Question

#Quality Question
@Differentiation
If ,$x =  cos2\theta + 2cos\theta \\ y = sin2\theta - 2sin\theta$
the calculate dy/dx​

Answer 1

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▪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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Answer 2

Answer:

Refer the attachment ..