How do we derive the expansion for sin 3x?

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

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

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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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