Interactive guide · Angle sum

Angle sum

Drag α, β or the result around the circle, or type the angles. The panel applies the sum and difference formulas and checks that they give the same value as the direct calculation. Tap any formula in the guide to see its proof.

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αβ (the arc shows the turn from α)α ± βcos(α ± β)sin(α ± β)
°
°
From 0° to 360°. You can also type radians using π, e.g. π/3.
Result
α + β = 75°
= 5π/12 rad ≈1.3090 rad
αβα + β
sin√2/21/2(√6+√2)/4
cos√2/2√3/2(√6−√2)/4
tan1√3/32+√3
Using the sum formulas
sin(α + β) = sin α · cos β + cos α · sin β
= √2/2 · √3/2 + √2/2 · 1/2
= (√6+√2)/4 ≈ 0.9659matches sin(75°)
cos(α + β) = cos α · cos β − sin α · sin β
= √2/2 · √3/2 − √2/2 · 1/2
= (√6−√2)/4 ≈ 0.2588matches cos(75°)
tan(α + β) = (tan α + tan β) / (1 − tan α · tan β)
= (1 + √3/3) / (1 − 1 · √3/3)
= 2+√3 ≈ 3.7321
1 · The idea

What adding angles means

Adding α (alpha) and β (beta) means turning α from the x-axis first and then turning β further. Subtracting means turning β back. The sum formulas give sin, cos and tan of the final angle using only the values of α and β.

Common mistake: sine doesn't distributesin(α + β) is almost never sin α + sin β. With the current angles:
sin 45° + sin 30° = 1.2071  ≠  sin(75°) = 0.9659
2 · The formulas

Sum and difference

Read ± with the top sign for a sum and the bottom one for a difference; ∓ is the other way round. In the cosine the sign flips.

Sine

sin⁡(α±β)=sin⁡α cos⁡β±cos⁡α sin⁡β\operatorname{sin}(\alpha \pm \beta) = \operatorname{sin}\alpha\,\operatorname{cos}\beta \pm \operatorname{cos}\alpha\,\operatorname{sin}\beta

Same sign as the operation. Sine and cosine are mixed.

See proof

Cosine

cos⁡(α±β)=cos⁡α cos⁡β∓sin⁡α sin⁡β\operatorname{cos}(\alpha \pm \beta) = \operatorname{cos}\alpha\,\operatorname{cos}\beta \mp \operatorname{sin}\alpha\,\operatorname{sin}\beta

Opposite sign. cos goes with cos and sin with sin.

See proof

Tangent

tan⁡(α±β)=(tan⁡α±tan⁡β)/(1∓tan⁡α tan⁡β)\operatorname{tan}(\alpha \pm \beta) = (\operatorname{tan}\alpha \pm \operatorname{tan}\beta) \mathbin{/} (1 \mp \operatorname{tan}\alpha\,\operatorname{tan}\beta)

Comes from dividing the two above. Undefined if the denominator is 0.

See proof
3 · Worked example

sin 75° without a calculator

75° isn't a key angle, but it is the sum of two: 45° + 30°.

  1. 1Split into key angles75°=45°+30°75\degree = 45\degree + 30\degree
  2. 2Apply the sine formulasin⁡75°=sin⁡45°⋅cos⁡30°+cos⁡45°⋅sin⁡30°\operatorname{sin} 75\degree = \operatorname{sin} 45\degree \cdot \operatorname{cos} 30\degree + \operatorname{cos} 45\degree \cdot \operatorname{sin} 30\degree
  3. 3Substitute the table values=(22)(32)+(22)(12)= \left(\tfrac{\sqrt{2}}{2}\right)\left(\tfrac{\sqrt{3}}{2}\right) + \left(\tfrac{\sqrt{2}}{2}\right)\left(\tfrac{1}{2}\right)
  4. 4Multiply and combine=64+24=6+24≈0.9659= \tfrac{\sqrt{6}}{4} + \tfrac{\sqrt{2}}{4} = \tfrac{\sqrt{6} + \sqrt{2}}{4} \approx 0.9659
4 · Consequences

Other important identities

They all come from the sum formulas. The check uses the current angles.

