Question

Find $\dfrac{dy}{dx}$ , if $x=2\cos \theta -\cos 2\theta $ and $y=2\sin \theta -\sin 2\theta $ .
(a) $\tan \dfrac{3\theta }{2}$
(b) $-\tan \dfrac{3\theta }{2}$
(c) $\cot \dfrac{3\theta }{2}$
(d) $-\cot \dfrac{3\theta }{2}$

Answer 1

Hint: First, we will find derivation of x with respect to $\theta $ . Also, y with respect to $\theta $ . By doing this, we will get $\dfrac{dx}{d\theta },\dfrac{dy}{d\theta }$ . Then we will divide $\dfrac{dy}{d\theta }$ by $\dfrac{dx}{d\theta }$ and on solving we will get the required answer. Also, formula needed to solve the equation as $\cos a-\cos b=2\sin \left( \dfrac{a+b}{2} \right)\sin \left( \dfrac{a-b}{2} \right)$ , $\sin a-\sin b=2\cos \left( \dfrac{a+b}{2} \right)\sin \left( \dfrac{a-b}{2} \right)$ , and $\dfrac{\sin \theta }{\cos \theta }=\tan \theta $ .

Complete step-by-step answer:
Here, we are given two functions i.e. $x=2\cos \theta -\cos 2\theta $ and $y=2\sin \theta -\sin 2\theta $ . So, first we will find derivative of x function. If suppose we have function let’s say $x=\sin 3\theta $ , then here there are two variables i.e. x and $\theta $ so, we will find $\dfrac{dx}{d\theta }$ . So, we will first use derivative of sine function which is cos function and then derivative of $3\theta $ which is equal to 3. Thus, using the same concept we will find first $\dfrac{dx}{d\theta },\dfrac{dy}{d\theta }$ .
We will find $\dfrac{dx}{d\theta }$ . We will get as
$\dfrac{dx}{d\theta }=2\dfrac{d}{d\theta }\left( \cos \theta \right)-\dfrac{d}{d\theta }\left( \cos 2\theta \right)$
Now, we know that derivative of cos function is $-\sin $ function. So, we will use the chain rule, given by derivative of function f(g(x))=f’(x).g’(x) and we get as
$\dfrac{dx}{d\theta }=-2\sin \theta -\left( -\sin 2\theta \right)\cdot \dfrac{d}{d\theta }\left( 2\theta \right)$
On further solving, we get as
$\dfrac{dx}{d\theta }=-2\sin \theta +2\sin 2\theta $ …………………………….(1)
Similarly, we will find $\dfrac{dy}{d\theta }$ .
$\dfrac{dy}{d\theta }=2\dfrac{d}{d\theta }\left( \sin \theta \right)-\dfrac{d}{d\theta }\left( \sin 2\theta \right)$
Now, we know that derivative of sin function is cos function. So, we can write it as
$\dfrac{dy}{d\theta }=2\cos \theta -\cos 2\theta \cdot \dfrac{d}{d\theta }\left( 2\theta \right)$
On further solving, we get as
$\dfrac{dy}{d\theta }=2\cos -2\cos 2\theta $ …………………..(2)
 Now, we have to find $\dfrac{dy}{dx}$ . So, dividing equation (2) by (1) we get as
$\dfrac{\dfrac{dy}{d\theta }}{\dfrac{dx}{d\theta }}=\dfrac{2\cos \theta -2\cos 2\theta }{-2\sin \theta +2\sin 2\theta }$
Now, we will take 2 common and cancelling it and on rearranging the terms, we get as
$\dfrac{dy}{dx}=\dfrac{\cos \theta -\cos 2\theta }{\sin 2\theta -\sin \theta }$
We can see that in the numerator there is a cos minus cos term. So, to solve this there formula as $2\sin a\sin b=\cos \left( a-b \right)-\cos \left( a+b \right)$ where a is $2\theta $ and b is $\theta $ . This formula can also be written as $\cos a-\cos b=2\sin \left( \dfrac{a+b}{2} \right)\sin \left( \dfrac{a-b}{2} \right)$ . Similarly, in denominator we will use $\sin a-\sin b=2\cos \left( \dfrac{a+b}{2} \right)\sin \left( \dfrac{a-b}{2} \right)$ .
On putting the values, we get equation as
$\dfrac{dy}{dx}=\dfrac{2\sin \left( \dfrac{2\theta +\theta }{2} \right)\sin \left( \dfrac{2\theta -\theta }{2} \right)}{2\cos \left( \dfrac{2\theta +\theta }{2} \right)\sin \left( \dfrac{2\theta -\theta }{2} \right)}$
On solving, we can see that $\sin \left( \dfrac{2\theta -\theta }{2} \right)$ is common in numerator and denominator. So, on cancelling we will get as
$\dfrac{dy}{dx}=\dfrac{2\sin \left( \dfrac{2\theta +\theta }{2} \right)}{2\cos \left( \dfrac{2\theta +\theta }{2} \right)}$
$\dfrac{dy}{dx}=\dfrac{2\sin \left( \dfrac{3\theta }{2} \right)}{2\cos \left( \dfrac{3\theta }{2} \right)}$
We know that $\dfrac{\sin \theta }{\cos \theta }=\tan \theta $ and on cancelling constant number 2, we will get
$\dfrac{dy}{dx}=\tan \left( \dfrac{3\theta }{2} \right)$
Hence, option (a) is the correct answer.

Note:Remember that while using the formula $\cos a-\cos b=2\sin \left( \dfrac{a+b}{2} \right)\sin \left( \dfrac{a-b}{2} \right)$ we can take a as $\theta $ and b as $2\theta $ instead of taking vice versa. By this we will get equation as $\dfrac{dy}{dx}=\dfrac{2\sin \left( \dfrac{\theta +2\theta }{2} \right)\sin \left( \dfrac{\theta -2\theta }{2} \right)}{2\cos \left( \dfrac{\theta +2\theta }{2} \right)\sin \left( \dfrac{\theta -2\theta }{2} \right)}$ . On solving this, we know that $\left( \sin \left( -\theta \right) \right)=-\sin \theta ,\cos \left( -\theta \right)=\cos \theta $ . So, at last on substituting proper signs and solving we will get the same answer.