Sin3x : Formula, Derivation, Proof, and Examples

#Trigonometry
TL;DR
Sin3x equals 3sin⁡x−4sin⁡3x3\sin x - 4\sin^3 x, the triple-angle identity that writes the sine of a tripled angle in terms of the sine of the single angle alone. This article covers the formula, its step-by-step derivation, the rearranged sin⁡3x\sin^3 x form, six worked examples, the mistakes students make, and where the triple-angle identity is used.
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Madhukar VLast updated on August 31, 20268 min read

The Identity That Folds Three Angles Into One

There is something quietly surprising about sin3x: the sine of an angle three times as large can be written using nothing but the sine of the original angle. No cosine, no extra terms, just sin⁡x\sin x raised and scaled. That compactness is why the triple-angle identity keeps appearing in equation-solving, integration, and signal work, and why it is worth knowing cold.

The formula is:

sin⁡3x=3sin⁡x−4sin⁡3x\sin 3x = 3\sin x - 4\sin^3 x

Read it carefully: the sine of the tripled angle is three times sin⁡x\sin x, minus four times sin⁡x\sin x cubed. The cube is what makes this a triple-angle identity rather than a simple multiple, and it is the piece most first attempts leave out.

What Is The Sin3x Formula?

The sin3x formula is the triple-angle identity:

sin⁡3x=3sin⁡x−4sin⁡3x\sin 3x = 3\sin x - 4\sin^3 x

Every symbol is plain:

  • xx is any angle, in degrees or radians.

  • 3x3x is that angle tripled.

  • sin⁡3x\sin^3 x means (sin⁡x)3(\sin x)^3, the sine of xx cubed, not the sine of x3x^3.

The identity is often rearranged to isolate the cube, which is how the "power-reduction" form appears:

sin⁡3x=3sin⁡x−sin⁡3x4\sin^3 x = \frac{3\sin x - \sin 3x}{4}

This second form is the one calculus uses to integrate sin⁡3x\sin^3 x, since a sum of plain sines is far easier to integrate than a cube. Both belong to the wider set of trigonometric identities, and sin3x sits beside its partners cos3x and tan3x in the triple-angle family.

How Do You Derive The Sin3x Formula?

The derivation writes 3x3x as 2x+x2x + x and then leans on the double-angle identities. Start from the sine angle-sum identity:

sin⁡3x=sin⁡(2x+x)=sin⁡2xcos⁡x+cos⁡2xsin⁡x\sin 3x = \sin(2x + x) = \sin 2x \cos x + \cos 2x \sin x

Now substitute the sin double angle formula sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos x and the cos 2x identity cos⁡2x=1−2sin⁡2x\cos 2x = 1 - 2\sin^2 x:

sin⁡3x=(2sin⁡xcos⁡x)cos⁡x+(1−2sin⁡2x)sin⁡x\sin 3x = (2\sin x \cos x)\cos x + (1 - 2\sin^2 x)\sin x

=2sin⁡xcos⁡2x+sin⁡x−2sin⁡3x= 2\sin x \cos^2 x + \sin x - 2\sin^3 x

Replace cos⁡2x\cos^2 x with 1−sin⁡2x1 - \sin^2 x, since that is how the cosine is expressed purely in sine:

=2sin⁡x(1−sin⁡2x)+sin⁡x−2sin⁡3x= 2\sin x(1 - \sin^2 x) + \sin x - 2\sin^3 x

=2sin⁡x−2sin⁡3x+sin⁡x−2sin⁡3x= 2\sin x - 2\sin^3 x + \sin x - 2\sin^3 x

Collect like terms:

sin⁡3x=3sin⁡x−4sin⁡3x\sin 3x = 3\sin x - 4\sin^3 x

That is the full proof. Why does everything reduce to sine alone? Because the cos⁡2x=1−2sin⁡2x\cos 2x = 1 - 2\sin^2 x substitution converts the last cosine into sine, so no cosine survives the algebra. The building blocks come from the sum and difference identities.

Examples Of Sin3x

Example 1

Find sin⁡3x\sin 3x if sin⁡x=12\sin x = \dfrac{1}{2}.

Substitute sin⁡x=12\sin x = \dfrac{1}{2} into the formula:

sin⁡3x=3⋅12−4⋅(12)3=32−4⋅18\sin 3x = 3 \cdot \frac{1}{2} - 4 \cdot \left(\frac{1}{2}\right)^3 = \frac{3}{2} - 4 \cdot \frac{1}{8}

=32−12=1= \frac{3}{2} - \frac{1}{2} = 1

Since sin⁡x=12\sin x = \dfrac{1}{2} at x=30°x = 30°, this confirms sin⁡90°=1\sin 90° = 1.

Final answer: sin⁡3x=1\sin 3x = 1.

Example 2

Evaluate sin⁡90°\sin 90° by tripling 30°30°. First the wrong instinct, then the fix.