Double angle

sin⁡2α=2sin⁡α cos⁡α\operatorname{sin} 2\alpha = 2\operatorname{sin}\alpha\,\operatorname{cos}\alpha
cos⁡2α=cos⁡2α−sin⁡2α\operatorname{cos} 2\alpha = \operatorname{cos}^2\alpha - \operatorname{sin}^2\alpha
tan⁡2α=2tan⁡α/(1−tan⁡2α)\operatorname{tan} 2\alpha = 2\operatorname{tan}\alpha \mathbin{/} (1 - \operatorname{tan}^2\alpha)

It's the sum with β = α.

α = 45° → 2α = 90°: sin 2α = 1.0000 (direct 1.0000), cos 2α = 0.0000 (direct 0.0000), tan 2α is undefined
See proof

Half angle

sin⁡(α/2)=±(1−cos⁡α)/2\operatorname{sin}(\alpha/2) = \pm\sqrt{(1 - \operatorname{cos}\alpha)/2}
cos⁡(α/2)=±(1+cos⁡α)/2\operatorname{cos}(\alpha/2) = \pm\sqrt{(1 + \operatorname{cos}\alpha)/2}

Solved from the double-angle cosine. The sign depends on the quadrant of α/2.

α/2 = 22.5°: √((1 − 0.7071)/2) = 0.3827 → sin = 0.3827; √((1 + 0.7071)/2) = 0.9239 → cos = 0.9239
See proof

Double-angle cosine variants

cos⁡2α=2cos⁡2α−1\operatorname{cos} 2\alpha = 2\operatorname{cos}^2\alpha - 1
cos⁡2α=1−2sin⁡2α\operatorname{cos} 2\alpha = 1 - 2\operatorname{sin}^2\alpha

Replacing sin² with 1 − cos² (or vice versa).

2·0.7071² − 1 = 0.0000 · 1 − 2·0.7071² = 0.0000
See proof

Sum-to-product

sin⁡α+sin⁡β=2sin⁡α+β2 cos⁡α−β2\operatorname{sin}\alpha + \operatorname{sin}\beta = 2\operatorname{sin}\tfrac{\alpha+\beta}{2}\,\operatorname{cos}\tfrac{\alpha-\beta}{2}
cos⁡α+cos⁡β=2cos⁡α+β2 cos⁡α−β2\operatorname{cos}\alpha + \operatorname{cos}\beta = 2\operatorname{cos}\tfrac{\alpha+\beta}{2}\,\operatorname{cos}\tfrac{\alpha-\beta}{2}

Turn a sum into a product, handy for simplifying.

sin α + sin β = 1.2071 = 1.2071 · cos α + cos β = 1.5731 = 1.5731
See proof
5 · What it's for

Where they're used

Exact valuesFind sin, cos or tan of 15°, 75°, 105°… by combining key angles.
Chained rotationsTurning α and then β is turning α + β; this is how you show rotation matrices multiply.
WavesAdding two sounds of similar frequency produces "beats": sum-to-product explains them.
Simplifying and solvingReducing expressions, solving trig equations and computing integrals like ∫ sin²x dx with the double angle.
Proof

Sine of a sum and a difference

sin⁡(α±β)=sin⁡α cos⁡β±cos⁡α sin⁡β\operatorname{sin}(\alpha \pm \beta) = \operatorname{sin}\alpha\,\operatorname{cos}\beta \pm \operatorname{cos}\alpha\,\operatorname{sin}\beta
  1. sin⁡x=cos⁡(90°−x)\operatorname{sin} x = \operatorname{cos}(90\degree - x)

    Complementary angles (adding up to 90°): in a right triangle, the side opposite one acute angle is adjacent to the other. So the sine of one is the cosine of the other.

  2. sin⁡(α+β)=cos⁡(90°−(α+β))=cos⁡((90°−α)−β)\operatorname{sin}(\alpha + \beta) = \operatorname{cos}(90\degree - (\alpha + \beta)) = \operatorname{cos}(\hl{(90\degree - \alpha)} - \beta)

    Apply that to α + β and regroup the bracket.

