Question

If $x = a\sin 2\theta (1 + \cos 2\theta ),\,y = b\cos 2\theta (1 - \cos 2\theta )$ , then $\dfrac{{dy}}{{dx}} = $
A. $\dfrac{{b\tan \theta }}{a}$
B. $\dfrac{{a\tan \theta }}{b}$
C. $\dfrac{a}{{b\tan \theta }}$
D. $\dfrac{b}{{a\tan \theta }}$

Answer 1

Hint: The two variables $x,y$ are in terms of two trigonometric functions $\sin \theta ,\,\cos \theta $ so we will first simplify the right-hand side of the two variables by using trigonometric identities. Then we will differentiate each variable individually with respect to $\theta $ and then we will finally get the value of $\dfrac{{dy}}{{dx}}$ .

Complete step-by-step answer:
We are given,
$
  x = a\sin 2\theta (1 + \cos 2\theta )\,\,\,\,\,\,...(1) \\
  y = b\cos 2\theta (1 - \cos 2\theta )\,\,\,\,\,...(2) \\
 $
Now, we know the trigonometric identities $\sin 2\theta = 2\sin \theta \cos \theta ,\,\cos 2\theta = 2{\cos ^2}\theta - 1$
Putting these values in $(1)$ , we get:
$
  x = a2\sin \theta \cos \theta (1 + (2{\cos ^2}\theta - 1)) \\
   \Rightarrow x = 2a\sin \theta \cos \theta (2{\cos ^2}\theta ) \\
   \Rightarrow x = 4a\sin \theta {\cos ^3}\theta \\
 $
Now differentiating both sides with respect to $\theta $ :
$
  \dfrac{{dx}}{{d\theta }} = \dfrac{{d(4a\sin \theta {{\cos }^3}\theta )}}{{d\theta }} \\
   \Rightarrow \dfrac{{dx}}{{d\theta }} = 4a[\sin \theta \dfrac{{d{{\cos }^3}\theta }}{d\theta } + {\cos ^3}\theta \dfrac{{d\sin \theta }}{{d\theta }}]\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\because \dfrac{{du(x)v(x)}}{{dx}} = u(x)\dfrac{{dv(x)}}{{dx}} + v(x)\dfrac{{du(x)}}{{dx}} \\
   \Rightarrow \dfrac{{dx}}{{d\theta }} = 4a[\sin \theta ( - 3{\cos ^2}\theta \sin \theta ) + {\cos ^3}\theta \cos \theta ]\,\,\,\,\,\,\,\,\,\,\because \dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}},\dfrac{{d\sin x}}{{dx}} = \cos x,\dfrac{{d\cos x}}{{dx}} = - \sin x \\
   \Rightarrow \dfrac{{dx}}{{d\theta }} = 4a{\cos ^2}\theta ({\cos ^2}\theta - 3{\sin ^2}\theta )\,\,\,\,...(3) \\
 $
We also know the trigonometric identities $\sin 2\theta = 2\sin \theta \cos \theta ,\,\cos 2\theta = 1 - 2{\sin ^2}\theta $
Putting these values in $(2)$ , we get –
$
  y = b(1 - 2{\sin ^2}\theta )(1 - (1 - 2{\sin ^2}\theta )) \\
   \Rightarrow y = b(1 - 2{\sin ^2}\theta )(2{\sin ^2}\theta ) \\
   \Rightarrow y = 2b({\sin ^2}\theta - 2{\sin ^4}\theta ) \\
 $
Differentiating both sides with respect to $\theta $:
\[
  \dfrac{{dy}}{{d\theta }} = \dfrac{{d2b({{\sin }^2}\theta - 2{{\sin }^4}\theta )}}{{d\theta }} \\
   \Rightarrow \dfrac{{dy}}{{d\theta }} = 2b(\dfrac{{d{{\sin }^2}\theta }}{{d\theta }} - 2\dfrac{{d{{\sin }^4}\theta }}{{d\theta }}) \\
   \Rightarrow \dfrac{{dy}}{{d\theta }} = 2b[2\sin \theta \cos \theta - 2(4{\sin ^3}\theta \cos \theta )]\,\,\,\,\,\,\,\because \dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}},\dfrac{{d\sin x}}{{dx}} = \cos x \\
   \Rightarrow \dfrac{{dy}}{{d\theta }} = 4b\sin \theta \cos \theta (1 - 4{\sin ^2}\theta )\,\,\,\,\,...(4) \\
 \]
Now, to find $\dfrac{{dy}}{{dx}}$ , divide $(4)$ by $(3)$ :
$
  \dfrac{{\dfrac{{dy}}{{d\theta }}}}{{\dfrac{{dx}}{{d\theta }}}} = \dfrac{{4b\sin \theta \cos \theta (1 - 4{{\sin }^2}\theta )}}{{4a{{\cos }^2}\theta ({{\cos }^2}\theta - 3{{\sin }^2}\theta )}} \\
   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{b\sin \theta \cos \theta (1 - 4{{\sin }^2}\theta )}}{{a{{\cos }^2}\theta (1 - {{\sin }^2}\theta - 3{{\sin }^2}\theta )}}\,\,\,\,\,\because {\cos ^2}\theta + {\sin ^2}\theta = 1 \\
   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{b\sin \theta (1 - 4{{\sin }^2}\theta )}}{{a\cos \theta (1 - 4{{\sin }^2}\theta )}} \\
   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{b\tan \theta }}{a}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\because \dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta \\
 $

The correct option is option A

Note: We have two values of $\cos 2\theta $ in this solution. One is $\cos 2\theta = 2{\cos ^2}\theta - 1$ and the other is $1 - 2{\sin ^2}\theta $ , they both are derived from $\cos 2\theta = {\cos ^2} - {\sin ^2}\theta $ . They both are correct and we can use any one of them according to our requirements. Here, we have used both forms according to their relevance in simplifying the function.