Question

If sin B = 1/2 then evaluate 3cos B - 4cos^3 B .​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm \: sinB = \dfrac{1}{2}$

can be rewritten as

$\rm \: sinB = sin30\degree$

$\rm\implies \:B = 30\degree$

Now, Consider

$\rm \: 3cosB - 4 {cos}^{3}B$

On substituting the value of B, we get

$\rm \: = \: 3cos30\degree - 4 {cos}^{3}30\degree$

$\rm \: = \: cos30\degree \bigg(3 - 4 {cos}^{2}30\degree \bigg)$

$\rm \: = \: \dfrac{ \sqrt{3} }{2} \bigg(3-4 \times {\bigg[\dfrac{ \sqrt{3} }{2} \bigg]}^{2} \bigg)$

$\rm \: = \: \dfrac{ \sqrt{3} }{2} \bigg(3-4 \times\dfrac{3}{4} \bigg)$

$\rm \: = \: \dfrac{ \sqrt{3} }{2} \bigg(3-3 \bigg)$

$\rm \: = \: \dfrac{ \sqrt{3} }{2} \times 0$

$\rm \: = \: 0$

Hence,

$\rm\implies \:\boxed{\rm{  \:\rm \: 3cosB - 4 {cos}^{3}B = 0 \: \: }}$

$\rule{190pt}{2pt}$

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$