Question

sin 5x - 2 sin 3x + sin x/cos 5x-cos x​

Answer 1

To evaluate :-

$\bf \:\dfrac{sin5x - 2sin3x + sinx}{cos5x - cosx}$

Identity used :-

$\bf \:❥︎ \: sinx + siny = 2sin(\dfrac{x + y}{2} )cos(\dfrac{x - y}{2} )$

$\bf \:❥︎ \: cosx - cosy = - 2sin(\dfrac{x + y}{2} )sin(\dfrac{x - y}{2} )$

$\bf \:❥︎ \: sin2x = 2sinx \: cosx$

$\bf \:❥︎ \: 1 - cosx = 2 {sin}^{2} \dfrac{x}{2}$

Solution :-

$\bf \:\dfrac{sin5x - 2sin3x + sinx}{cos5x - cosx}$

$\bf \:\dfrac{sin5x + sinx - 2sin3x}{cos5x - cosx}$

Using above (1) and (2) Identity, we get

$\bf\implies \:\dfrac{2sin(\dfrac{5x + x}{2})cos(\dfrac{5x - x}{2} ) + 2sin3x }{ - 2sin(\dfrac{5x + x}{2})sin(\dfrac{5x - x}{2} ) }$

$\bf\implies \:\dfrac{2sin3x \: cos2x - 2sin3x}{ - 2sin3x \: sin2x}$

$\bf\implies \:\dfrac{ - 2sin3x(1 - cos2x)}{ - 2sin3x \: sin2x}$

$\bf\implies \:\dfrac{1 - cos2x}{sin2x}$

$\bf\implies \:\dfrac{2 {sin}^{2}x }{2sinx \: cosx}$

$\bf\implies \:\dfrac{sinx}{cosx}$

$\bf\implies \:tanx$

$\bf \:\implies \:\:\dfrac{sin5x - 2sin3x + sinx}{cos5x - cosx} = tanx$