The tempting shortcut is sin⁡3x=3sin⁡x\sin 3x = 3\sin x, giving sin⁡90°=3sin⁡30°=3⋅12=32\sin 90° = 3\sin 30° = 3 \cdot \dfrac{1}{2} = \dfrac{3}{2}.

That is impossible: sine never exceeds 11. The error is dropping the −4sin⁡3x-4\sin^3 x term and treating the triple angle as a simple triple.

The correct route keeps the cube:

sin⁡90°=3sin⁡30°−4sin⁡330°=32−4⋅18=32−12=1\sin 90° = 3\sin 30° - 4\sin^3 30° = \frac{3}{2} - 4 \cdot \frac{1}{8} = \frac{3}{2} - \frac{1}{2} = 1

Final answer: sin⁡90°=1\sin 90° = 1.

Example 3

Express sin⁡3x\sin^3 x in terms of sin⁡x\sin x and sin⁡3x\sin 3x.

Rearrange the formula to isolate the cube. Starting from sin⁡3x=3sin⁡x−4sin⁡3x\sin 3x = 3\sin x - 4\sin^3 x:

4sin⁡3x=3sin⁡x−sin⁡3x4\sin^3 x = 3\sin x - \sin 3x

sin⁡3x=3sin⁡x−sin⁡3x4\sin^3 x = \frac{3\sin x - \sin 3x}{4}

Final answer: sin⁡3x=3sin⁡x−sin⁡3x4\sin^3 x = \dfrac{3\sin x - \sin 3x}{4}.

Example 4

Evaluate sin⁡270°\sin 270° using x=90°x = 90°.

With x=90°x = 90°, sin⁡x=1\sin x = 1:

sin⁡270°=3sin⁡90°−4sin⁡390°=3(1)−4(1)3=3−4=−1\sin 270° = 3\sin 90° - 4\sin^3 90° = 3(1) - 4(1)^3 = 3 - 4 = -1

Final answer: sin⁡270°=−1\sin 270° = -1, matching the known value.

Example 5

Integrate ∫sin⁡3x,dx\displaystyle\int \sin^3 x , dx using the power-reduction form.

Replace sin⁡3x\sin^3 x with 3sin⁡x−sin⁡3x4\dfrac{3\sin x - \sin 3x}{4} from Example 3:

∫sin⁡3x,dx=14∫(3sin⁡x−sin⁡3x),dx\int \sin^3 x , dx = \frac{1}{4}\int (3\sin x - \sin 3x), dx

Integrate each term:

=14(−3cos⁡x+cos⁡3x3)+C= \frac{1}{4}\left(-3\cos x + \frac{\cos 3x}{3}\right) + C

=−3cos⁡x4+cos⁡3x12+C= -\frac{3\cos x}{4} + \frac{\cos 3x}{12} + C

Final answer: ∫sin⁡3x,dx=−3cos⁡x4+cos⁡3x12+C\displaystyle\int \sin^3 x , dx = -\frac{3\cos x}{4} + \frac{\cos 3x}{12} + C.

Example 6

Given sin⁡x=35\sin x = \dfrac{3}{5} with xx in the first quadrant, find sin⁡3x\sin 3x.

Substitute sin⁡x=35\sin x = \dfrac{3}{5}:

sin⁡3x=3⋅35−4⋅(35)3=95−4⋅27125\sin 3x = 3 \cdot \frac{3}{5} - 4 \cdot \left(\frac{3}{5}\right)^3 = \frac{9}{5} - 4 \cdot \frac{27}{125}

=225125−108125=117125= \frac{225}{125} - \frac{108}{125} = \frac{117}{125}

Final answer: sin⁡3x=117125\sin 3x = \dfrac{117}{125}.

Where Is The Sin3x Formula Used?

The identity earns its keep whenever a tripled angle needs to be handled in terms of the single angle, or a cube of sine needs to be broken down. Why prefer the formula to a calculator? Because it turns an unwieldy expression into something you can integrate, differentiate, or solve symbolically, which a decimal cannot do.

  • Integration. The power-reduction form sin⁡3x=3sin⁡x−sin⁡3x4\sin^3 x = \dfrac{3\sin x - \sin 3x}{4} is the standard way to integrate sin⁡3x\sin^3 x, as Example 5 showed, and the same trick extends to differentiation of trigonometric functions.

  • Solving trigonometric equations. Equations mixing sin⁡3x\sin 3x and sin⁡x\sin x collapse to a single-variable cubic in sin⁡x\sin x once the identity is applied.

  • Signal processing. A pure tone driven through a non-linear device produces a third harmonic, and the sin⁡3x\sin 3x term is the mathematics of that harmonic appearing.

Common Mistakes With Sin3x

Mistake 1: Writing sin⁡3x=3sin⁡x\sin 3x = 3\sin x

Where it slips in: Tripling the angle and reflexively tripling the sine.

Don't do this: sin⁡3x=3sin⁡x\sin 3x = 3\sin x.

