Sin3x Formula in Terms of Sinx, Cosx [With Proof]

Sin3x is the trigonometric sine function of a triple angle 3x. The formula of sin3x in terms of sinx is given by sin3x = 3sinx – 4sin³x. That is, the formula of sine 3x can be expressed as 3 sine x – 4 sine cube x.

The sin3x identity in terms of sin x is given as follows:

sin3x = 3sinx – 4sin³x.

Sin3x Formula in Terms of sinx

The formula of sin3x is sin3x = 3sinx – 4sin3x.

Proof:

The following steps have to be followed in order to prove the above formula of sin3x in terms of sinx.

STEP 1:

First we write 3x=2x+x, and then apply the formula sin(a+b) = sina cosb + cosa sinb. Therefore, we deduce that

sin 3x = sin(2x+x)

⇒ sin3x = sin2x cosx + cos2x sinx.

STEP 2:

In the above, let us now apply the trigonometric formulae given below.

• sin2x = 2sinx cosx
• cos2x = 1-2sin²x

So we obtain that

sin3x = 2sinx cosx cosx + (1-2sin²x) sinx

⇒ sin3x = 2sinx cos²x + sinx – 2sin³x

⇒ sin3x = 2sinx (1- sin²x) + sinx – 2sin³x as we know that sin²x+cos²x=1.

⇒ sin3x = 2sinx – 2sin³x + sinx – 2sin³x

⇒ sin3x = 3sinx – 4sin³x.

CONCLUSION: So the sin3x formula in terms of sinx is given by sin3x = 3sinx – 4sin³x.

Sin3x Formula in Terms of cosx

The following formulas will be used in order to express sin3x in terms of cosx.

• sin2x = 2sinx cosx.
• cos2x = 2cos²x -1.

As sin3x = sin(2x+x), we have that

sin3x = sin2x cosx + cos2x sinx.

⇒ sin3x = 2sinx cosx cosx + cos2x sinx.

⇒ sin3x = 2sinx cos²x + (2cos²x -1) sinx

⇒ sin3x = sinx (2cos²x + 2cos²x -1)
⇒ sin3x = $\sqrt{1-\cos^2x}$ (4cos²x -1) as sin²x = 1-cos²x.

So the formula of sin3x in terms of cosx is given by sin3x = $\sqrt{1-\cos^2x}$ (4cos²x -1).

Question-Answer

Question 1: Find the value of sin135 degree.

Answer:

Applying the sin3x formula sin3x = 3sinx – 4sin³x with x=45°, we get that

sin135° = sin(3⋅45)°

⇒ sin135° = 3sin45° – 4sin³45°

⇒ sin135° = 3 × $(\dfrac{1}{\sqrt{2}})$ – 4 × $(\dfrac{1}{\sqrt{2}})^3$

⇒ sin135° = $\dfrac{3}{\sqrt{2}}$ – $\dfrac{4}{2\sqrt{2}}$ = $\dfrac{3}{\sqrt{2}}$ – $\dfrac{2}{\sqrt{2}}$

⇒ sin135° = $\dfrac{1}{\sqrt{2}}$.

So the value of sin135 degree is equal to 1/√2. That is, sin135° is equal to 1/√2 in fraction form.

Question 1: Find the value of sin270 degree.

Answer:

Apply sin3x formula sin3x = 3sinx – 4sin³x with x=90°. Therefore,

sin270° = sin(3⋅90)°

⇒ sin270° = 3sin90° – 4sin³90°

⇒ sin270° = 3sin90° – 4sin³90° as we know sin90° = 1.

⇒ sin270° = 3×1 – 4×1³

⇒ sin270° = 3-4 = -1.

So the value of sin270 degree is equal to -1.

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