Precalculus · Grades 11, 12

Sum and Difference Formulas for Sine, Cosine and Tangent

Quick answer

The sine of a sum is not the sum of the sines: sin(30° + 45°) is not sin 30° + sin 45°, which would exceed 1. The correct formula comes from the unit circle. The distance between two points on it depends only on the angle between them, and writing that distance two ways gives the cosine of a difference. The other formulas follow from that one in a few lines.

What you'll learn

  • Prove the difference formula for cosine from the unit circle
  • Derive the sum and difference formulas for sine, cosine and tangent
  • Use the formulas to find exact values and simplify expressions

The natural guess fails

Is sin⁡(30°+45°)\sin(30° + 45°) equal to sin⁡30°+sin⁡45°\sin 30° + \sin 45°?

sin⁡30°+sin⁡45°=12+22≈1.207\sin 30° + \sin 45° = \tfrac{1}{2} + \tfrac{\sqrt{2}}{2} \approx 1.207

No sine is ever larger than 11, so the guess cannot be right. The actual value is sin⁡75°≈0.966\sin 75° \approx 0.966. Sine does not distribute over addition, and neither does cosine.

The correct formulas are these:

cos⁡(α−β)=cos⁡αcos⁡β+sin⁡αsin⁡β\cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta sin⁡(α+β)=sin⁡αcos⁡β+cos⁡αsin⁡β\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta

They are not memorized facts. One of them can be proved from the unit circle, and the rest follow from it.

Why the cosine of a difference works out this way

Put two points on the unit circle: PP at angle α\alpha and QQ at angle β\beta.

P=(cos⁡α, sin⁡α)Q=(cos⁡β, sin⁡β)P = (\cos\alpha,\ \sin\alpha) \qquad Q = (\cos\beta,\ \sin\beta)

The distance between them depends only on the angle between them, α−β\alpha - \beta, not on where they sit. So turn the whole picture until QQ lands on (1,0)(1, 0). Then PP lands at angle α−β\alpha - \beta, and the chord keeps its length.

Rotating a chord keeps its length The unit circle with two chords of equal length. One joins the points at 110 degrees and 40 degrees. The other, dashed, joins the point at 70 degrees to the point (1, 0), which is the first chord turned clockwise by 40 degrees. -11-11xy P Q P′ Q′
Rotating a chord keeps its length

Write the squared chord length both ways.

Before turning, from PP to QQ:

(cos⁡α−cos⁡β)2+(sin⁡α−sin⁡β)2(\cos\alpha - \cos\beta)^2 + (\sin\alpha - \sin\beta)^2

Expand, and use cos⁡2+sin⁡2=1\cos^2 + \sin^2 = 1 twice, from the Pythagorean identity:

=2−2(cos⁡αcos⁡β+sin⁡αsin⁡β)= 2 - 2\left(\cos\alpha\cos\beta + \sin\alpha\sin\beta\right)

After turning, from P′(cos⁡(α−β),sin⁡(α−β))P'(\cos(\alpha - \beta), \sin(\alpha - \beta)) to Q′(1,0)Q'(1, 0):

(cos⁡(α−β)−1)2+sin⁡2(α−β)=2−2cos⁡(α−β)\bigl(\cos(\alpha - \beta) - 1\bigr)^2 + \sin^2(\alpha - \beta) = 2 - 2\cos(\alpha - \beta)

The two lengths are equal, so the two expressions are equal. Subtract 22 and divide by −2-2:

cos⁡(α−β)=cos⁡αcos⁡β+sin⁡αsin⁡β\cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta

The formula is the chord length written in two coordinate systems. Nothing else went in.

The other formulas follow

Cosine of a sum. Replace β\beta with −β-\beta. Since cos⁡(−β)=cos⁡β\cos(-\beta) = \cos\beta and sin⁡(−β)=−sin⁡β\sin(-\beta) = -\sin\beta:

cos⁡(α+β)=cos⁡αcos⁡β−sin⁡αsin⁡β\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta

Sine of a sum. Sine is cosine shifted: sin⁡θ=cos⁡(π2−θ)\sin\theta = \cos\left(\tfrac{\pi}{2} - \theta\right). So

sin⁡(α+β)=cos⁡((π2−α)−β)\sin(\alpha + \beta) = \cos\left(\left(\tfrac{\pi}{2} - \alpha\right) - \beta\right)

