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# The value of the trigonometric function $\dfrac{2\cos {{40}^{\circ }}-\cos {{20}^{\circ }}}{\sin {{20}^{\circ }}}$ is

Answer
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Hint:Simplify the numerator of the given expression and use the trigonometric formulae like$\left( \cos A-\cos B \right)$ and $\left( \operatorname{sinA}-\sin B \right)$ to solve the expression in numerator.

Complete step by step answer:
We are given the expression$\dfrac{2\cos {{40}^{\circ }}-\cos {{20}^{\circ }}}{\sin {{20}^{\circ }}}$.
Let us first solve the numerator$\left( 2\cos {{40}^{\circ }}-\cos {{20}^{\circ }} \right)$.
The numerator can also be written as,
\begin{align} & \cos {{40}^{\circ }}+\cos {{40}^{\circ }}-\cos {{20}^{\circ }} \\ & \Rightarrow \cos {{40}^{\circ }}+\cos {{40}^{\circ }}-\cos {{20}^{\circ }} \\ & \Rightarrow \cos {{40}^{\circ }}+\left( \cos {{40}^{\circ }}-\cos {{20}^{\circ }} \right) \\ & \Rightarrow \cos {{40}^{\circ }}-\left( \cos {{20}^{\circ }}-cos{{40}^{\circ }} \right)......(1) \\ \end{align}
Taking negative sign common from brackets,
We know the formula,
\begin{align} & \cos A-\cos B=-2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right) \\ & \cos A-\cos B=2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{B-A}{2} \right) \\ \end{align}
Let us put$A={{20}^{\circ }}$and$B={{40}^{\circ }}$.
\begin{align} & \therefore \cos {{20}^{\circ }}-\cos {{40}^{\circ }}=2\sin \left( \dfrac{20+40}{2} \right)\sin \left( \dfrac{40-20}{2} \right) \\ & \Rightarrow 2\sin \left( \dfrac{60}{2} \right)\sin \left( \dfrac{20}{2} \right)=2\sin {{30}^{\circ }}\sin {{10}^{\circ }} \\ \end{align}
Substituting value of$\left( \cos {{20}^{\circ }}-\cos {{40}^{\circ }} \right)$in equation (1),
$\Rightarrow \cos {{40}^{\circ }}-2\sin {{30}^{\circ }}\sin {{10}^{\circ }}........(2)$
We know the value of$\sin {{30}^{\circ }}={}^{1}/{}_{2}$from the trigonometric table.
$\therefore 2\sin {{30}^{\circ }}=1$, substitute it in equation (2).
$\therefore \cos {{40}^{\circ }}-\sin {{10}^{\circ }}$
By putting its denominator and substituting$\left( \cos {{40}^{\circ }}-\sin {{10}^{\circ }} \right)$ in place of$\left( 2\cos {{40}^{\circ }}-\cos {{20}^{\circ }} \right)$,
$\Rightarrow \dfrac{\cos {{40}^{\circ }}-\sin {{10}^{\circ }}}{\sin {{20}^{\circ }}}......(3)$
Let us take the cosine function of$\cos {{40}^{\circ }}$, [we know 90-50=40]
$\cos \left( 90-\theta \right)=\sin {{\theta }^{\circ }}$
$\therefore$Put$\theta ={{50}^{\circ }}$.
$\therefore \cos \left( 90-50 \right)=\sin {{50}^{\circ }}$
$\therefore$Equation (3) becomes$\dfrac{\sin {{50}^{\circ }}-\sin {{10}^{\circ }}}{\sin {{20}^{\circ }}}$.
We know the formula for$sinA-sinB=2cos\left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)$.
Let us put$A={{50}^{\circ }}$and$B={{10}^{\circ }}$.
\begin{align} & \sin {{50}^{\circ }}-\sin {{10}^{\circ }}=2\cos \left( \dfrac{50+10}{2} \right)\sin \left( \dfrac{50-10}{2} \right) \\ & \Rightarrow 2\cos \dfrac{60}{2}\sin \dfrac{40}{2} \\ & \Rightarrow 2\cos 30\sin 20 \\ & \therefore \dfrac{\sin {{50}^{\circ }}-\sin {{10}^{\circ }}}{\sin {{20}^{\circ }}}=\dfrac{2\cos {{30}^{\circ }}\sin {{20}^{\circ }}}{\sin {{20}^{\circ }}} \\ \end{align}
Cancel out the like term$\left( \sin {{20}^{\circ }} \right)$from the numerator and denominator.
Hence, we get$2\cos {{30}^{\circ }}.$
We know the value of$\cos {{30}^{\circ }}={}^{\sqrt{3}}/{}_{2}$.
Therefore, the expression =$2\cos {{30}^{\circ }}=2\times \dfrac{\sqrt{3}}{2}=\sqrt{3}$
Hence the value of$\dfrac{2\cos {{40}^{\circ }}-\cos {{20}^{\circ }}}{\sin {{20}^{\circ }}}=\sqrt{3}.$

Note:
To solve problems like these you should remember the trigonometric identities of sine and cosine functions along with the trigonometric table. In this expression, most are substitution of the formulae of sine and cosine functions and their simplification. Therefore be thorough with the formulae.