Question

$\underline{\underline{ \bf Question}}$
Prove that :-
$\boxed{ \bf \dfrac { \cos\theta+ \cos2\theta+ \cos3 \theta + cos 4\theta}{\sin\theta+ \sin2\theta+ \sin3 \theta + sin4\theta} = \cot\dfrac{5}{2} \theta}$
Topic : Compound and Multiple Angles ( 11 th)
Relevant Answers Needed !​

Answer 1

$\large\underline{\sf{Solution-}}$

Consider LHS

$\rm \: \dfrac { \cos\theta+ \cos2\theta+ \cos3 \theta + cos 4\theta}{\sin\theta+ \sin2\theta+ \sin3 \theta + sin4\theta}$

can be re-arranged as

$\rm \: = \: \dfrac { (\cos4\theta+ \cos\theta)+( \cos3 \theta + cos 2\theta)}{(\sin4\theta+ \sin\theta)+ (\sin3 \theta + sin2\theta)}$

We know,

$\boxed{\tt{ cosx + cosy = 2cos\bigg[\dfrac{x + y}{2} \bigg]cos\bigg[\dfrac{x - y}{2} \bigg] \: }}$

and

$\boxed{\tt{sinx + siny = 2sin\bigg[\dfrac{x + y}{2} \bigg]cos\bigg[\dfrac{x - y}{2} \bigg] \: }}$

So, using these Identities, we get

$\rm \: = \: \dfrac{2cos\bigg[\dfrac{4 \theta + \theta }{2} \bigg]cos\bigg[\dfrac{4 \theta - \theta }{2} \bigg] + 2cos\bigg[\dfrac{3 \theta + 2 \theta }{2} \bigg]cos\bigg[\dfrac{3 \theta - 2 \theta }{2} \bigg]}{2sin\bigg[\dfrac{4 \theta + \theta }{2} \bigg]cos\bigg[\dfrac{4 \theta - \theta }{2} \bigg] + 2sin\bigg[\dfrac{3 \theta + 2 \theta }{2} \bigg]cos\bigg[\dfrac{3 \theta - 2 \theta }{2} \bigg]}$

$\rm \: = \: \dfrac{2cos\bigg[\dfrac{5 \theta }{2} \bigg]cos\bigg[\dfrac{3 \theta }{2} \bigg] + 2cos\bigg[\dfrac{5 \theta }{2} \bigg]cos\bigg[\dfrac{ \theta }{2} \bigg]}{2sin\bigg[\dfrac{5 \theta }{2} \bigg]cos\bigg[\dfrac{3 \theta }{2} \bigg] + 2sin\bigg[\dfrac{5 \theta }{2} \bigg]cos\bigg[\dfrac{ \theta }{2} \bigg]}$

$\rm \: = \: \dfrac{ \cancel2 \: cos\bigg[\dfrac{5 \theta }{2} \bigg] \cancel{\bigg(cos\bigg[\dfrac{3 \theta }{2} \bigg] + cos\bigg[\dfrac{ \theta }{2} \bigg]\bigg)} \: }{ \cancel2 \: sin\bigg[\dfrac{5 \theta }{2} \bigg] \cancel{\bigg(cos\bigg[\dfrac{3 \theta }{2} \bigg] + cos\bigg[\dfrac{ \theta }{2} \bigg]\bigg)}}$

$\rm \: = \: \dfrac{cos\bigg[\dfrac{5 \theta }{2} \bigg]}{sin\bigg[\dfrac{5 \theta }{2} \bigg]}$

$\rm \: = \: cot\bigg(\dfrac{5 \theta }{2} \bigg)$

Hence,

$\boxed{\tt{ \dfrac { \cos\theta+ \cos2\theta+ \cos3 \theta + cos 4\theta}{\sin\theta+ \sin2\theta+ \sin3 \theta + sin4\theta} = \cot\dfrac{5 \theta }{2}}}$

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ADDITIONAL INFORMATION

$\boxed{\tt{ cosx - cosy = - 2sin\bigg[\dfrac{x + y}{2} \bigg]sin\bigg[\dfrac{x - y}{2} \bigg] \: }}$

$\boxed{\tt{sinx - siny = 2cos\bigg[\dfrac{x + y}{2} \bigg]sin\bigg[\dfrac{x - y}{2} \bigg] \: }}$

