Sin(a+b) Formula : Proof, Uses, and Examples

#Trigonometry
TL;DR
The sin(a+b) formula is sin⁡(a+b)=sin⁡acos⁡b+cos⁡asin⁡b\sin(a+b) = \sin a \cos b + \cos a \sin b, the angle-sum identity for sine. This article covers its geometric proof, why it is not equal to sin⁡a+sin⁡b\sin a + \sin b, how it builds the double-angle and sin⁡(a−b)\sin(a-b) formulas, six worked examples, and the mistakes students make with it.
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Bhanzu TeamLast updated on August 14, 20269 min read

The Rule Hiding In Two Overlapping Waves

Add two sound waves out of step and the combined height follows the sin(a+b) formula. When two oscillations share a frequency but differ in phase, the crest of the sum lands wherever the angle-sum identity says it does. Audio mixing, radio, and alternating-current analysis all run on this single expansion.

The sin(a+b) formula answers a precise question: what is the sine of a sum of two angles, written using the sines and cosines of the separate angles? That expansion is the seed of most other trigonometric identities, which is why it heads the sum and difference identities.

What Is The Sin(a+b) Formula?

Sin(a+b) is the sine of the sum of two angles aa and bb, and the formula expands it into single-angle pieces:

sin⁡(a+b)=sin⁡acos⁡b+cos⁡asin⁡b\sin(a+b) = \sin a \cos b + \cos a \sin b

Read it carefully: the right-hand side mixes sine and cosine. It is not two sines added, and not two cosines - each term pairs the sine of one angle with the cosine of the other. The identity holds for every aa and bb, positive or negative, which is what makes it an identity. It sits among the compound-angle rules in the wider trigonometric identities set, and its cosine partner is cos(a−b).

Why Isn't Sin(a+b) Equal To Sin a + Sin b?

This is the single most-asked question about the formula, and the answer is worth internalising. Sine is not linear - you cannot distribute it across an addition the way you distribute multiplication.

The fastest proof is a bounds check. Take a=30∘a = 30^\circ and b=45∘b = 45^\circ. Then sin⁡30∘+sin⁡45∘=0.5+0.707=1.207\sin 30^\circ + \sin 45^\circ = 0.5 + 0.707 = 1.207. But every sine lives in [−1,1][-1, 1], so a value of 1.2071.207 is impossible for any single sine. Meanwhile the real formula gives sin⁡75∘≈0.966\sin 75^\circ \approx 0.966, safely inside the range. The mismatch proves sin⁡(a+b)≠sin⁡a+sin⁡b\sin(a+b) \neq \sin a + \sin b.

How Do You Prove The Sin(a+b) Formula?

The classic proof builds a diagram of stacked angles and reads ratios off it. Start with a ray OXOX. Rotate it counter-clockwise by angle aa to OYOY, then by a further angle bb to OZOZ, so the total angle from OXOX to OZOZ is a+ba + b. Take a point PP on OZOZ and drop perpendiculars to build right triangles.

The construction gives sin⁡(a+b)=PQOP\sin(a+b) = \dfrac{PQ}{OP}, where PQPQ is PP's height above OXOX. Splitting PQPQ into two measurable pieces along the intermediate ray, one line per step:

sin⁡(a+b)=PT+TQOP=PTOP+TQOP\sin(a+b) = \frac{PT + TQ}{OP} = \frac{PT}{OP} + \frac{TQ}{OP}

Each piece resolves into a product of a sine and a cosine through the right triangles in the figure:

PTOP=cos⁡asin⁡b,TQOP=sin⁡acos⁡b\frac{PT}{OP} = \cos a \sin b, \qquad \frac{TQ}{OP} = \sin a \cos b

Adding them gives the identity:

sin⁡(a+b)=sin⁡acos⁡b+cos⁡asin⁡b\sin(a+b) = \sin a \cos b + \cos a \sin b

The proof rests entirely on the sine and cosine ratios inside right triangles, from the trigonometric ratios toolkit.

What Can The Sin(a+b) Formula Do?

Beyond evaluating awkward angles, the formula generates other core identities, which is the real reason it is taught first.

