Sum and Difference Identities

Learn the sum and difference formulas for sine, cosine and tangent, find exact values like sin 75° and combine angles from different quadrants.

Advanced30 min18 problems
Practice Problems

Definition

The sum and difference identities give the sine, cosine and tangent of A+BA + B or A−BA - B from the values of AA and BB alone.

Sine

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

Cosine

cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos(A + B) = \cos A \cos B - \sin A \sin B
cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos(A - B) = \cos A \cos B + \sin A \sin B

Tangent

tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}
tan⁡(A−B)=tan⁡A−tan⁡B1+tan⁡Atan⁡B\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}
How to remember the signs:
  • Sine keeps the sign: ++ in A+BA + B gives ++ in the formula, and it mixes sine with cosine.
  • Cosine flips the sign: A+BA + B gives a minus, A−BA - B gives a plus, and it pairs cosine with cosine, sine with sine.
  • Tangent: the numerator keeps the sign, the denominator flips it.
Important: sin⁡(A+B)\sin(A + B) is not sin⁡A+sin⁡B\sin A + \sin B. For example sin⁡(30∘+60∘)=sin⁡90∘=1\sin(30^\circ + 60^\circ) = \sin 90^\circ = 1, but sin⁡30∘+sin⁡60∘≈1.366\sin 30^\circ + \sin 60^\circ \approx 1.366.
With the special angles 30∘30^\circ, 45∘45^\circ and 60∘60^\circ these formulas give exact values for new angles such as 15∘=45∘−30∘15^\circ = 45^\circ - 30^\circ, 75∘=45∘+30∘75^\circ = 45^\circ + 30^\circ, 105∘=60∘+45∘105^\circ = 60^\circ + 45^\circ and 165∘=120∘+45∘165^\circ = 120^\circ + 45^\circ. If you set B=AB = A, you get the double angle identities.

Try it now

Which expression is equal to sin⁡(A+B)\sin(A + B)?

Worked Examples

Find the exact value of sin⁡75∘\sin 75^\circ.
1

Write 75∘75^\circ as a sum of special angles

75∘=45∘+30∘75^\circ = 45^\circ + 30^\circ → A=45∘A = 45^\circ, B=30∘B = 30^\circ
2

Apply the sine sum identity

sin⁡75∘=sin⁡45∘cos⁡30∘+cos⁡45∘sin⁡30∘\sin 75^\circ = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ → Two products of known values
3

Substitute the exact values

22⋅32+22⋅12=64+24\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} → 6+24\frac{\sqrt{6} + \sqrt{2}}{4}
4

Check with a calculator

6+24≈2.4495+1.41424≈0.9659\frac{\sqrt{6} + \sqrt{2}}{4} \approx \frac{2.4495 + 1.4142}{4} \approx 0.9659 and sin⁡75∘≈0.9659\sin 75^\circ \approx 0.9659 → The values agree

Common Mistakes

Writing sin⁡(A+B)=sin⁡A+sin⁡B\sin(A + B) = \sin A + \sin B

Why it's wrong: Sine is not a linear function, so you cannot split it over a sum. With A=30∘A = 30^\circ and B=60∘B = 60^\circ: sin⁡90∘=1\sin 90^\circ = 1, but sin⁡30∘+sin⁡60∘≈1.366\sin 30^\circ + \sin 60^\circ \approx 1.366, which is not even a possible sine value.

Correct: sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A + B) = \sin A \cos B + \cos A \sin B. Move the sliders in the graph: the red curve sin⁡x+sin⁡b\sin x + \sin b leaves the other two.

Keeping the sign in the cosine formula: cos⁡(A+B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos(A + B) = \cos A \cos B + \sin A \sin B

Why it's wrong: Cosine flips the sign. Test it with A=B=45∘A = B = 45^\circ: cos⁡90∘=0\cos 90^\circ = 0, but cos⁡245∘+sin⁡245∘=1\cos^2 45^\circ + \sin^2 45^\circ = 1.

Correct: cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos(A + B) = \cos A \cos B - \sin A \sin B and cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos(A - B) = \cos A \cos B + \sin A \sin B.

Taking every missing value as positive, e.g. cos⁡A=45\cos A = \frac{4}{5} when sin⁡A=35\sin A = \frac{3}{5} and AA is in Quadrant II

Why it's wrong: The Pythagorean identity only gives the size of the value. The quadrant decides the sign, and a wrong sign changes the whole result.

Correct: Use the quadrant: in Quadrant II cosine is negative, so cos⁡A=−45\cos A = -\frac{4}{5}. Remember: Quadrant I all positive, II only sine, III only tangent, IV only cosine.

Using 1+tan⁡Atan⁡B1 + \tan A \tan B in the denominator of tan⁡(A+B)\tan(A + B)

Why it's wrong: In the tangent formula the numerator keeps the sign but the denominator flips it, just like in the cosine formula.

