Question

Find the value of $\dfrac{dy}{dx}$, if $x = 2 \cos \theta - \cos 2\theta $ and $y = 2 \sin \theta - \sin 2\theta $.
1). $\tan \dfrac{{3\theta }}{2}$
2). $ - \tan \dfrac{{3\theta }}{2}$
3). $\cot \dfrac{{3\theta }}{2}$
4). $ - \cot \dfrac{{3\theta }}{2}$

Answer 1

Hint: We will differentiate x and y separately with respect to $\theta $ and after that we will divide them in such a way that we get the value of $\dfrac{dy}{dx}$. Then, by using appropriate trigonometric identities we will solve the function and reach the answer.

Complete step-by-step solution:
Given: $x = \cos \theta - \cos 2\theta $
$y = \sin \theta - \sin 2\theta $
We will differentiate x and y separately with respect to $\theta$ because they both are in terms of $\theta $. So, we cannot directly differentiate them with each other.
So, first We will differentiate the equation x with respect to $\theta $
$x = \cos \theta - \cos 2\theta $
$\Rightarrow \dfrac{{dx}}{{d\theta }} = - 2\sin \theta - ( - \sin 2\theta ) \times 2$
$\Rightarrow \dfrac{{dx}}{{d\theta }} = - 2\sin \theta + 2\sin 2\theta $
Now, we will differentiate the equation y with respect to $\theta $.
$y = 2\sin \theta - \sin 2\theta $
$\dfrac{{dy}}{{d\theta }} = 2\cos \theta - 2\cos 2\theta $
So, now we will divide them in order to find $\dfrac{dy}{dx}$.
$\dfrac{{dy}}{{dx}} = \dfrac{{2\cos \theta - 2\cos 2\theta }}{{ - 2\sin \theta + 2\sin 2\theta }}$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\cos \theta - \cos 2\theta }}{{ - \sin \theta + \sin 2\theta }}$
Now, by using trigonometric formula $\cos A - \cos B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)$ and $\sin A - \sin B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)$, we will rewrite the above equation.
$\dfrac{{dy}}{{dx}} = \dfrac{{2\sin \left( {\dfrac{{\theta + 2\theta }}{2}} \right)\sin \left( {\dfrac{{2\theta - \theta }}{2}} \right)}}{{2\cos \left( {\dfrac{{\theta + 2\theta }}{2}} \right)\sin \left( {\dfrac{{2\theta - \theta }}{2}} \right)}}$
Now, after cutting the same terms from the above equation. We get,
$\dfrac{{dy}}{{dx}} = \dfrac{{\sin \dfrac{{3\theta }}{2}}}{{\cos \dfrac{{3\theta }}{2}}}$
 Value of Sin / cos is equal to tan. So,
$\dfrac{{dy}}{{dx}} = \tan \dfrac{{3\theta }}{2}$
The value of $\dfrac{dy}{dx}$ is $\tan \dfrac{{3\theta }}{2}$
So, option (1) is the correct answer.

Note: To find $\dfrac{dy}{dx}$ of the equation which are not in the terms of x and y, we have to first differentiate the equation in whichever form they are (which in this case is $\theta $. So, we differentiate x and y with respect to $\theta $) and then divide them in such a way that they give us $\dfrac{dy}{dx}$. We also have to simplify the function using the trigonometric identities before differentiating them.