  • Finds non-standard angles. Angles like 75∘75^\circ or 105∘105^\circ split into sums of known angles (30∘+45∘30^\circ + 45^\circ, 60∘+45∘60^\circ + 45^\circ), so their exact sines come straight from the formula.

  • Builds the double-angle rule. Setting b=ab = a turns sin⁡(a+b)\sin(a+b) into sin⁡2a=2sin⁡acos⁡a\sin 2a = 2\sin a \cos a, the 2 sin a cos a identity.

  • Builds the difference rule. Replacing bb with −b-b gives sin⁡(a−b)=sin⁡acos⁡b−cos⁡asin⁡b\sin(a-b) = \sin a \cos b - \cos a \sin b, the sin(a−b) formula.

One important not-the-same note: sin⁡(a+b)\sin(a+b) is a compound-angle formula, while sin A + sin B is a sum-to-product formula equal to 2sin⁡A+B2cos⁡A−B22\sin\frac{A+B}{2}\cos\frac{A-B}{2}. They answer different questions — one expands a single sine of a sum, the other combines two separate sines — so keep them apart.

Examples Of The Sin(a+b) Formula

Example 1

Evaluate sin⁡(30∘+60∘)\sin(30^\circ + 60^\circ) with the formula, then check.

sin⁡(30∘+60∘)=sin⁡30∘cos⁡60∘+cos⁡30∘sin⁡60∘\sin(30^\circ + 60^\circ) = \sin 30^\circ \cos 60^\circ + \cos 30^\circ \sin 60^\circ

=(12)(12)+(32)(32)=14+34=1= \left(\frac{1}{2}\right)\left(\frac{1}{2}\right) + \left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) = \frac{1}{4} + \frac{3}{4} = 1

Check: sin⁡90∘=1\sin 90^\circ = 1.

Final answer: sin⁡90∘=1\sin 90^\circ = 1.

Example 2

Find sin⁡75∘\sin 75^\circ using 75∘=30∘+45∘75^\circ = 30^\circ + 45^\circ. First instinct, then the correct route.

The tempting shortcut is to distribute the sine: sin⁡75∘=sin⁡30∘+sin⁡45∘=0.5+0.707=1.207\sin 75^\circ = \sin 30^\circ + \sin 45^\circ = 0.5 + 0.707 = 1.207.

Test that against what sine can do. No sine ever exceeds 11, yet 1.2071.207 is greater than 11. The answer is impossible, so distributing the sine is wrong.

Use the formula, which mixes sine and cosine:

sin⁡75∘=sin⁡30∘cos⁡45∘+cos⁡30∘sin⁡45∘\sin 75^\circ = \sin 30^\circ \cos 45^\circ + \cos 30^\circ \sin 45^\circ

=(12)(22)+(32)(22)=24+64=6+24= \left(\frac{1}{2}\right)\left(\frac{\sqrt{2}}{2}\right) + \left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{2}}{2}\right) = \frac{\sqrt{2}}{4} + \frac{\sqrt{6}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}

Numerically that is about 0.9660.966, inside [−1,1][-1, 1], as it must be.

Final answer: sin⁡75∘=6+24≈0.966\sin 75^\circ = \dfrac{\sqrt{6} + \sqrt{2}}{4} \approx 0.966.

Example 3

Derive the double-angle formula by setting b=ab = a.

sin⁡(a+a)=sin⁡acos⁡a+cos⁡asin⁡a\sin(a + a) = \sin a \cos a + \cos a \sin a

sin⁡2a=2sin⁡acos⁡a\sin 2a = 2\sin a \cos a

Final answer: sin⁡2a=2sin⁡acos⁡a\sin 2a = 2\sin a \cos a, produced directly from the angle-sum formula.

Example 4

Derive sin⁡(a−b)\sin(a-b) by replacing bb with −b-b.