  3. sin⁡(α+β)=cos⁡(90°−α) cos⁡β+sin⁡(90°−α) sin⁡β\operatorname{sin}(\alpha + \beta) = \operatorname{cos}(90\degree - \alpha)\,\operatorname{cos}\beta + \operatorname{sin}(90\degree - \alpha)\,\operatorname{sin}\beta

    Use the cosine-of-a-difference formula with 90° − α in place of α.

  4. sin⁡(α+β)=sin⁡α cos⁡β+cos⁡α sin⁡β\operatorname{sin}(\alpha + \beta) = \hl{\operatorname{sin}\alpha}\,\operatorname{cos}\beta + \hl{\operatorname{cos}\alpha}\,\operatorname{sin}\beta

    cos(90° − α) = sin α and sin(90° − α) = cos α (complementary again).

  5. sin⁡(α−β)=sin⁡α cos⁡β−cos⁡α sin⁡β\operatorname{sin}(\alpha - \beta) = \operatorname{sin}\alpha\,\operatorname{cos}\beta - \operatorname{cos}\alpha\,\operatorname{sin}\beta

    For the difference replace β with −β: cos(−β) = cos β and sin(−β) = −sin β.

What it's for

Find exact sines of non-key angles (15°, 75°, 105°…) and add waves in physics.

Proof

Cosine of a sum and a difference

cos⁡(α±β)=cos⁡α cos⁡β∓sin⁡α sin⁡β\operatorname{cos}(\alpha \pm \beta) = \operatorname{cos}\alpha\,\operatorname{cos}\beta \mp \operatorname{sin}\alpha\,\operatorname{sin}\beta
  1. A=(cos⁡α,sin⁡α),B=(cos⁡β,sin⁡β)A = (\operatorname{cos}\alpha, \operatorname{sin}\alpha),\quad B = (\operatorname{cos}\beta, \operatorname{sin}\beta)

    Two points on the unit circle, at angles α and β. The angle between them is α − β. We'll compute the distance d between A and B in two ways.

  2. d2=(cos⁡α−cos⁡β)2+(sin⁡α−sin⁡β)2d^2 = (\operatorname{cos}\alpha - \operatorname{cos}\beta)^2 + (\operatorname{sin}\alpha - \operatorname{sin}\beta)^2

    Distance formula between two points: (x₂ − x₁)² + (y₂ − y₁)².

  3. d2=cos⁡2α−2cos⁡α cos⁡β+cos⁡2β+sin⁡2α−2sin⁡α sin⁡β+sin⁡2βd^2 = \operatorname{cos}^2\alpha - 2\operatorname{cos}\alpha\,\operatorname{cos}\beta + \operatorname{cos}^2\beta + \operatorname{sin}^2\alpha - 2\operatorname{sin}\alpha\,\operatorname{sin}\beta + \operatorname{sin}^2\beta

    Expand each square with (a − b)² = a² − 2ab + b².

  4. d2=(sin⁡2α+cos⁡2α)+(sin⁡2β+cos⁡2β)−2(cos⁡α cos⁡β+sin⁡α sin⁡β)d^2 = \hl{(\operatorname{sin}^2\alpha + \operatorname{cos}^2\alpha)} + \hl{(\operatorname{sin}^2\beta + \operatorname{cos}^2\beta)} - 2(\operatorname{cos}\alpha\,\operatorname{cos}\beta + \operatorname{sin}\alpha\,\operatorname{sin}\beta)

    Group the squares of each angle.

  5. d2=2−2(cos⁡α cos⁡β+sin⁡α sin⁡β)d^2 = 2 - 2(\operatorname{cos}\alpha\,\operatorname{cos}\beta + \operatorname{sin}\alpha\,\operatorname{sin}\beta)

    Each coloured bracket equals 1 because sin² + cos² = 1.

  6. A′=(cos⁡(α−β),sin⁡(α−β)),B′=(1,0)A' = (\operatorname{cos}(\alpha - \beta), \operatorname{sin}(\alpha - \beta)),\quad B' = (1, 0)

    Second way: rotate the figure until B lies on the x-axis. Rotating doesn't change distances, so A′B′ is also d.