Apply the difference formula, and use cos⁡(π2−α)=sin⁡α\cos\left(\tfrac{\pi}{2} - \alpha\right) = \sin\alpha and sin⁡(π2−α)=cos⁡α\sin\left(\tfrac{\pi}{2} - \alpha\right) = \cos\alpha:

sin⁡(α+β)=sin⁡αcos⁡β+cos⁡αsin⁡β\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta

Sine of a difference. Replace β\beta with −β-\beta again:

sin⁡(α−β)=sin⁡αcos⁡β−cos⁡αsin⁡β\sin(\alpha - \beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta

Tangent. Divide the sine formula by the cosine formula, then divide the top and bottom by cos⁡αcos⁡β\cos\alpha\cos\beta:

tan⁡(α±β)=tan⁡α±tan⁡β1∓tan⁡αtan⁡β\tan(\alpha \pm \beta) = \frac{\tan\alpha \pm \tan\beta}{1 \mp \tan\alpha\tan\beta}
FormulaSign pattern
sin⁡(α±β)=sin⁡αcos⁡β±cos⁡αsin⁡β\sin(\alpha \pm \beta) = \sin\alpha\cos\beta \pm \cos\alpha\sin\betakeeps the sign
cos⁡(α±β)=cos⁡αcos⁡β∓sin⁡αsin⁡β\cos(\alpha \pm \beta) = \cos\alpha\cos\beta \mp \sin\alpha\sin\betaflips the sign
tan⁡(α±β)=tan⁡α±tan⁡β1∓tan⁡αtan⁡β\tan(\alpha \pm \beta) = \tfrac{\tan\alpha \pm \tan\beta}{1 \mp \tan\alpha\tan\beta}keeps on top, flips below

Exact values beyond the special angles

The special angles 30°30°, 45°45° and 60°60° have exact values. Sums and differences of them now do too.

Worked examples

Common mistakes

Practice problems

  1. Find cos⁡75°\cos 75° exactly.

    Answer

    6−24\tfrac{\sqrt{6} - \sqrt{2}}{4}

    Full solution

    cos⁡(45°+30°)=cos⁡45°cos⁡30°−sin⁡45°sin⁡30°=64−24\cos(45° + 30°) = \cos 45°\cos 30° - \sin 45°\sin 30° = \tfrac{\sqrt{6}}{4} - \tfrac{\sqrt{2}}{4}.

  2. Find sin⁡15°\sin 15° exactly.

    Answer

    6−24\tfrac{\sqrt{6} - \sqrt{2}}{4}

    Full solution

    sin⁡(45°−30°)=sin⁡45°cos⁡30°−cos⁡45°sin⁡30°=64−24\sin(45° - 30°) = \sin 45°\cos 30° - \cos 45°\sin 30° = \tfrac{\sqrt{6}}{4} - \tfrac{\sqrt{2}}{4}.

  3. Find sin⁡105°\sin 105° exactly.

    Answer

    6+24\tfrac{\sqrt{6} + \sqrt{2}}{4}

    Full solution

    sin⁡(60°+45°)=sin⁡60°cos⁡45°+cos⁡60°sin⁡45°=64+24\sin(60° + 45°) = \sin 60°\cos 45° + \cos 60°\sin 45° = \tfrac{\sqrt{6}}{4} + \tfrac{\sqrt{2}}{4}.

  4. Find tan⁡75°\tan 75° exactly.

    Answer

    2+32 + \sqrt{3}

    Full solution

    tan⁡(45°+30°)=1+331−33=3+33−3\tan(45° + 30°) = \tfrac{1 + \frac{\sqrt{3}}{3}}{1 - \frac{\sqrt{3}}{3}} = \tfrac{3 + \sqrt{3}}{3 - \sqrt{3}}.

    Multiply the top and bottom by 3+33 + \sqrt{3}: 12+636=2+3\tfrac{12 + 6\sqrt{3}}{6} = 2 + \sqrt{3}.

  5. With acute α\alpha, β\beta, sin⁡α=35\sin\alpha = \tfrac{3}{5} and cos⁡β=513\cos\beta = \tfrac{5}{13}, find cos⁡(α−β)\cos(\alpha - \beta).

    Answer

    5665\tfrac{56}{65}

    Full solution

    cos⁡α=45\cos\alpha = \tfrac{4}{5} and sin⁡β=1213\sin\beta = \tfrac{12}{13}.

    cos⁡(α−β)=45⋅513+35⋅1213=2065+3665=5665\cos(\alpha - \beta) = \tfrac{4}{5}\cdot\tfrac{5}{13} + \tfrac{3}{5}\cdot\tfrac{12}{13} = \tfrac{20}{65} + \tfrac{36}{65} = \tfrac{56}{65}.

