

We know that :

$\mathsf{cos3x=4cos^3x-3cosx}\\\\\mathsf{Put\:\dfrac{\alpha}{3}=x\:in\:the\:equation}\\\\\implies \mathsf{cos\alpha=4cos^3(\dfrac{\alpha}{3})-3cos\dfrac{\alpha}{3}}\\\\\implies \mathsf{cos\alpha=4(\dfrac{1}{2})^3-3\times \dfrac{1}{2}}\\\\\implies \mathsf{cos\alpha=4\times \dfrac{1}{8}-\dfrac{3}{2}}\\\\\implies \mathsf{cos\alpha=\dfrac{1}{2}-\dfrac{3}{2}}\\\\\implies \mathsf{cos\alpha=\dfrac{1-3}{2}}\\\\\implies \mathsf{cos\alpha=\dfrac{-2}{2}}\\\\\implies \mathsf{cos\alpha=-1}$

We know that :

$\mathsf{sin^2\alpha+cos^2\alpha=1}\\\\\implies \mathsf{sin^2\alpha+(-1)^2=1}\\\\\implies \mathsf{sin^2\alpha+1=1}\\\\\implies \mathsf{sin^2\alpha=1-1=0}\\\\\implies \mathsf{sin\alpha=0}$

The value will be 0 .

Step-by-step explanation:

This one is called the triple angle formula . ( I learnt today :p )

$\mathtt{cos3A=4cos^3A-3cosA}$