Question

prove that sin5x-2sin3x+sinx/
cos5x-cosx =tanx​

Answer 1

EXPLANATION.

⇒ sin5x - 2sin3x + sinx/cos5x - cosx = tanx.

As we know that,

Formula of :

⇒ sin(C) + sin(D) = 2sin(C + D)/2.cos(C - D)/2.

⇒ cos(C) - cos(D) = 2sin(C + D)/2.sin(D - C)/2.

Using this formula in equation, we get.

$\sf \implies \dfrac{sin5x + sinx - 2sin3x}{cos5x - cosx.}$

$\sf \implies \dfrac{2sin \bigg(\dfrac{5x + x}{2}\bigg).cos \bigg(\dfrac{5x - x}{2} \bigg) - 2sin3x }{2sin \bigg(\dfrac{5x + x}{2}\bigg).sin \bigg(\dfrac{x - 5x}{2} \bigg)}$

$\sf \implies \dfrac{2sin(3x).cos(2x) - 2sin3x}{2sin(3x) .sin(-2x)}$

$\sf \implies \dfrac{2sin3x [cos(2x) - 1]}{- 2sin3x .sin(2x)}$

$\sf \implies \dfrac{cos(2x) - 1}{-sin(2x)}$

As we know that,

Formula of :

⇒ cos2θ = 1 - 2sin²θ.

⇒ sin2θ = 2sinθ.cosθ.

Using this formula in equation, we get.

$\sf \implies \dfrac{[1 - 2sin^{2}x] - 1 }{-2sinx.cosx}$

$\sf \implies \dfrac{-2sin^{2} x}{-2sinx.cosx} = \dfrac{sinx}{cosx} = tanx$

                                                                                                                       

MORE INFORMATION.

Trigonometric ratios of multiple angles.

(1) = sin2θ = 2sinθ.cosθ = 2tanθ/1 + tan²θ.

(2) = cos2θ = cos²θ - sin²θ = 2cos²θ - 1 = 1 - 2sin²θ = 1 - tan²θ/1 + tan²θ.

(3) = tan2θ = 2tanθ/1 - tan²θ.

(4) = sin3θ = 3sinθ - 4cos³θ.

(5) = cos3θ = 4cos³θ - 3cosθ.

(6) = tan3θ = 3tanθ - tan³θ/1 - 3tan²θ.