$\boxed{\tt{ 2sinx \: cosy \: = \: sin(x + y) + sin(x - y) \: }}$

$\boxed{\tt{ 2cosx \: cosy \: = \: cos(x + y) + cos(x - y) \: }}$

$\boxed{\tt{ 2sinx \: siny \: = \: cos(x - y) - cos(x + y) \: }}$

Answer 2

Answer:

Question :-

$\mapsto \bf \dfrac{\cos\theta + \cos2\theta + \cos3\theta + \cos4\theta}{\sin\theta + \sin2\theta + \sin3\theta + \sin4\theta} =\: cot\bigg[\dfrac{5\theta}{2}\bigg]$

Solution :-

$\bigstar\: \: \sf\bold{\purple{L.H.S =\: \dfrac{cos\theta + cos2\theta + cos3\theta + cos4\theta}{sin\theta + sin2\theta + sin3\theta + sin4\theta}}}$

We can also write this :

$\implies \sf \dfrac{(cos 4\theta + cos\theta) + (cos 3\theta + cos 2\theta)}{(sin 4\theta + sin\theta) + (sin 3\theta + sin 2\theta)}$

As we know that :

$\clubsuit\: \: \sf\bold{\pink{cosA + cosB =\: 2cos\bigg\lgroup \dfrac{A + B}{2}\bigg\rgroup cos\bigg\lgroup \dfrac{A - B}{2}\bigg\rgroup}}$

$\clubsuit\: \: \sf\bold{\pink{sinA + sinB =\: 2sin\bigg\lgroup \dfrac{A + B}{2}\bigg\rgroup cos\bigg\lgroup \dfrac{A - B}{2}\bigg\rgroup}}$

So, according to the formula we get :

$\implies \sf \dfrac{2 cos\bigg(\dfrac{4\theta + \theta}{2}\bigg) cos\bigg(\dfrac{4\theta - \theta}{2}\bigg) + 2cos\bigg(\dfrac{3\theta + 2\theta}{2}\bigg) cos\bigg(\dfrac{3\theta - 2\theta}{2}\bigg)}{2sin\bigg(\dfrac{4\theta + \theta}{2}\bigg) cos\bigg(\dfrac{4\theta - \theta}{2}\bigg) + 2sin\bigg(\dfrac{3\theta + 2\theta}{2}\bigg) cos\bigg(\dfrac{3\theta - 2\theta}{2}\bigg)}$

$\implies \sf \dfrac{2 cos\bigg(\dfrac{5\theta}{2}\bigg) cos\bigg(\dfrac{3\theta}{2}\bigg) + 2cos\bigg(\dfrac{5\theta}{2}\bigg) cos\bigg(\dfrac{\theta}{2}\bigg)}{2 sin\bigg(\dfrac{5\theta}{2}\bigg) cos\bigg(\dfrac{3\theta}{2}\bigg) + 2sin\bigg(\dfrac{5\theta}{2}\bigg) cos\bigg(\dfrac{\theta}{2}\bigg)}$

By taking common we get,

$\implies \sf \dfrac{2 cos\bigg(\dfrac{5\theta}{2}\bigg) \bigg\{ \cancel{cos\bigg(\dfrac{3\theta}{2}\bigg) + cos\bigg(\dfrac{\theta}{2}\bigg)} \bigg\} }{2 sin\bigg(\dfrac{5\theta}{2}\bigg) \bigg\{ \cancel{cos\bigg(\dfrac{3\theta}{2}\bigg) + cos\bigg(\dfrac{\theta}{2}\bigg)} \bigg\} }$

$\implies \sf \dfrac{\cancel{2} cos\bigg(\dfrac{5\theta}{2}\bigg)}{\cancel{2} sin \bigg(\dfrac{5\theta}{2}\bigg)}$

$\implies \sf \dfrac{cos \bigg(\dfrac{5\theta}{2}\bigg)}{sin \bigg(\dfrac{5\theta}{2}\bigg)}$

As we know that :

$\clubsuit$ $\sf\bold{\pink{\dfrac{cos\theta}{sin\theta} =\: cot\theta}}$

So, by using this identities we get,

$\implies \sf\bold{\red{cot \bigg[\dfrac{5\theta}{2}\bigg]}}$

Again,

$\bigstar\: \: \sf\bold{\purple{R.H.S =\: cot\bigg[\dfrac{5\theta}{2}\bigg]}}$

$\implies \sf\bold{\red{cot\bigg[\dfrac{5\theta}{2}\bigg]}}$

Hence,

$\dashrightarrow \sf\bold{L.H.S =\: R.H.S}$

$\sf\boxed{\bold{\green{HENCE\: PROVED}}}$

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