Question

x=cos2theta+2costheta, y=sin2theta-2sintheta find dy/ dx​

Answer 1

$x = cos2 \theta + 2cos \theta \\ \frac{dx}{d \theta} = - 2sin 2\theta - 2sin \theta \\ \\ y = sin2 \theta - 2sin \theta \\ \frac{dy}{d \theta} = 2cos2 \theta - 2cos \theta \\ \\ \frac{dy}{dx} = \frac{dy}{d \theta} \times \frac{d \theta}{dx} \\ \\ \frac{dy}{dx} = \frac{2cos2 \theta - 2cos \theta}{ - 2sin2 \theta - 2sin \theta} \\ \\ \frac{dy}{dx} = \frac{cos2 \theta - cos \theta}{ - sin2 \theta - sin \theta} \\ \\ \frac{dy}{dx} = \frac{ - 2sin( \frac{3 \theta}{2}) sin( \frac{ \theta}{2} )}{ - 2sin( \frac{3 \theta}{2})cos( \frac{ \theta}{2}) } \\ \\ \frac{dy}{dx} = tan \frac{ \theta}{2}$

Answer 2

▪Given :-

$\bf x = cos2\theta + 2cos\theta \\ \bf y = sin2\theta - 2sin\theta$

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▪To Calculate :-

$\bf\huge \green{ \frac{dy}{dx} }$

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▪Formulae Used :-

$\bf \maltese \: \: \: \frac{d}{dx} cos \: \theta = - sin\theta \\ \\ \bf \maltese \: \: \: \frac{d}{dx}sin \theta = cos\theta\\\\ \bf \maltese \: \: \:\small sin \: x + sin \: y = 2sin \bigg( \frac{x + y}{2} \bigg)cos \bigg(\frac{x - y}{2} \bigg) \\ \\ \bf \maltese \: \: \: \small cos \: x - cos \: y = - 2sin \bigg( \frac{x + y}{2} \bigg)sin \bigg( \frac{x - y}{2} \bigg)$

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▪Solution :-

We have ,

$\bf x = cos2\theta + 2cos\theta$

So,

$\sf \dfrac{dx}{d \theta} = \dfrac{d}{d \theta} \bigg( cos2\theta + 2cos\theta \bigg) \\ \\ = \sf - sin2\theta(2) - 2sin\theta \\ \\ \implies \purple{ \underline{ \boxed{ \bf \frac{dx}{d \theta} = - 2(sin2\theta + sin\theta)}}}$

Also,

$\bf y = sin2\theta - 2sin\theta$

So,

$\sf \dfrac{dy}{d \theta} = \dfrac{d}{d \theta} \bigg( sin2\theta - 2sin\theta \bigg) \\ \\ \sf = cos2\theta(2) + 2cos \theta \\ \\ \implies \purple{ \underline{ \boxed{ \bf \frac{dy}{d \theta} = 2(cos2\theta - cos\theta)}}}$

Now ,

$\bf \dfrac{dy}{dx} = \dfrac{dy/d \theta}{dx/d \theta} \\ \\ = \sf \frac{2(cos2 \theta - cos\theta)}{ - 2(sin2 \theta + sin\theta)} \\ \\ \sf = \dfrac{cos2 \theta - cos \theta }{ - (sin2 \theta + sin \theta)} \\ \\ \sf = \dfrac{ - 2sin \bigg( \dfrac{2 \theta + \theta}{2} \bigg)sin \bigg( \dfrac{2 \theta - \theta}{2} \bigg)}{ - \bigg \{2sin \bigg( \dfrac{2 \theta + \theta }{2} \bigg)cos\bigg( \dfrac{2 \theta - \theta}{2} \bigg) \bigg\}} \\ \\ \sf = \dfrac{ \cancel{sin \bigg( \dfrac{3 \theta}{2} \bigg)} \times sin \bigg( \dfrac{ \theta}{2} \bigg) }{ \cancel{sin \bigg( \dfrac{3 \theta}{2} \bigg)} \times cos \bigg( \dfrac{ \theta}{2} \bigg) } \\ \\ \sf = \dfrac{sin \dfrac{ \theta}{2} }{cos \dfrac{ \theta}{2} } \\ \\ \Large\implies \purple{ \underline{ \boxed{ \bf \frac{dy}{dx} = tan \dfrac{ \theta}{2} }}}$

$\Large \red{\mathfrak{  \text{W}hich \:   \: is  \:  \: the  \:  \: required} }\\ \huge \red{\mathfrak{ \text{ A}nswer.}}$

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