# Cosine of Sum

## Theorem

$\map \cos {a + b} = \cos a \cos b - \sin a \sin b$

where $\sin$ denotes the sine and $\cos$ denotes the cosine.

### Corollary

$\map \cos {a - b} = \cos a \cos b + \sin a \sin b$

## Proof 1

 $\ds \map \cos {a + b} + i \, \map \sin {a + b}$ $=$ $\ds e^{i \paren {a + b} }$ Euler's Formula $\ds$ $=$ $\ds e^{i a} e^{i b}$ Exponential of Sum $\ds$ $=$ $\ds \paren {\cos a + i \sin a} \paren {\cos b + i \sin b}$ Euler's Formula $\ds$ $=$ $\ds \paren {\cos a \cos b - \sin a \sin b} + i \paren {\sin a \cos b + \cos a \sin b}$ Complex Numbers form Field

The result follows by equating the real parts.

$\blacksquare$

## Proof 2

Recall the analytic definitions of sine and cosine:

$\ds \sin x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!}$
$\ds \cos x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!}$

Let:

 $\ds \map g a$ $=$ $\ds \map \sin {a + b} - \sin a \cos b - \cos a \sin b$ $\ds \map h a$ $=$ $\ds \map \cos {a + b} - \cos a \cos b + \sin a \sin b$

Let us differentiate these with respect to $a$, keeping $b$ constant.

Then from Derivative of Sine Function and Derivative of Cosine Function, we have:

 $\ds \map {g'} a$ $=$ $\ds \map \cos {a + b} - \cos a \cos b + \sin a \sin b = \map h a$ $\ds \map {h'} a$ $=$ $\ds -\map \sin {a + b} + \sin a \cos b + \cos a \sin b = -\map g a$

Hence:

 $\ds \map {D_a} {\paren {\map g a}^2 + \paren {\map h a}^2}$ $=$ $\ds 2 \map g a \map {g'} a + 2 \map h a \map {h'} a$ $\ds$ $=$ $\ds 0$

Thus from Derivative of Constant:

$\forall a \in \R: \map g a^2 + \map h a^2 = c$

In particular, it is true for $a = 0$, and so:

$\map g 0^2 + \map h 0^2 = 0$

So:

$\map g a^2 + \map h a^2 = 0$
$\map g a^2 \ge 0$

and:

$\map h a^2 \ge 0$

So it follows that:

$\map g a = 0$

and:

$\map h a = 0$

Hence the result.

$\blacksquare$

## Proof 3

 $\ds \cos a \cos b - \sin a \sin b$ $=$ $\ds \paren {\frac {e^{i a} + e^{-i a} } 2} \paren {\frac {e^{i b} + e^{-i b} } 2} - \sin a \sin b$ Euler's Cosine Identity $\ds$ $=$ $\ds \paren {\frac {e^{i a} + e^{-i a} } 2} \paren {\frac {e^{i b} + e^{-i b} } 2}  - \paren {\frac {e^{i a} - e^{-i a} } {2 i} } \paren {\frac {e^{i b} - e^{-i b} } {2 i} }$  Euler's Sine Identity $\ds$ $=$ $\ds \frac {e^{i a} e^{i b} + e^{i a} e^{-i b} + e^{-i a} e^{i b} + e^{-i a} e^{-i b} } 4$ $\ds$  $\, \ds - \,$ $\ds \frac {e^{i a} e^{i b} - e^{i a} e^{-i b} - e^{-i a} e^{i b} + e^{-i a} e^{-i b} } {4 i^2}$ $\ds$ $=$ $\ds \frac {e^{i a} e^{i b} + e^{i a} e^{-i b} + e^{-i a} e^{i b} + e^{-i a} e^{-i b} } 4$ $\ds$  $\, \ds + \,$ $\ds \frac {e^{i a} e^{i b} - e^{i a} e^{-i b} - e^{-i a} e^{i b} + e^{-i a} e^{-i b} } 4$ as $i^2 = -1$ $\ds$ $=$ $\ds \frac {e^{i a} e^{i b} + e^{-ia} e^{-ib} } 2$ $\ds$ $=$ $\ds \frac {e^{i \paren {a + b} } + e^{-i \paren {a + b} } } 2$ Exponential of Sum $\ds$ $=$ $\ds \map \cos {a + b}$ Euler's Cosine Identity

$\blacksquare$

## Proof 4

$AB$, $AC$, $AE$, and $AD$ are radii of the circle centered at $A$.

Let $\angle BAC = a$ and $\angle DAC = \angle BAE = b$.

By Euclid's First Postulate, we can construct line segments $BD$ and $CE$.

By Euclid's second common notion, $\angle DAB = \angle CAE$.

Thus by Triangle Side-Angle-Side Congruence, $\triangle DAB \cong \triangle CAE$.

Therefore, $DB = CE$.

We now assign Cartesian coordinates to the points $B$, $C$, $D$, and $E$:

 $\ds B$ $=$ $\ds \tuple {1, 0}$ $\ds C$ $=$ $\ds \tuple {\cos a, \sin a}$ $\ds D$ $=$ $\ds \tuple {\map \cos {a + b}, \map \sin {a + b} }$ $\ds E$ $=$ $\ds \tuple {\cos b, -\sin b}$ Cosine Function is Even and Sine Function is Odd

We use the definition of the distance function on the Euclidean space $\struct {\R^2, d}$ as defined by the Euclidean metric:

$\forall x, y \in \R^2: \map d {x, y} = \sqrt {\paren {x_1 - x_2}^2 + \paren {y_1 - y_2}^2}$

where $x = \tuple {x_1, y_1}, y = \tuple {x_2, y_2}$.