Correct: tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} and tan⁡(A−B)=tan⁡A−tan⁡B1+tan⁡Atan⁡B\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}.

Interactive Visual

Function graph

-300-200-1000100200300-2-1012xy(45, 0.97)

Tap or hover over the graph to read off values.

  • sin(a + b) = 0.9659
  • sin a · cos b + cos a · sin b = 0.9659
  • sin a + sin b = 1.2071

Tap a name to show or hide its graph.

Prove

sin⁡(x+90∘)=cos⁡x\sin(x + 90^\circ) = \cos x

  1. Start with the left side

    sin⁡(x+90∘)\sin(x + 90^\circ)

Check with numbers

Left side

sin⁡(x+90∘)\sin(x + 90^\circ)

= 0.866

Right side

cos⁡x\cos x

= 0.866

Both sides have the same value.

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Practice Problems

18 problems
Problem 1 of 18
Easy
Which expression is equal to sin⁡(A+B)\sin(A + B)?

Why It Matters

The sum and difference identities are the parent formulas of most other trigonometric identities:
  • Exact values: a calculator gives sin⁡75∘≈0.9659\sin 75^\circ \approx 0.9659, but the identity gives the exact value 6+24\frac{\sqrt{6} + \sqrt{2}}{4}, which is what exams and proofs need.
  • More identities: double angle, half angle and the cofunction rules all follow from them (set B=AB = A, or A=90∘A = 90^\circ).
  • Waves and sound: physicists rewrite 3sin⁡t+4cos⁡t3\sin t + 4\cos t as one wave 5sin⁡(t+φ)5\sin(t + \varphi) to read off its amplitude.
  • Rotations: every computer game and robot arm turns points with x′=xcos⁡β−ysin⁡βx' = x\cos\beta - y\sin\beta and y′=xsin⁡β+ycos⁡βy' = x\sin\beta + y\cos\beta, which is exactly the sum identity.
  • Angles between lines: the tangent difference identity gives the angle between two roads, ramps or roof lines from their slopes.

Real World Applications

Combining Two Waves

In physics and electronics, a sum like $3\sin t + 4\cos t$ describes two overlapping waves. The sine sum identity shows that it is one single wave $R\sin(t + \varphi)$, and $R$ is its amplitude (how loud or how strong it is).

Example:

Rsin⁡(t+φ)=Rcos⁡φsin⁡t+Rsin⁡φcos⁡tR\sin(t + \varphi) = R\cos\varphi \sin t + R\sin\varphi \cos t. Comparing with 3sin⁡t+4cos⁡t3\sin t + 4\cos t gives Rcos⁡φ=3R\cos\varphi = 3 and Rsin⁡φ=4R\sin\varphi = 4.

1Try It Yourself

A signal is 3sin⁡t+4cos⁡t3\sin t + 4\cos t. You want to write it as Rsin⁡(t+φ)R\sin(t + \varphi) with R>0R > 0.

What is the amplitude RR?

Step 1: Write the mathematical expression

Use R2=(Rcos⁡φ)2+(Rsin⁡φ)2R^2 = (R\cos\varphi)^2 + (R\sin\varphi)^2:

Rotating Points in Games and Robotics

A point at distance $r$ and angle $\alpha$ has coordinates $x = r\cos\alpha$, $y = r\sin\alpha$. Turning it by $\beta$ gives the angle $\alpha + \beta$, and the sum identities turn that into the rotation formulas $x' = x\cos\beta - y\sin\beta$ and $y' = x\sin\beta + y\cos\beta$.

Example:

Rotating (1,0)(1, 0) by 90∘90^\circ: x′=1⋅0−0⋅1=0x' = 1 \cdot 0 - 0 \cdot 1 = 0 and y′=1⋅1+0⋅0=1y' = 1 \cdot 1 + 0 \cdot 0 = 1, so the point moves to (0,1)(0, 1).

2Try It Yourself

A robot arm holds a tool at (4,3)(4, 3). It turns by an angle β\beta with cos⁡β=0.8\cos\beta = 0.8 and sin⁡β=0.6\sin\beta = 0.6.

What is the new yy-coordinate of the tool?

Step 1: Write the mathematical expression

Use y′=xsin⁡β+ycos⁡βy' = x\sin\beta + y\cos\beta:

The Angle Between Two Roads

A line with slope $m$ makes an angle $\theta$ with the $x$-axis where $\tan\theta = m$. The tangent difference identity gives the angle between two lines straight from their slopes, which planners use for road junctions, ramps and roof lines.

Example:

Lines with slopes 11 and 00: tan⁡(α−β)=1−01+0=1\tan(\alpha - \beta) = \frac{1 - 0}{1 + 0} = 1, so they meet at 45∘45^\circ.

3Try It Yourself

On a map, one road has slope 33 and a second road has slope 12\frac{1}{2}.

What is the angle between the roads, in degrees?

