**What is the formula of sin3x**

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Sin3x formula derivation : Sin3x formula in terms of sinx expansion means its a triple angle identity of Sine Trigonometric function which can be written as sin^{3} x. In words we can say that Sin3x is the whole cube of the sine function which is equal to sin^{3} x. This is a important formula because this sin3x formula can be used to solve various trigonometry questions or problems.

In this article, we will discuss the Sin3x formula in terms of sinx and what is the concepts of sin3x and sin^3x integration. We will understand this sin3x formula graph and application by solving some good questions, so we can understand the concept of this Sin3x ka formula.

**What is Sin3x Trigonometry?**

We know that Sin3x is a very important trigonometric identity. Its expansion can be derivate in terms of sin x. You must know that the graph of sin3x formula or sin3x function is very much similar to that of sin x trigonometric function. We know that the period of sin x is 2π ( it means that sinx function complete its one cycle after every 2π radians). The period of sin3x is 2π/3 (what does it mean ? it make any sense) it means that sin3x function complete its one cycle after every 2π/3 radians.

**Proof of Sin3x Formula**

The Sin3x formula in terms of sinx is given by, sin3x = 3 sinx – 4 sin^{3}x. This sin3x formula can be written as sine three x (sin3x) is equal to the 3 sin x – 4 sin3x. Now, we will derive the sin3x trigonometric formula using the angle addition trigonometric identity.

To drive the sin3x formula in terms of sinx trigonometric function, we will use the angle addition trigonometry formula of the sine function. Now we can rewrite the angle 3x as 3x = 2x + x. After this step we will use some trigonometric identities to prove the sin3x formula :

- sin (a + b) = sin a cos b + cos a sin b
- sin 2x = 2 sin x cos x
- cos 2x = 1 – 2sin
^{2}x - sin
^{2}x + cos^{2}x = 1

- sin3x = sin (2x + x)
- = sin2x cosx + cos2x sinx [Because sin (a + b) = sin a cos b + cos a sin b]
- = (2 sin x cos x) cos x + (1 – 2sin
^{2}x) sin x - = 2cos
^{2}x sin x – 2sin^{3}x + sin x - = 2 (1 – sin
^{2}x) sin x – 2sin^{3}x + sin x [Because sin^{2}x + cos^{2}x = 1 ⇒ cos^{2}x = 1 – sin^{2}x] - = 2 sin x – 2sin
^{3}x – 2sin^{3}x + sin x - = 2 sin x + sin x – 2sin
^{3}x – 2sin^{3}x - = 3 sin x – 4 sin
^{3}x - = 3 sinx – 4 sin^3x

So Sin3x is equal to 3 sinx – 4 sin^3x = 3 sinx – 4 sin 3x = 3 sinx – 4 sin^{3}x. Please follow this link of Explanation of Mathematics formula – **Click**

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