  7. d2=(cos⁡(α−β)−1)2+sin⁡2(α−β)=2−2cos⁡(α−β)d^2 = (\operatorname{cos}(\alpha - \beta) - 1)^2 + \operatorname{sin}^2(\alpha - \beta) = 2 - 2\operatorname{cos}(\alpha - \beta)

    Same distance formula. Expanding, sin² + cos² gives 1 again.

  8. 2−2cos⁡(α−β)=2−2(cos⁡α cos⁡β+sin⁡α sin⁡β)2 - 2\hl{\operatorname{cos}(\alpha - \beta)} = 2 - 2\hl{(\operatorname{cos}\alpha\,\operatorname{cos}\beta + \operatorname{sin}\alpha\,\operatorname{sin}\beta)}

    Both expressions are the same distance, so they're equal.

  9. cos⁡(α−β)=cos⁡α cos⁡β+sin⁡α sin⁡β\operatorname{cos}(\alpha - \beta) = \operatorname{cos}\alpha\,\operatorname{cos}\beta + \operatorname{sin}\alpha\,\operatorname{sin}\beta

    Subtract 2 from both sides and divide by −2. That's the difference formula.

  10. cos⁡(α+β)=cos⁡α cos⁡(−β)+sin⁡α sin⁡(−β)=cos⁡α cos⁡β−sin⁡α sin⁡β\operatorname{cos}(\alpha + \beta) = \operatorname{cos}\alpha\,\operatorname{cos}(\hl{-\beta}) + \operatorname{sin}\alpha\,\operatorname{sin}(\hl{-\beta}) = \operatorname{cos}\alpha\,\operatorname{cos}\beta - \operatorname{sin}\alpha\,\operatorname{sin}\beta

    For the sum write α + β = α − (−β). A negative angle reflects the point in the x-axis: cos(−β) = cos β and sin(−β) = −sin β. That's where the minus sign comes from.

What it's for

It's the base formula: sine, tangent, double angle and half angle all come from it.

Proof

Tangent of a sum and a difference

tan⁡(α±β)=tan⁡α±tan⁡β1∓tan⁡α tan⁡β\operatorname{tan}(\alpha \pm \beta) = \dfrac{\operatorname{tan}\alpha \pm \operatorname{tan}\beta}{1 \mp \operatorname{tan}\alpha\,\operatorname{tan}\beta}
  1. tan⁡(α±β)=sin⁡(α±β)cos⁡(α±β)\operatorname{tan}(\alpha \pm \beta) = \dfrac{\operatorname{sin}(\alpha \pm \beta)}{\operatorname{cos}(\alpha \pm \beta)}

    Tangent is sine over cosine.

  2. tan⁡(α±β)=sin⁡α cos⁡β±cos⁡α sin⁡βcos⁡α cos⁡β∓sin⁡α sin⁡β\operatorname{tan}(\alpha \pm \beta) = \dfrac{\operatorname{sin}\alpha\,\operatorname{cos}\beta \pm \operatorname{cos}\alpha\,\operatorname{sin}\beta}{\operatorname{cos}\alpha\,\operatorname{cos}\beta \mp \operatorname{sin}\alpha\,\operatorname{sin}\beta}

    Substitute the sine and cosine formulas.

  3. tan⁡(α±β)=sin⁡α cos⁡βcos⁡α cos⁡β±cos⁡α sin⁡βcos⁡α cos⁡βcos⁡α cos⁡βcos⁡α cos⁡β∓sin⁡α sin⁡βcos⁡α cos⁡β\operatorname{tan}(\alpha \pm \beta) = \dfrac{\dfrac{\operatorname{sin}\alpha\,\operatorname{cos}\beta}{\hl{\operatorname{cos}\alpha\,\operatorname{cos}\beta}} \pm \dfrac{\operatorname{cos}\alpha\,\operatorname{sin}\beta}{\hl{\operatorname{cos}\alpha\,\operatorname{cos}\beta}}}{\dfrac{\operatorname{cos}\alpha\,\operatorname{cos}\beta}{\hl{\operatorname{cos}\alpha\,\operatorname{cos}\beta}} \mp \dfrac{\operatorname{sin}\alpha\,\operatorname{sin}\beta}{\hl{\operatorname{cos}\alpha\,\operatorname{cos}\beta}}}

    Divide top and bottom by cos α cos β. The value doesn't change: it's like multiplying by 1.