Use cos⁡(−b)=cos⁡b\cos(-b) = \cos b (cosine is even) and sin⁡(−b)=−sin⁡b\sin(-b) = -\sin b (sine is odd):

sin⁡(a−b)=sin⁡acos⁡(−b)+cos⁡asin⁡(−b)\sin(a - b) = \sin a \cos(-b) + \cos a \sin(-b)

sin⁡(a−b)=sin⁡acos⁡b−cos⁡asin⁡b\sin(a - b) = \sin a \cos b - \cos a \sin b

Final answer: sin⁡(a−b)=sin⁡acos⁡b−cos⁡asin⁡b\sin(a-b) = \sin a \cos b - \cos a \sin b.

Example 5

Given sin⁡a=35\sin a = \frac{3}{5} and cos⁡b=513\cos b = \frac{5}{13}, with both aa and bb acute, find sin⁡(a+b)\sin(a+b).

From the right-angled triangle ratios, cos⁡a=45\cos a = \frac{4}{5} and sin⁡b=1213\sin b = \frac{12}{13}. Then:

sin⁡(a+b)=sin⁡acos⁡b+cos⁡asin⁡b=(35)(513)+(45)(1213)\sin(a+b) = \sin a \cos b + \cos a \sin b = \left(\frac{3}{5}\right)\left(\frac{5}{13}\right) + \left(\frac{4}{5}\right)\left(\frac{12}{13}\right)

=1565+4865=6365= \frac{15}{65} + \frac{48}{65} = \frac{63}{65}

Final answer: sin⁡(a+b)=6365\sin(a+b) = \dfrac{63}{65}.

Example 6

Set up sin⁡(a+b+c)\sin(a + b + c) using the formula.

Group the first two angles and apply the rule twice:

sin⁡((a+b)+c)=sin⁡(a+b)cos⁡c+cos⁡(a+b)sin⁡c\sin\big((a+b) + c\big) = \sin(a+b)\cos c + \cos(a+b)\sin c

Then expand sin⁡(a+b)\sin(a+b) and cos⁡(a+b)\cos(a+b) inside.

Final answer: sin⁡(a+b+c)=(sin⁡acos⁡b+cos⁡asin⁡b)cos⁡c+(cos⁡acos⁡b−sin⁡asin⁡b)sin⁡c\sin(a+b+c) = \big(\sin a \cos b + \cos a \sin b\big)\cos c + \big(\cos a \cos b - \sin a \sin b\big)\sin c.

Where Does The Sin(a+b) Formula Earn Its Keep? - "Combining two waves into one"

The sin(a+b) formula exists to describe what happens when two rotations, or two waves, combine. That combination is everywhere signals meet.

  • Wave superposition. Two signals of the same frequency but different phase add to a single wave; the angle-sum identity is how the combined amplitude and phase are computed, the basis of interference and beat effects.

  • Alternating current. Voltages and currents that are out of phase add through sine-of-a-sum, which is why the identity underlies AC power analysis.

  • Rotations in graphics and robotics. Composing two rotations means adding their angles, and the coordinates of the result come from the sine and cosine angle-sum formulas.

Why must the formula mix sine and cosine rather than staying pure sine? Because rotating by aa then by bb moves a point both up and sideways, and recovering the final height needs the sideways travel too - the cosine terms carry exactly that horizontal contribution. The full family of these expansions is catalogued at Wolfram MathWorld.

Common Mistakes With The Sin(a+b) Formula

Mistake 1: Distributing The Sine Across The Sum

Where it slips in: The first time a student meets a sine of a sum and reaches for algebra habits.

Don't do this: Writing sin⁡(a+b)=sin⁡a+sin⁡b\sin(a+b) = \sin a + \sin b.

The correct way: sin⁡(a+b)=sin⁡acos⁡b+cos⁡asin⁡b\sin(a+b) = \sin a \cos b + \cos a \sin b. The bounds check settles it instantly: sin⁡a+sin⁡b\sin a + \sin b can exceed 11, but a real sine never does. The memorizer who carries the distributive law over from algebra applies a rule that trigonometric functions simply do not obey.

Mistake 2: Losing The Cosine Terms

Where it slips in: Recalling the formula from memory under time pressure.

