Question

If the trigonometric identity $\cos A=\dfrac{\sqrt{3}}{2}$ , then find the value of $\tan 3A=?$

Answer 1

Hint: We know the value of $\cos A$, find value of $\cos 3A$ using formula $\cos 3A=4{{\cos }^{3}}A-3\cos A$.
Then find $\sin 3A$ using identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, next find the value of $\tan 3A$ which is equal to $\dfrac{\sin 3A}{\cos 3A}$ .

Complete step-by-step answer:
In the question we are given the value of $\cos A=\dfrac{\sqrt{3}}{2}.$
As we know the value of $\cos A$, now we will find out the value of $\cos 3A$, let 3A = 2A + A, so we get
$\cos 3A=\cos \left( 2A+A \right)$
Now by applying the identity, $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$ the above expression can be written as,
$\cos \left( 3A \right)=\cos 2A\operatorname{cosA}-\sin 2A\sin A$
Now we will use the identity, $\cos 2A=2{{\cos }^{2}}A-1,\sin 2A=2\sin \cos A$ so above expression can be written as,
$\cos \left( 3A \right)=\left( 2{{\cos }^{2}}A-1 \right)\cos A-2{{\sin }^{2}}A\cos A$
Using the identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ ,we can write ${{\sin }^{2}}A$ as $\left( 1-{{\cos }^{2}}A \right)$
By substituting in above expression
$\cos \left( 3A \right)=\left( 2{{\cos }^{2}}A-1 \right)\cos A-2\cos A\left( 1-{{\cos }^{2}}A \right)$
Now opening the brackets, the above equation can be written as,
$\begin{aligned} & \cos 3A=2{{\cos }^{3}}A-\cos A-2\cos A+2{{\cos }^{3}}A \\\ & \Rightarrow \cos 3A=4{{\cos }^{3}}A-3\cos A \\\ \end{aligned}$
Now substituting the given value $\cos A=\dfrac{\sqrt{3}}{2}$, the above expression can be written as,
$\begin{aligned} & \cos 3A=4{{\left( \dfrac{\sqrt{3}}{2} \right)}^{3}}-3\times \dfrac{\sqrt{3}}{2} \\\ & \cos 3A=\left( \dfrac{3\sqrt{3}}{2} \right)-\dfrac{3\sqrt{3}}{2}=0........(i) \\\ \end{aligned}$
Now we know,
$\tan 3A=\dfrac{\sin 3A}{\cos 3A}$
Converting this into cosine terms using the identity, ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\Rightarrow \sin \theta =\sqrt{1-{{\cos }^{2}}\theta }$, so above expression can be written as,
$\tan 3A=\dfrac{\sqrt{1-{{\cos }^{2}}3A}}{\cos 3A}$
Substituting the values from equation (i), we get
$\tan 3A=\dfrac{\sqrt{1-0}}{0}=\dfrac{1}{0}$
Hence, $\tan 3A$ has the value of $\dfrac{1}{0}$ which means $\tan 3A$ is undefined.
The value of $\tan 3A$ is undefined.

Note: We can do by another method as we know that $\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}$ so we can say that
$\begin{aligned} & \cos A=\cos {{30}^{\circ }} \\\ & A={{30}^{\circ }} \\\ \end{aligned}$
Then value of,
$3A=3\times {{30}^{\circ }}={{90}^{\circ }}.$
Then value of,
$\tan 3a=\tan {{90}^{\circ }}$
This is undefined.
Another approach is using the direct formula, $\cos 3A=4{{\cos }^{3}}A-3\cos A$.