  4. tan⁡(α±β)=tan⁡α±tan⁡β1∓tan⁡α tan⁡β\operatorname{tan}(\alpha \pm \beta) = \dfrac{\operatorname{tan}\alpha \pm \operatorname{tan}\beta}{1 \mp \operatorname{tan}\alpha\,\operatorname{tan}\beta}

    Simplify each fraction: sin α/cos α = tan α, sin β/cos β = tan β, and anything equal on top and bottom gives 1.

What it's for

Find the angle between two lines from their slopes.

Proof

Double angle

sin⁡2α=2sin⁡α cos⁡α\operatorname{sin} 2\alpha = 2\operatorname{sin}\alpha\,\operatorname{cos}\alpha
  1. sin⁡(α+β)=sin⁡α cos⁡β+cos⁡α sin⁡β\operatorname{sin}(\alpha + \beta) = \operatorname{sin}\alpha\,\operatorname{cos}\beta + \operatorname{cos}\alpha\,\operatorname{sin}\beta

    Start from the sum formula.

  2. sin⁡(α+α)=sin⁡α cos⁡α+cos⁡α sin⁡α\operatorname{sin}(\alpha + \hl{\alpha}) = \operatorname{sin}\alpha\,\operatorname{cos}\hl{\alpha} + \operatorname{cos}\alpha\,\operatorname{sin}\hl{\alpha}

    Set β = α (add an angle to itself).

  3. sin⁡2α=2sin⁡α cos⁡α\operatorname{sin} 2\alpha = 2\operatorname{sin}\alpha\,\operatorname{cos}\alpha

    Both terms are equal.

  4. cos⁡2α=cos⁡α cos⁡α−sin⁡α sin⁡α=cos⁡2α−sin⁡2α\operatorname{cos} 2\alpha = \operatorname{cos}\alpha\,\operatorname{cos}\alpha - \operatorname{sin}\alpha\,\operatorname{sin}\alpha = \operatorname{cos}^2\alpha - \operatorname{sin}^2\alpha

    Same trick with the cosine formula.

  5. tan⁡2α=tan⁡α+tan⁡α1−tan⁡α tan⁡α=2tan⁡α1−tan⁡2α\operatorname{tan} 2\alpha = \dfrac{\operatorname{tan}\alpha + \operatorname{tan}\alpha}{1 - \operatorname{tan}\alpha\,\operatorname{tan}\alpha} = \dfrac{2\operatorname{tan}\alpha}{1 - \operatorname{tan}^2\alpha}

    And with the tangent formula.

What it's for

Simplify expressions and compute integrals like ∫ sin x cos x dx.

Proof

Double-angle cosine variants

cos⁡2α=2cos⁡2α−1=1−2sin⁡2α\operatorname{cos} 2\alpha = 2\operatorname{cos}^2\alpha - 1 = 1 - 2\operatorname{sin}^2\alpha
  1. cos⁡2α=cos⁡2α−sin⁡2α\operatorname{cos} 2\alpha = \operatorname{cos}^2\alpha - \operatorname{sin}^2\alpha

    Cosine of the double angle.

  2. sin⁡2α=1−cos⁡2α\operatorname{sin}^2\alpha = 1 - \operatorname{cos}^2\alpha

    Solved from sin²α + cos²α = 1.

  3. cos⁡2α=cos⁡2α−(1−cos⁡2α)=2cos⁡2α−1\operatorname{cos} 2\alpha = \operatorname{cos}^2\alpha - \hl{(1 - \operatorname{cos}^2\alpha)} = 2\operatorname{cos}^2\alpha - 1

    Substitute sin²α and collect the cos²α terms.