Don't do this: Writing sin⁡(a+b)=sin⁡asin⁡b+cos⁡acos⁡b\sin(a+b) = \sin a \sin b + \cos a \cos b, swapping in the cosine pattern.

The correct way: Each term pairs a sine with a cosine of the other angle: sin⁡acos⁡b+cos⁡asin⁡b\sin a \cos b + \cos a \sin b. The all-sines or all-cosines version belongs to different identities. The second-guesser who half-remembers the cosine formula often bleeds its pattern into this one.

Mistake 3: Flipping The Sign In Sin(a−b)

Where it slips in: Moving from the sum formula to the difference formula.

Don't do this: Writing sin⁡(a−b)=sin⁡acos⁡b+cos⁡asin⁡b\sin(a-b) = \sin a \cos b + \cos a \sin b, keeping the plus sign.

The correct way: The difference formula flips only the second sign: sin⁡(a−b)=sin⁡acos⁡b−cos⁡asin⁡b\sin(a-b) = \sin a \cos b - \cos a \sin b. A dropped sign here is the same kind of quiet error that corrupts a wave-combination in signal processing - feed two phases into a mixer with a sign flipped, and the output interferes destructively instead of constructively, turning a clean tone into silence. Track the sign deliberately.

Key Takeaways

  • The sin(a+b) formula is sin⁡(a+b)=sin⁡acos⁡b+cos⁡asin⁡b\sin(a+b) = \sin a \cos b + \cos a \sin b, the angle-sum identity for sine.

  • It mixes sine and cosine; it is not sin⁡a+sin⁡b\sin a + \sin b.

  • Setting b=ab = a gives sin⁡2a=2sin⁡acos⁡a\sin 2a = 2\sin a \cos a; replacing bb with −b-b gives the difference formula.

  • It evaluates non-standard angles like 75∘75^\circ by splitting them into known sums.

  • sin⁡(a+b)\sin(a+b) (compound angle) is different from sin⁡A+sin⁡B\sin A + \sin B (sum-to-product).

To work through the sin(a+b) formula with a teacher, explore Bhanzu's trigonometry tutor sessions, a high school math tutor for identity practice, or live math classes online with peers from 20+ countries.

A Practical Next Step

Practice these to solidify your understanding: evaluate sin⁡105∘\sin 105^\circ by splitting it into 60∘+45∘60^\circ + 45^\circ, then derive sin⁡(a−b)\sin(a-b) from scratch by substituting −b-b. If the sine-cosine mixing trips you up, come back to the proof section above. Want a live Bhanzu trainer to build the identity family with you? Book a free demo class.

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Frequently Asked Questions

What is the sin(a+b) formula?
sin⁡(a+b)=sin⁡acos⁡b+cos⁡asin⁡b\sin(a+b) = \sin a \cos b + \cos a \sin b. It expands the sine of a sum of two angles into single-angle sines and cosines.
Is sin(a+b) equal to sin a + sin b?
No. Sine does not distribute across addition. The correct expansion mixes sine and cosine, and sin⁡a+sin⁡b\sin a + \sin b can exceed the sine's [−1,1][-1,1] range, which alone proves they differ.
How do you find sin 15° with this formula?
Write 15∘=45∘−30∘15^\circ = 45^\circ - 30^\circ and use the difference version: sin⁡15∘=sin⁡45∘cos⁡30∘−cos⁡45∘sin⁡30∘=6−24\sin 15^\circ = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ = \dfrac{\sqrt{6} - \sqrt{2}}{4}.
How is sin(a+b) related to the double-angle formula?
Set b=ab = a. Then sin⁡(a+a)=sin⁡2a=2sin⁡acos⁡a\sin(a+a) = \sin 2a = 2\sin a \cos a, so the double-angle rule is a special case.
What is the difference between sin(a+b) and sin A + sin B?
sin⁡(a+b)\sin(a+b) is one sine of a summed angle; sin⁡A+sin⁡B\sin A + \sin B is two sines added, which factors as 2sin⁡A+B2cos⁡A−B22\sin\frac{A+B}{2}\cos\frac{A-B}{2}. They are different formulas for different situations.
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