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Question: Find the value of \(\dfrac{dy}{dx}\), if \(x = 2 \cos \theta - \cos 2\theta \) and \(y = 2 \sin \the...

Find the value of dydx\dfrac{dy}{dx}, if x=2cos⁡θ−cos⁡2θx = 2 \cos \theta - \cos 2\theta and y=2sin⁡θ−sin⁡2θy = 2 \sin \theta - \sin 2\theta .
1). tan⁡3θ2\tan \dfrac{{3\theta }}{2}
2). −tan⁡3θ2 - \tan \dfrac{{3\theta }}{2}
3). cot⁡3θ2\cot \dfrac{{3\theta }}{2}
4). −cot⁡3θ2 - \cot \dfrac{{3\theta }}{2}

Explanation

Solution

We will differentiate x and y separately with respect to θ\theta and after that we will divide them in such a way that we get the value of dydx\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⁡θ−cos⁡2θx = \cos \theta - \cos 2\theta
y=sin⁡θ−sin⁡2θy = \sin \theta - \sin 2\theta
We will differentiate x and y separately with respect to θ\theta because they both are in terms of θ\theta . So, we cannot directly differentiate them with each other.
So, first We will differentiate the equation x with respect to θ\theta
x=cos⁡θ−cos⁡2θx = \cos \theta - \cos 2\theta
⇒dxdθ=−2sin⁡θ−(−sin⁡2θ)×2\Rightarrow \dfrac{{dx}}{{d\theta }} = - 2\sin \theta - ( - \sin 2\theta ) \times 2
⇒dxdθ=−2sin⁡θ+2sin⁡2θ\Rightarrow \dfrac{{dx}}{{d\theta }} = - 2\sin \theta + 2\sin 2\theta
Now, we will differentiate the equation y with respect to θ\theta .
y=2sin⁡θ−sin⁡2θy = 2\sin \theta - \sin 2\theta
dydθ=2cos⁡θ−2cos⁡2θ\dfrac{{dy}}{{d\theta }} = 2\cos \theta - 2\cos 2\theta
So, now we will divide them in order to find dydx\dfrac{dy}{dx}.
dydx=2cos⁡θ−2cos⁡2θ−2sin⁡θ+2sin⁡2θ\dfrac{{dy}}{{dx}} = \dfrac{{2\cos \theta - 2\cos 2\theta }}{{ - 2\sin \theta + 2\sin 2\theta }}
⇒dydx=cos⁡θ−cos⁡2θ−sin⁡θ+sin⁡2θ\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\cos \theta - \cos 2\theta }}{{ - \sin \theta + \sin 2\theta }}
Now, by using trigonometric formula cos⁡A−cos⁡B=2sin⁡(A+B2)sin⁡(A−B2)\cos A - \cos B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right) and sin⁡A−sin⁡B=2cos⁡(A+B2)sin⁡(A−B2)\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.
dydx=2sin⁡(θ+2θ2)sin⁡(2θ−θ2)2cos⁡(θ+2θ2)sin⁡(2θ−θ2)\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,
dydx=sin⁡3θ2cos⁡3θ2\dfrac{{dy}}{{dx}} = \dfrac{{\sin \dfrac{{3\theta }}{2}}}{{\cos \dfrac{{3\theta }}{2}}}
Value of Sin / cos is equal to tan. So,
dydx=tan⁡3θ2\dfrac{{dy}}{{dx}} = \tan \dfrac{{3\theta }}{2}
The value of dydx\dfrac{dy}{dx} is tan⁡3θ2\tan \dfrac{{3\theta }}{2}
So, option (1) is the correct answer.

Note: To find dydx\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 θ\theta . So, we differentiate x and y with respect to θ\theta ) and then divide them in such a way that they give us dydx\dfrac{dy}{dx}. We also have to simplify the function using the trigonometric identities before differentiating them.