  4. cos⁡2α=1−sin⁡2α\operatorname{cos}^2\alpha = 1 - \operatorname{sin}^2\alpha

    Now solve for the cosine.

  5. cos⁡2α=(1−sin⁡2α)−sin⁡2α=1−2sin⁡2α\operatorname{cos} 2\alpha = \hl{(1 - \operatorname{sin}^2\alpha)} - \operatorname{sin}^2\alpha = 1 - 2\operatorname{sin}^2\alpha

    Substitute and collect the sin²α terms.

What it's for

Pick the version with only sine or only cosine, depending on what you're given.

Proof

Half angle

sin⁡α2=±1−cos⁡α2\operatorname{sin}\dfrac{\alpha}{2} = \pm\sqrt{\dfrac{1 - \operatorname{cos}\alpha}{2}}
  1. cos⁡2x=1−2sin⁡2x\operatorname{cos} 2x = 1 - 2\operatorname{sin}^2 x

    A variant of the double-angle cosine, valid for any angle x.

  2. x=α2  →  cos⁡α=1−2sin⁡2α2x = \dfrac{\alpha}{2} \;\to\; \operatorname{cos}\alpha = 1 - 2\operatorname{sin}^2\dfrac{\alpha}{2}

    Let x be half of α. Then 2x = α.

  3. 2sin⁡2α2=1−cos⁡α2\operatorname{sin}^2\dfrac{\alpha}{2} = 1 - \operatorname{cos}\alpha

    Move terms from one side to the other.

  4. sin⁡α2=±1−cos⁡α2\operatorname{sin}\dfrac{\alpha}{2} = \pm\sqrt{\dfrac{1 - \operatorname{cos}\alpha}{2}}

    Divide by 2 and take the square root. The root has two signs: pick the one for the quadrant where α/2 lies.

  5. cos⁡α2=±1+cos⁡α2\operatorname{cos}\dfrac{\alpha}{2} = \pm\sqrt{\dfrac{1 + \operatorname{cos}\alpha}{2}}

    Same, starting from cos 2x = 2cos²x − 1.

What it's for

Find exact values like sin 15° or cos 22.5° from 30° or 45°.

Proof

Sum-to-product

sin⁡α+sin⁡β=2sin⁡α+β2 cos⁡α−β2\operatorname{sin}\alpha + \operatorname{sin}\beta = 2\operatorname{sin}\dfrac{\alpha + \beta}{2}\,\operatorname{cos}\dfrac{\alpha - \beta}{2}
  1. sin⁡(x+y)=sin⁡x cos⁡y+cos⁡x sin⁡y\operatorname{sin}(x + y) = \operatorname{sin} x\,\operatorname{cos} y + \operatorname{cos} x\,\operatorname{sin} y

    Sum formula.

  2. sin⁡(x−y)=sin⁡x cos⁡y−cos⁡x sin⁡y\operatorname{sin}(x - y) = \operatorname{sin} x\,\operatorname{cos} y - \operatorname{cos} x\,\operatorname{sin} y

    Difference formula.

  3. sin⁡(x+y)+sin⁡(x−y)=2sin⁡x cos⁡y\operatorname{sin}(x + y) + \operatorname{sin}(x - y) = 2\operatorname{sin} x\,\operatorname{cos} y

    Add them: the cos x sin y terms cancel.

  4. α=x+y,β=x−y  →  x=α+β2,y=α−β2\alpha = x + y,\quad \beta = x - y \;\to\; x = \dfrac{\alpha + \beta}{2},\quad y = \dfrac{\alpha - \beta}{2}

    Rename. Adding and subtracting the two equations gives x and y.

  5. sin⁡α+sin⁡β=2sin⁡α+β2 cos⁡α−β2\operatorname{sin}\alpha + \operatorname{sin}\beta = 2\operatorname{sin}\dfrac{\alpha + \beta}{2}\,\operatorname{cos}\dfrac{\alpha - \beta}{2}

    Substitute. Doing the same with the cosine formulas gives cos α + cos β.

What it's for

A product is 0 if any factor is 0, which makes equations easier to solve. It also explains the "beats" when two sounds are added.

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