Question

(cos x + cos 3x + cos 5x + cos 7x)/(sinx + sin 3x + sin 5x + sin 7x)=cot 4x

Answer 1

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

Consider LHS

$\rm \: \frac{cosx + cos3x + cos5x + cos7x}{sinx + sin3x + sin5x + sin7x}$

can be re-arranged as

$\rm \: =  \: \dfrac{(cos7x + cosx) + (cos5x + cos3x)}{(sin7x + sinx) + (sin5x + sin3x)}$

We know,

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

and

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

So, on using these results, we get

$\rm \: =  \: \dfrac{2cos\bigg[\dfrac{7x + x}{2} \bigg]cos\bigg[\dfrac{7x - x}{2} \bigg] \: + \: 2cos\bigg[\dfrac{5x + 3x}{2} \bigg]cos\bigg[\dfrac{5x - 3x}{2} \bigg]}{2sin\bigg[\dfrac{7x + x}{2} \bigg]cos\bigg[\dfrac{7x - x}{2} \bigg] \: + \: 2sin\bigg[\dfrac{5x + 3x}{2} \bigg]cos\bigg[\dfrac{5x - 3x}{2} \bigg]}$

$\rm \: =  \: \dfrac{2cos\bigg[\dfrac{8x}{2} \bigg]cos\bigg[\dfrac{6x}{2} \bigg] \: + \: 2cos\bigg[\dfrac{8x}{2} \bigg]cos\bigg[\dfrac{2x}{2} \bigg]}{2sin\bigg[\dfrac{8x}{2} \bigg]cos\bigg[\dfrac{6x}{2} \bigg] \: + \: 2sin\bigg[\dfrac{8x}{2} \bigg]cos\bigg[\dfrac{2x}{2} \bigg]}$

$\rm \: =  \: \dfrac{2cos4xcos3x + 2cos4xcosx}{2sin4xcos3x + 2sin4xcosx}$

$\rm \: =  \: \dfrac{2cos4x(cos3x + cosx)}{2sin4x(cos3x +cosx)}$

$\rm \: =  \: \frac{cos4x}{sin4x}$

$\rm \: =  \: cot4x$

Hence,

$\rm\implies\boxed{\sf{\rm \: \frac{cosx + cos3x + cos5x + cos7x}{sinx + sin3x + sin5x + sin7x} = cot4x \: }}$

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Additional Information :-

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

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

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

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

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