How do we derive the expansion for sin 3x?
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I think by expansion you mean sin 3x in terms of sin x.
We start with the relations sin (x + y) = sin x* cos y + cos x*sin y and cos (x + y) = cos x* cos y – sin x*sin y
sin 3x = sin (2x + x) = sin 2x* cos x + cos 2x*sin x
=> sin (x + x)* cos x + cos (x +x)*sin x
=> [sin x* cos x + cos x*sin x]* cos x + [cos x* cos x – sin x*sin x]*sin x
=> 2*sin x * (cos x) ^2 + (cos x) ^2 * sin x – (sin x) ^3
=> 3*sin x (cos x) ^2 – (sin x) ^3
now (cos x) ^2 + (sin x) ^2 = 1 or (cos x) ^2 = 1 – (sin x) ^2
=> 3* sin x*(1 – (sin x) ^2) - (sin x) ^3
=> 3* sin x – 3(sin x) ^2) - (sin x) ^3
=> 3*sin x – 4*(sin x) ^3
Therefore we get sin 3x = 3*sin x – 4*(sin x) ^3
sin(2x+x) = sin2x*cosx + sin x*cos 2x
sin 2x = 2sinx*cos x
cos 2x = 1 - 2(sin x)^2
sin 3x = 2sin x*cos x*sin x + sin x*[1-2(sin x)^2]
sin 3x = 2(cos x)^2*sinx + sin x - 2 (sin x)^3
From the fundamental formula of trigonometry, we'll have:
(cos x)^2 = 1 - (sin x)^2
We'll try to express sin 3x in terms of sin x:
sin 3x = 2[1 - (sin x)^2]*sin x + sin x - 2(sin x)^3
sin 3x = 2sin x - 2(sin x)^3 + sin x - 2(sin x)^3
We'll combine like terms:
sin 3x = 3sin x - 4(sin x)^3
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