The correct way: sin⁡3x=3sin⁡x−4sin⁡3x\sin 3x = 3\sin x - 4\sin^3 x. The −4sin⁡3x-4\sin^3 x term is essential, and dropping it can produce values above 11, which sine can never reach.

The range test catches this instantly: if a "sine" comes out larger than 11, the cubed term went missing. The memoriser who learned "triple it" without the cube is exactly who this trap catches.

Mistake 2: Misreading sin⁡3x\sin^3 x

Where it slips in: Confusing sin⁡3x\sin^3 x with sin⁡x3\sin x^3 or sin⁡3x\sin 3x.

Don't do this: Treating sin⁡3x\sin^3 x as sin⁡(x3)\sin(x^3).

The correct way: sin⁡3x=(sin⁡x)3\sin^3 x = (\sin x)^3, the sine value cubed. It is not the sine of xx cubed, and it is not sin⁡3x\sin 3x. These three read almost alike but are entirely different quantities.

Mistake 3: Sign Slip When Collecting Terms

Where it slips in: During the derivation, when expanding 2sin⁡x(1−sin⁡2x)2\sin x(1 - \sin^2 x).

Don't do this: Writing 2sin⁡x−2sin⁡3x+sin⁡x+2sin⁡3x2\sin x - 2\sin^3 x + \sin x + 2\sin^3 x and cancelling the cubes to get 3sin⁡x3\sin x.

The correct way: Both sin⁡3x\sin^3 x terms are negative: −2sin⁡3x−2sin⁡3x=−4sin⁡3x-2\sin^3 x - 2\sin^3 x = -4\sin^3 x. The second-guesser who redoes the algebra without tracking signs often lands on the wrong coefficient.

A dropped or flipped harmonic term is not only a homework issue. In audio engineering, the third harmonic that sin⁡3x\sin 3x describes is what gives total harmonic distortion its character, and mis-modelling it changes how a circuit is judged to sound.

Key Takeaways

  • The sin3x formula is sin⁡3x=3sin⁡x−4sin⁡3x\sin 3x = 3\sin x - 4\sin^3 x, a triple-angle identity in sine alone.

  • It is derived by writing 3x=2x+x3x = 2x + x and substituting the double-angle identities.

  • The rearranged form sin⁡3x=3sin⁡x−sin⁡3x4\sin^3 x = \dfrac{3\sin x - \sin 3x}{4} is the key to integrating sin⁡3x\sin^3 x.

  • The most common error is writing sin⁡3x=3sin⁡x\sin 3x = 3\sin x, which drops the cube and breaks the range.

To work through triple-angle identities with a teacher, explore Bhanzu's trigonometry tutor sessions, a high school math tutor for board and entrance prep, or live math classes online with peers from 20+ countries.

A Practical Next Step

Practice these to solidify your understanding: find sin⁡3x\sin 3x when sin⁡x=12\sin x = \dfrac{1}{\sqrt{2}}, then use the power-reduction form to integrate sin⁡32x\sin^3 2x. If you get stuck, come back to the derivation and rebuild it from 3x=2x+x3x = 2x + x. Want a live Bhanzu trainer to walk through the proof with you? Book a free demo class.

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Frequently Asked Questions

What is the sin3x formula?
sin⁡3x=3sin⁡x−4sin⁡3x\sin 3x = 3\sin x - 4\sin^3 x. It expresses the sine of a tripled angle using only the sine of the single angle.
Is sin3x the same as 3 sin x?
No. sin⁡3x=3sin⁡x−4sin⁡3x\sin 3x = 3\sin x - 4\sin^3 x. The cubed term cannot be dropped, and leaving it out can give impossible values greater than 11.
What is the formula for sin³x?
sin⁡3x=3sin⁡x−sin⁡3x4\sin^3 x = \dfrac{3\sin x - \sin 3x}{4}. It is the sin3x identity rearranged to isolate the cube, and it is used to integrate sin⁡3x\sin^3 x.
What is the period of sin3x?
2π3\dfrac{2\pi}{3}. Because the angle is tripled, the graph completes three cycles in the space sin⁡x\sin x takes for one.
How is the sin3x formula derived?
Write 3x=2x+x3x = 2x + x, apply the sine angle-sum identity, then substitute sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos x and cos⁡2x=1−2sin⁡2x\cos 2x = 1 - 2\sin^2 x and simplify to sine alone.
✍️ Written By
Madhukar V
Madhukar V
Math Columnist & Analyst
I see mathematics as more than just numbers and formulas—it is a way of thinking, questioning, and understanding the world around us. I work at the intersection of mathematics, EdTech, technology, and AI-powered learning, I enjoy exploring how mathematical ideas can be made simple, meaningful, and relatable. Through my articles, I aim to present mathematics with curiosity and a practical perspective, helping readers see the beauty and relevance behind every concept in a Bhanzu way.
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