Thus:

$DB \cong CE \iff \map d {D, B} = \map d {C, E}$

So, plugging in the coordinates of $B, C, D, E$, we get:

 $\ds \paren {\map \cos {a + b} } - 1)^2 + \map {\sin^2} {a + b}$ $=$ $\ds \paren {\cos a - \cos b}^2 + \paren {\sin a + \sin b}^2$ $\ds \leadsto \ \$ $\ds \cos^2 \left({a + b}\right) + \sin^2 \left({a + b}\right)$  $\ds$ multiplying out left hand side $\ds {} - 2 \, \map \cos {a + b} + 1$ $=$ $\ds \paren {\cos a - \cos b}^2 + \paren {\sin a + \sin b}^2$ $\ds \leadsto \ \$ $\ds 1 - 2 \, \map \cos {a + b} + 1$ $=$ $\ds \paren {\cos a - \cos b}^2 + \paren {\sin a + \sin b}^2$ Sum of Squares of Sine and Cosine $\ds \leadsto \ \$ $\ds 2 - 2 \, \map \cos {a + b}$ $=$ $\ds \cos^2 a - 2 \cos a \cos b + \cos^2 b$ multiplying out right hand side $\ds$  $\, \ds + \,$ $\ds \sin^2 a + 2 \sin a \sin b + \sin^2 b$ $\ds \leadsto \ \$ $\ds 2 - 2 \, \map \cos {a + b}$ $=$ $\ds 2 - 2 \cos a \cos b + 2 \sin a \sin b$ Sum of Squares of Sine and Cosine $\ds \leadsto \ \$ $\ds \map \cos {a + b}$ $=$ $\ds \cos a \cos b - \sin a \sin b$

$\blacksquare$

## Proof 5

We begin by enclosing a right-angled triangle $BEF$ with hypotenuse $BF$ of length $1$, inside Rectangle $ABCD$.

Let $\angle EBF = a$ and $\angle ABE = b$.

Therefore:

 $\ds BF$ $=$ $\ds 1$ Given $\ds BE$ $=$ $\ds \cos a$ Definition of Cosine of Angle $\ds EF$ $=$ $\ds \sin a$ Definition of Sine of Angle $\ds AB$ $=$ $\ds \cos a \cos b$ $\ds AE$ $=$ $\ds \cos a \sin b$ $\ds ED$ $=$ $\ds \sin a \cos b$ $\ds DF$ $=$ $\ds \sin a \sin b$ $\ds \map \cos {a + b }$ $=$ $\ds FC$ $\ds$ $=$ $\ds AB - DF$ $\ds$ $=$ $\ds \cos a \cos b - \sin a \sin b$

$\blacksquare$

## Proof 6

 $\ds \map \cos {a + b}$ $=$ $\ds \map \cos {a - \paren {-b} }$ $\ds$ $=$ $\ds \cos a \map \cos {-b} + \sin a \map \sin {-b}$ Cosine of Difference $\ds$ $=$ $\ds \cos a \cos b + \sin a \paren {-\sin b}$ Cosine Function is Even, Sine Function is Odd $\ds$ $=$ $\ds \cos a \cos b - \sin a \sin b$ simplifying

$\blacksquare$

## Proof 7

Let two triangles $\triangle ABC$ and $\triangle ABD$ be inscribed in a circle in the same semicircle:.

By Thales' Theorem, these are both right triangles with:

$\angle ACB = \angle ADB = 90 \degrees$

Let $AB = 1$.

Join $DC$.

By construction, $\Box ABCD$ is a cyclic quadrilateral.

Let:

$\angle CAB = \alpha$
$\angle DBA = \beta$
$\angle DAC = \gamma$

From the construction above, we have the following:

$\cos \alpha = AC$
$\cos \beta = BD$
$\sin \alpha = CB$
$\sin \beta = AD$
$\gamma = \dfrac \pi 2 - \alpha - \beta$
$CD = 2 r \map \sin {\gamma}$

Since $2r \mathop = 1$:

$CD = \map \sin {\gamma}$

Therefore:

 $\ds CD$ $=$ $\ds \map \sin {\gamma}$ $\ds$ $=$ $\ds \map \cos {\dfrac \pi 2 - \gamma}$ Cosine of Complement equals Sine $\ds$ $=$ $\ds \map \cos {\alpha + \beta}$ $\gamma$ is complementary to $\alpha + \beta$
$AC \times BD = AD \times BC + AB \times CD$

Substituting:

$\cos \alpha \times \cos \beta = \sin \beta \times \sin \alpha + 1 \times \map \cos {\alpha + \beta}$

Rearranging:

$\cos \tuple {\alpha + \beta} = \cos \alpha \cos \beta - \sin \alpha \sin \beta$

By Equivalence of Definitions of Cosine of Angle, the definition of cosine from the circle, from the triangle and as a real function are equivalent.

It follows that all real numbers $x$ and $y$ correspond to values of $\alpha$ and $\beta$ for which the proof above applies, with one exception.

The exception occurs when both $\alpha$ and $\beta$ are equal to $\dfrac {\pi} 2$.

But then the result is simply:

$\cos {\pi} = \cos \dfrac {\pi} 2 \cos \dfrac {\pi} 2 - \sin \dfrac {\pi} 2 \sin \dfrac {\pi} 2$
$-1 = 0 \cdot 0 - 1 \cdot 1$

The result follows.

$\blacksquare$

## Historical Note

The Cosine of Sum formula and its corollary were proved by François Viète in about $1579$.