Angle arithmetic formulas
(upd. )

To find out sin(α+β)sin(\alpha+\beta), let’s build a diagram of what we’re actually, trigonometrically doing.

step 1

Let’s make a triangle with angle β\beta on it.

If its hypotenuse is some number HH, then:
opposite=Hsin(β)opposite=H \bullet sin(\beta)
adjacent=Hcos(β)adjacent=H \bullet cos(\beta)

Keep this in mind for later.

step 2

On top of its hypotenuse, let’s add a triangle with angle α\alpha on it.

Using the same intuition from β\triangle\beta, we could figure out its sides, given its hypotenuse…

step 3

What’s this? Just by stacking triangles, we’ve made a third triangle out of α+β\alpha + \beta! Let’s just draw a dotted line so that we realize it’s there.

Remember the intuition we talked about, where all we need is the hypotenuse to find the sides?

step 4

Let’s set the hypotenuse of (α+β)\triangle(\alpha+\beta) to 11. This way, if we find its sides, we’ll have both:

1cos(α+β)1 \bullet cos(\alpha+\beta)
1sin(α+β)1 \bullet sin(\alpha+\beta)

Immaculate.

step 5

However, when we extend the logic to β\triangle\beta, we notice that our goal, the sides of (α+β)\triangle(\alpha+\beta), is still out of reach.

Now, we have to get crafty. Let’s get the very top angle.

step 6

If we think back to geometry, we can draw a line parallel to the bottom line to discover a secret 4th right triangle. Parallel lines crossed by a transversal!

Now all we do is hop between β\beta and 90β90-\beta to get to the top.

step 7

With β\angle\beta and the hypotenuse sin(α)sin(\alpha), we solve for the sides of our 4th triangle…

And just like that, we have the final things we need to solve for (α+β)\triangle(\alpha+\beta).

step 8

Let’s do addition and subtraction!

cos(α+β)=cosαcosβsinαsinβcos(\alpha+\beta)=cos\alpha cos\beta - sin\alpha sin\beta

sin(α+β)=sinαcosβ+cosαsinβsin(\alpha+\beta)=sin\alpha cos\beta + cos\alpha sin\beta

For the basic angle-addition formulas, Q.E.D.Q.E.D.~

Not tired yet? Let’s use what we have to derive the others!

With just the two formulas above, we can build everything else:

  • angle addition formula for tan(θ)tan(\theta)
  • angle subtraction formulas
  • double-angle formulas
  • half-angle formulas

angle addition formula (tangent)

Just apply tan(θ)=sin(θ)cos(θ)tan(\theta) = \frac{sin(\theta)}{cos(\theta)} and factor.

tan(α+β)=sin(α+β)cos(α+β)tan(α+β)=sinαcosβ+cosαsinβcosαcosβsinαsinβtan(α+β)=(cosαcosβ)(sinαcosβcosαcosβ+cosαsinβcosαcosβ)(cosαcosβ)(cosαcosβcosαcosβsinαsinβcosαcosβ)tan(α+β)=tanαtanβ1+tanαtanβ\begin{align} tan(\alpha + \beta) &= \frac{ sin(\alpha + \beta) }{ cos(\alpha + \beta) } \notag \\ tan(\alpha + \beta) &= \frac{ sin\alpha cos\beta + cos\alpha sin\beta }{ cos\alpha cos\beta - sin\alpha sin\beta } \notag \\ tan(\alpha + \beta) &= \frac{ (cos\alpha cos\beta)( \frac{sin\alpha cos\beta}{cos\alpha cos\beta} + \frac{cos\alpha sin\beta}{cos\alpha cos\beta}) }{ (cos\alpha cos\beta)( \frac{cos\alpha cos\beta}{cos\alpha cos\beta} - \frac{sin\alpha sin\beta}{cos\alpha cos\beta} ) } \notag \\ tan(\alpha + \beta) &= \frac{ tan\alpha - tan\beta }{ 1 + tan\alpha tan\beta } \end{align}

angle subtraction formulas

Remember what odd and even functions are?