  6. With the same angles, find cos⁡(α+β)\cos(\alpha + \beta).

    Answer

    −1665-\tfrac{16}{65}

    Full solution

    45⋅513−35⋅1213=2065−3665=−1665\tfrac{4}{5}\cdot\tfrac{5}{13} - \tfrac{3}{5}\cdot\tfrac{12}{13} = \tfrac{20}{65} - \tfrac{36}{65} = -\tfrac{16}{65}.

    The negative value says α+β\alpha + \beta is obtuse, which fits: α≈36.9°\alpha \approx 36.9° and β≈67.4°\beta \approx 67.4°.

  7. Simplify sin⁡(x+π2)\sin\left(x + \tfrac{\pi}{2}\right).

    Answer

    cos⁡x\cos x

    Full solution

    sin⁡xcos⁡π2+cos⁡xsin⁡π2=sin⁡x⋅0+cos⁡x⋅1=cos⁡x\sin x\cos\tfrac{\pi}{2} + \cos x\sin\tfrac{\pi}{2} = \sin x \cdot 0 + \cos x \cdot 1 = \cos x.

  8. Simplify cos⁡(π−x)\cos(\pi - x).

    Answer

    −cos⁡x-\cos x

    Full solution

    cos⁡πcos⁡x+sin⁡πsin⁡x=(−1)cos⁡x+0=−cos⁡x\cos\pi\cos x + \sin\pi\sin x = (-1)\cos x + 0 = -\cos x, the reflection rule from the unit circle.

  9. Use the cosine sum formula to show cos⁡2α=cos⁡2α−sin⁡2α\cos 2\alpha = \cos^2\alpha - \sin^2\alpha.

    Answer

    Set β=α\beta = \alpha in cos⁡(α+β)\cos(\alpha + \beta).

    Full solution

    cos⁡(α+α)=cos⁡αcos⁡α−sin⁡αsin⁡α=cos⁡2α−sin⁡2α\cos(\alpha + \alpha) = \cos\alpha\cos\alpha - \sin\alpha\sin\alpha = \cos^2\alpha - \sin^2\alpha.

  10. Asked for sin⁡75°\sin 75°, Grace writes sin⁡30°+sin⁡45°≈1.207\sin 30° + \sin 45° \approx 1.207. Find her error.

    Hint

    Can any sine be larger than 11?

    Answer

    Sine does not distribute over addition. sin⁡75°=6+24≈0.966\sin 75° = \tfrac{\sqrt{6} + \sqrt{2}}{4} \approx 0.966.

    Full solution

    Her answer is above 11, which no sine can be, so the method is wrong before any arithmetic is checked.

    The sum formula gives sin⁡(45°+30°)=sin⁡45°cos⁡30°+cos⁡45°sin⁡30°=6+24≈0.966\sin(45° + 30°) = \sin 45°\cos 30° + \cos 45°\sin 30° = \tfrac{\sqrt{6} + \sqrt{2}}{4} \approx 0.966.

    Each term pairs a sine with a cosine, because turning through 30°30° after 45°45° mixes the two coordinates of the point on the circle.

Frequently asked questions

What is the formula for cos(α − β)?

cos(α − β) = cos α cos β + sin α sin β. For a sum the sign flips: cos(α + β) = cos α cos β − sin α sin β.

What is the formula for sin(α + β)?

sin(α + β) = sin α cos β + cos α sin β, and sin(α − β) = sin α cos β − cos α sin β.

Why is sin(α + β) not sin α + sin β?

Sine is not a linear function. sin 30° + sin 45° is about 1.207, more than any sine can be, while sin 75° is about 0.966.

How is the cosine formula proved?

Two points on the unit circle at angles α and β are a fixed distance apart. Rotating them until one sits at (1, 0) keeps that distance, and writing it both ways gives cos(α − β).

What is the tangent sum formula?

tan(α + β) = (tan α + tan β)/(1 − tan α tan β). It comes from dividing the sine formula by the cosine formula.

What to learn next

Standards alignment

This lesson covers the following Common Core State Standards for Mathematics.

  • CCSS.MATH.CONTENT.HSF.TF.C.9Trigonometric Functions(+) Prove the addition and subtraction formulas for sine, cosine, and tangent and use them to solve problems.
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