Question

Prove that
$\sf{sinx\:=\:2^n\:sin\left(\dfrac{x}{2^n}\right)\:cos\left(\dfrac{x}{2}\right) \:cos\left(\dfrac{x}{2^2}\right) \cdots\cdots cos\left(\dfrac{x}{2^n}\right)$

Answer 1

Given :   Sinx  = 2ⁿ sin(x/2ⁿ)Cos(x/2)Cos(x/2²) .....................Cos(x/2ⁿ)

To find : prove the given identity

Solution:

Sinx  = 2ⁿ sin(x/2ⁿ)Cos(x/2)Cos(x/2²) .....................Cos(x/2ⁿ)

LHS = Sinx

using Sin2θ = 2SinθCosθ

= 2Sin(x/2)Cos(x/2)

= 2* 2 Sin(x/2²)Cos(x/2²)Cos(x/2)

= 2² Sin(x/2²)Cos(x/2).Cos(x/2²)

= 2² (2 Sin(x/2³)Cos(x/2³) Cos(x/2).Cos(x/2²)

= 2³Sin(x/2³)Cos(x/2).Cos(x/2²)Cos(x/2³)

Similarly this way  

=  2ⁿ sin(x/2ⁿ)Cos(x/2)Cos(x/2²) .....................Cos(x/2ⁿ)

QED

or taking form RHS to LHS

RHS = 2ⁿ sin(x/2ⁿ)Cos(x/2)Cos(x/2²) .....................Cos(x/2ⁿ)

= 2 * 2ⁿ⁻¹  sin(x/2ⁿ)Cos(x/2)Cos(x/2²) .....................Cos(x/2ⁿ)

= 2ⁿ⁻¹ (  2  sin(x/2ⁿ) Cos(x/2ⁿ)  ( Cos(x/2)Cos(x/2²) .....................Cos(x/2ⁿ⁻¹) )

= 2ⁿ⁻¹ Sin(x/2ⁿ⁻¹)  ( Cos(x/2)Cos(x/2²) .....................Cos(x/2ⁿ⁻¹) )

= 2ⁿ⁻² ( 2Sin(x/2ⁿ⁻¹)Cos(x/2ⁿ⁻¹) ) ( Cos(x/2)Cos(x/2²) .....................Cos(x/2ⁿ⁻²) )

= 2ⁿ⁻² Sin(x/2ⁿ⁻²)  ( Cos(x/2)Cos(x/2²) .....................Cos(x/2ⁿ⁻²))

and so on

= 2ⁿ-⁽ⁿ⁻¹⁾ Sin(x/2ⁿ⁻⁽⁻¹⁾) Cos(x/2)

= 2Sin(x/2)Cos(x/2)

= Sinx

QED

Hence Proved

Sinx  = 2ⁿ sin(x/2ⁿ)Cos(x/2)Cos(x/2²) .....................Cos(x/2ⁿ)

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Answer 2

Answer:

hope it is helpful to you