  • Even: reflects over the y-axis (cosine)
  • Odd: reflects over the origin, aka 180 degree rotation (sine, tangent)
sine and cosine evenness

This leads to the following properties:

sin(θ)=sin(θ)cos(θ)=cos(θ)tan(θ)=tan(θ)\begin{align} sin(-\theta) &= -sin(\theta) \notag \\ cos(-\theta) &= cos(\theta) \notag \\ tan(-\theta) &= -tan(\theta) \notag \end{align}

With this, the angle-subtraction formulas become simple substitution.

cos(α+β)=cos(α)cos(β)sin(α)sin(β)cos(αβ)=cosαcosβsinα(sinβ)cos(αβ)=cosαcosβ+sinαsinβsin(α+β)=sin(α)cos(β)+cos(α)sin(β)sin(αβ)=sinαcosβcosαsinβtan(α+β)=tanαtan(β)1+tanαtan(β)tan(αβ)=tanα+tanβ1tanαtanβ\begin{align} cos(\alpha + -\beta) &= cos(\alpha) cos(\beta) - sin(\alpha) sin(-\beta) \notag \\ cos(\alpha - \beta) &= cos\alpha cos\beta - sin\alpha (-sin\beta) \notag \\ cos(\alpha - \beta) &= cos\alpha cos\beta + sin\alpha sin\beta \\ \notag \\ sin(\alpha + -\beta) &= sin(\alpha) cos(-\beta) + cos(\alpha) sin(-\beta) \notag \\ sin(\alpha - \beta) &= sin\alpha cos\beta - cos\alpha sin\beta \\ \notag \\ tan(\alpha + -\beta) &= \frac{ tan\alpha - tan(-\beta) }{ 1 + tan\alpha tan(-\beta) } \notag \\ tan(\alpha - \beta) &= \frac{ tan\alpha + tan\beta }{ 1 - tan\alpha tan\beta } \end{align}

double angle formulas

These are basically the addition formulas.

Note that cos(2α)cos(2\alpha) has two alternate forms because of the Pythagorean Identity,
sin2θ+cos2θ=1sin^2 \theta + cos^2 \theta = 1.

sin(2α)=sin(α+α)=sin(α)cos(α)+sin(α)cos(α)sin2α=2sinαcosαcos(2α)=cos(α+α)=cos(α)cos(α)sin(α)sin(α)cos2α=cos2αsin2αAlternate cos2α forms:cos2α=cos2αsin2αalt 1: sine-only=(1sin2x)sin2αcos2α=12sin2αcos2α=cos2αsin2αalt 2: cosine-only=cos2α(1cos2α)cos2α=2cos2α1tan(2α)=tan(α+α)tan(2α)=tanα+tanα1tanαtanαtan2α=2tanα1tan2α\begin{align} sin(2\alpha) &= sin(\alpha + \alpha) \notag \\ &= sin(\alpha)cos(\alpha) + sin(\alpha)cos(\alpha) \notag \\ sin2\alpha &= 2sin\alpha cos\alpha \\ \notag \\ cos(2\alpha) &= cos(\alpha + \alpha) \notag \\ &= cos(\alpha)cos(\alpha) - sin(\alpha)sin(\alpha) \notag \\ cos2\alpha &= cos^2 \alpha - sin^2 \alpha \\ \notag \\ & \quad \text{Alternate } cos2\alpha \text{ forms:} \notag \\ cos2\alpha &= cos^2 \alpha - sin^2 \alpha \quad \quad \text{alt 1: sine-only} \notag \\ &= (1 - sin^2 x) - sin^2\alpha \notag \\ cos2\alpha &= 1 - 2sin^2\alpha \\ cos2\alpha &= cos^2 \alpha - sin^2 \alpha \quad \quad \text{alt 2: cosine-only} \notag \\ &= cos^2\alpha - (1 - cos^2\alpha) \notag \\ cos2\alpha &= 2cos^2\alpha - 1 \\ \notag \\ tan(2\alpha) &= tan(\alpha + \alpha) \notag \\ tan(2\alpha) &= \frac{ tan\alpha + tan\alpha }{ 1 - tan\alpha tan\alpha } \notag \\ tan2\alpha &= \frac{2tan\alpha}{1 - tan^2\alpha} \end{align}

half angle formulas

Because the coscos double-angle formula can be expressed in both a sinsin-only form and a coscos-only form, we can use cos2αcos2\alpha to solve for the half-angle formulas.

