Question

If sinalpha = 1/2, prove that 3cosalpha-4cosquealpha=0

Answer 1

$\huge\star{\green{\underline{\mathfrak{Correct\:Question :-}}}}$

If $\mapsto\tt{sin\:\alpha =\dfrac{1}{2}}$.

Prove : $\mapsto\tt{sin\:\alpha =\dfrac{1}{2}}$.

$\huge\star{\green{\underline{\mathfrak{Answer :-}}}}$

Given :

$\mapsto\tt{sin\:\alpha =\dfrac{1}{2}}$

To Prove :

$\mapsto\tt{3\:cos\:\alpha - 4\:{cos}^{3}\:\alpha = 0}$

Solution :

$\mapsto\tt{sin\:\alpha =\dfrac{1}{2}}$

We know that, when angle is 30° then, sin 30° = 1/2.

So, here, $\tt{\alpha=30^{\circ}}$

Using this in the LHS,

$\mapsto\tt{3\:cos\:(30^{\circ})-4\:{cos}^{3}\:(30^{\circ})= 0}$

$\mapsto\tt{3\:\times\dfrac{\sqrt{3}}{2} - 4\:\times({\dfrac{\sqrt{3}}{2}})^{3}}$

$\mapsto\tt{3\:\times\dfrac{\sqrt{3}}{2} - 4\:\times\dfrac{3\sqrt{3}}{8}}$

$\mapsto\tt{\dfrac{3\sqrt{3}}{2}-\dfrac{3\sqrt{3}}{2}}$

$\huge\mapsto{\boxed{\tt{\green{LHS=RHS=0}}}}$

$\rule{200}4$

Answer 2

$\bf Given:\ \tt sin\ \alpha\ =\ \dfrac{1}{2}.\\\\\\\bf To\ prove:\ \tt 3\ cos\ \alpha\ -\ 4\ cos^3\ \alpha\ =\ 0.\\\\\\\bf Answer:\\\\\\\tt sin\ \alpha\ =\ \dfrac{1}{2}\\\\\\\sf We\ know\ that\ sin\ {30}^{\circ}\ =\ \dfrac{1}{2}.\\\\\\Hence, \alpha\ =\ {30}^{\circ}.\\\\\\\tt 3\ cos\ \alpha\ -\ 4\ cos^3\ \alpha\ =\ 0.\\\\\\\bf Left\ Hand\ Side:\\\\\\\tt 3\ cos\ \alpha\ -\ 4\ cos^3\ \alpha\\\\\\\implies\ 3\ cos\ {30}^{\circ}\ -\ 4\ cos^3\ {30}^{\circ}$

$\bf cos\ {30}^{\circ}\ =\ \dfrac{\sqrt{3}}{2}.\\\\\\\tt =\ \bigg(3\ \times\ \dfrac{\sqrt{3}}{2}\bigg)\ -\ \bigg(4\ \times\ \bigg[\dfrac{\sqrt{3}}{2}\bigg]^3\bigg)\\\\\\=\ \dfrac{3\sqrt{3}}{2}\ -\ 4\bigg(\dfrac{3\sqrt{3}}{8}\bigg)\\\\\\=\ \dfrac{3\sqrt{3}}{2}\ -\ \dfrac{3\sqrt{3}}{2}\\\\\\=\ 0\\\\\\=\ \bf Right\ Hand\ Side\ (0).$