Step 1: Write the mathematical expression

Use tan⁡(α−β)=tan⁡α−tan⁡β1+tan⁡αtan⁡β\tan(\alpha - \beta) = \frac{\tan\alpha - \tan\beta}{1 + \tan\alpha \tan\beta} with tan⁡α=3\tan\alpha = 3, tan⁡β=12\tan\beta = \frac{1}{2}:

Key Takeaways

  • 1sin⁡(A±B)=sin⁡Acos⁡B±cos⁡Asin⁡B\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B: sine keeps the sign and mixes sine with cosine
  • 2cos⁡(A±B)=cos⁡Acos⁡B∓sin⁡Asin⁡B\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B: cosine flips the sign
  • 3tan⁡(A±B)=tan⁡A±tan⁡B1∓tan⁡Atan⁡B\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}: the numerator keeps the sign, the denominator flips it
  • 4Exact values of 15∘15^\circ, 75∘75^\circ, 105∘105^\circ, 165∘165^\circ come from sums and differences of 30∘30^\circ, 45∘45^\circ, 60∘60^\circ (and 120∘120^\circ, 135∘135^\circ)
  • 5With angles in other quadrants, find the missing sine or cosine with the Pythagorean identity and take the sign from the quadrant
  • 6sin⁡(A+B)≠sin⁡A+sin⁡B\sin(A + B) \neq \sin A + \sin B; setting B=AB = A gives the double angle identities

Frequently Asked Questions

Because sine is not a linear function. Doubling an angle does not double its sine: sin⁡30∘=0.5\sin 30^\circ = 0.5, but sin⁡60∘≈0.866\sin 60^\circ \approx 0.866, not 11. The sum identity is the correct replacement for that wrong rule.
Because sine is not a linear function. Doubling an angle does not double its sine: sin⁡30∘=0.5\sin 30^\circ = 0.5, but sin⁡60∘≈0.866\sin 60^\circ \approx 0.866, not 11. The sum identity is the correct replacement for that wrong rule.
Set B=AB = A. Then sin⁡(2A)=sin⁡Acos⁡A+cos⁡Asin⁡A=2sin⁡Acos⁡A\sin(2A) = \sin A \cos A + \cos A \sin A = 2\sin A \cos A and cos⁡(2A)=cos⁡2A−sin⁡2A\cos(2A) = \cos^2 A - \sin^2 A. The tangent formula gives tan⁡(2A)=2tan⁡A1−tan⁡2A\tan(2A) = \frac{2\tan A}{1 - \tan^2 A}.
Any pair works if it gives the angle: 75∘=45∘+30∘75^\circ = 45^\circ + 30^\circ, 15∘=45∘−30∘15^\circ = 45^\circ - 30^\circ or 60∘−45∘60^\circ - 45^\circ, 105∘=60∘+45∘105^\circ = 60^\circ + 45^\circ, 165∘=120∘+45∘165^\circ = 120^\circ + 45^\circ or 135∘+30∘135^\circ + 30^\circ. Different pairs lead to the same final value.
When 1−tan⁡Atan⁡B=01 - \tan A \tan B = 0, that is when tan⁡Atan⁡B=1\tan A \tan B = 1. Then A+BA + B is 90∘90^\circ (or 270∘270^\circ), where tangent does not exist. Example: tan⁡30∘tan⁡60∘=1\tan 30^\circ \tan 60^\circ = 1 and 30∘+60∘=90∘30^\circ + 60^\circ = 90^\circ.

Glossary

Sum identity
A formula for a trigonometric function of A+BA + B, e.g. sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A + B) = \sin A \cos B + \cos A \sin B
Difference identity
A formula for a trigonometric function of A−BA - B, e.g. cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos(A - B) = \cos A \cos B + \sin A \sin B
Exact value
A value written with whole numbers, fractions and roots, like 6+24\frac{\sqrt{6} + \sqrt{2}}{4}, instead of a rounded decimal
Special angles
30∘30^\circ, 45∘45^\circ and 60∘60^\circ (and their relatives on the unit circle), whose sine, cosine and tangent are known exactly
Quadrant sign
The sign a trigonometric value has in each quadrant: I all positive, II sine, III tangent, IV cosine

Formula Card

Sine of a sum

sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A + B) = \sin A \cos B + \cos A \sin B

Keeps the sign, mixes sine and cosine

Sine of a difference

sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin(A - B) = \sin A \cos B - \cos A \sin B

Minus between the two products

Cosine of a sum

cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos(A + B) = \cos A \cos B - \sin A \sin B

The sign flips to minus

Cosine of a difference

cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos(A - B) = \cos A \cos B + \sin A \sin B

The sign flips to plus

Tangent of a sum

tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}

Numerator keeps, denominator flips the sign

Tangent of a difference

tan⁡(A−B)=tan⁡A−tan⁡B1+tan⁡Atan⁡B\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}

Numerator minus, denominator plus

Today's challengeOne problem, three tries

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