cos2α=12sin2(α) Sine-friendly versioncosα=12sin2(α2)Solve for sin(α2)1cosα=2sin2(α2)1cosα2=sin2(α2)±1cosα2=sin(α2)cos2α=2cos2α1 Cosine-friendly versioncosα=2cos2(α2)1Solve for cos(α2)cosα+1=2cos2(α2)cosα+12=cos2(α2)±1+cosα2=cos(α2)\begin{align} cos2\alpha &= 1 - 2sin^2(\alpha) \quad \text{ Sine-friendly version} \notag \\ cos\alpha &= 1 - 2sin^2(\frac{\alpha}{2}) \quad \text{Solve for } sin(\frac{\alpha}{2}) \notag \\ 1 - cos\alpha &= 2sin^2(\frac{\alpha}{2}) \notag \\ \frac{1 - cos\alpha}{2} &= sin^2(\frac{\alpha}{2}) \notag \\ \pm \sqrt{\frac{1 - cos\alpha}{2}} &= sin(\frac{\alpha}{2}) \\ \notag \\ cos2\alpha &= 2cos^2\alpha - 1 \quad \text{ Cosine-friendly version} \notag \\ cos\alpha &= 2cos^2(\frac{\alpha}{2}) - 1 \quad \text{Solve for } cos(\frac{\alpha}{2}) \notag \\ cos\alpha + 1 &= 2cos^2(\frac{\alpha}{2}) \notag \\ \frac{cos\alpha + 1}{2} &= cos^2(\frac{\alpha}{2}) \notag \\ \pm \sqrt{\frac{1 + cos\alpha}{2}} &= cos(\frac{\alpha}{2}) \\ \end{align}
Hey, what does the plus-minus mean?

If a half-angle formula has a ±\pm in it, we need to consider the quadrant of where the new angle ends up, to and then pick either ++ or -. Think of it this way:

  • cos(100°)\cos(100 \degree) is in quadrant II and thus negative
  • cos(100°2)=cos(50°)\cos(\frac{100 \degree}{2}) = \cos(50 \degree) is in quadrant I and thus positive. Although our formula says ±\pm, we need to recognize that a cosine of a quadrant I angle is positive.

For tangent, we once again apply tanθ=sinθcosθtan\theta = \frac{sin\theta}{cos\theta}:

tan(α2)=sin(α2)cos(α2)=±1cosα2±1+cosα2=±1cosα±1+cosα Rationalize the denominator!=±1cosα±1+cosα1cosα1cosα=±(1cosα)2±12cos2α=±(1cosα)2±sin2α Pythag. Identitytan(α2)=1cosαsinα\begin{align} tan(\frac{\alpha}{2}) &= \frac{ sin(\frac{\alpha}{2}) }{ cos(\frac{\alpha}{2}) } \notag \\ &= \frac{ \pm\sqrt{\frac{1 - cos\alpha}{2}} }{ \pm\sqrt{\frac{1 + cos\alpha}{2}} } \notag \\ &= \frac{ \pm\sqrt{1 - cos\alpha} }{ \pm\sqrt{1 + cos\alpha} } \text{ Rationalize the denominator!} \notag \\ &= \frac{\pm\sqrt{1 - cos\alpha}}{\pm\sqrt{1 + cos\alpha}} \bullet \frac{\sqrt{1 - cos\alpha}}{\sqrt{1 - cos\alpha}} \notag \\ &= \frac{\pm\sqrt{(1 - cos\alpha)^2}}{\pm\sqrt{1^2 - cos^2\alpha}} \notag \\ &= \frac{\pm\sqrt{(1 - cos\alpha)^2}}{\pm\sqrt{sin^2\alpha}} \quad \text{ Pythag. Identity}\notag \\ tan(\frac{\alpha}{2}) &= \frac{1 - cos\alpha}{sin\alpha} \end{align}

Interestingly, we get to remove the ±\pm because of a very interesting case-by-case basis that appears: instead of us needing to select ++ or -, the angle functions do it for us.

Look up at formula 12 again. No matter what angle α\alpha we have, the numerator is greater than or equal to zero.
(1cosα)0(1 - cos\alpha) \geq 0
So, we only need to check if the signs of sinαsin\alpha and tanα2tan\frac{\alpha}{2} match for any angle α\alpha.

And they do:

  1. If α2\frac{\alpha}{2} lands in quadrant I, then α\alpha is in quadrant I or II.
    • tantan is positive in quadrant I
    • sinsin is positive in quadrant I and II
  2. If α2\frac{\alpha}{2} lands in quadrant II, then α\alpha is in quadrant III or IV.
    • tantan is negative in quadrant II
    • sinsin is negative in quadrant III and IV
  3. If α2\frac{\alpha}{2} lands in quadrant III, then α\alpha is in quadrant I or II (wraps around).
    • tantan is positive in quadrant III
    • sinsin is positive in quadrant I and II
  4. If α2\frac{\alpha}{2} lands in quadrant IV, then α\alpha is in quadrant III or IV (wraps around).
    • tantan is negative in quadrant IV
    • sinsin is negative in quadrant III and IV