Question

If $\sin \theta =\dfrac{1}{2}$, show that $\left( 3\cos \theta -4{{\cos }^{3}}\theta \right)=0$.

Answer 1

Hint:Assume that in the given function: $\sin \theta =\dfrac{1}{2}$, 1 is the length of perpendicular and 2 is the length of hypotenuse of a right angle triangle. Use Pythagoras theorem given by: $\text{hypotenuse}{{\text{e}}^{\text{2}}}=\text{bas}{{\text{e}}^{\text{2}}}+\text{perpendicular}{{\text{r}}^{\text{2}}}$, to determine the length of the base of the right angle triangle. Now, find $\cos \theta $ by taking the ratio of base and hypotenuse. Finally, substitute the value of $\cos \theta $ in the expression: $\left( 3\cos \theta -4{{\cos }^{3}}\theta \right)$ to get the result.

Complete step-by-step answer:
We have been provided with the trigonometric ratio, $\sin \theta =\dfrac{1}{2}$.
We know that, $\sin \theta =\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}$. Therefore, on comparing it with the above provided ratio, we have, 1 as the length of perpendicular and 2 as the length of hypotenuse of a right angle triangle.

Now, using Pythagoras theorem: $\text{hypotenuse}{{\text{e}}^{\text{2}}}=\text{bas}{{\text{e}}^{\text{2}}}+\text{perpendicular}{{\text{r}}^{\text{2}}}$, we get,
\[\begin{align}
  & \text{bas}{{\text{e}}^{\text{2}}}=\text{hypotenuse}{{\text{e}}^{\text{2}}}-\text{perpendicular}{{\text{r}}^{\text{2}}} \\
 & \Rightarrow \text{base}=\sqrt{\text{hypotenuse}{{\text{e}}^{\text{2}}}-\text{perpendicular}{{\text{r}}^{\text{2}}}} \\
 & \Rightarrow \text{base}=\sqrt{{{2}^{\text{2}}}-{{1}^{\text{2}}}} \\
 & \Rightarrow \text{base}=\sqrt{4-1} \\
 & \Rightarrow \text{base}=\sqrt{3} \\
\end{align}\]
We know that, $\cos \theta =\dfrac{\text{Base}}{\text{Hypotenuse}}$.
$\Rightarrow \cos \theta =\dfrac{\sqrt{3}}{2}$
Here, we have to find the value of the expression: $\left( 3\cos \theta -4{{\cos }^{3}}\theta \right)$. Therefore, substituting the value of $\cos \theta $ in the expression, we get,
\[\begin{align}
  & =3\times \dfrac{\sqrt{3}}{2}-4{{\left( \dfrac{\sqrt{3}}{2} \right)}^{3}} \\
 & =\dfrac{3\sqrt{3}}{2}-4\times \dfrac{3\sqrt{3}}{8} \\
 & =\dfrac{3\sqrt{3}}{2}-\dfrac{3\sqrt{3}}{2} \\
 & =0 \\
\end{align}\]

Note: You may note that there is an alternate and easy method to solve this question. This requires the information relating to the values of trigonometric functions at some particular angles. We know that the value of $\sin {{30}^{\circ }}=\dfrac{1}{2}$, on comparing it with the information given in the question, we have $\theta ={{30}^{\circ }}$. Now, we know that, $\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}$. From here we may substitute the value of $\cos \theta $ in the expression: $\left( 3\cos \theta -4{{\cos }^{3}}\theta \right)$ to get the answer. In this way we do not have to use Pythagoras theorem.