# Sine of Sum

## Theorem

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

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

### Corollary

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

## 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

By equating the imaginary parts, the result follows.

$\blacksquare$

## 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

 $\ds \sin a \cos b + \cos a \sin b$ $=$ $\ds \paren {\frac {e^{i a} - e^{-i a} }{2 i} } \cos b + \cos a \paren {\frac {e^{i b} - e^{-i b} }{2 i} }$ Sine Exponential Formulation $\ds$ $=$ $\ds \paren {\frac {e^{i a} - e^{-i a} } {2 i} } \paren {\frac {e^{i b} + e^{-i b} } 2} + \paren {\frac {e^{i a} + e^{-i a} } 2} \paren {\frac {e^{i b} - e^{-i b} } {2 i} }$ Cosine Exponential Formulation $\ds$ $=$ $\ds \frac {e^{i a} e^{i b} + e^{i a} e^{-i b} - e^{-i a} e^{i b} - e^{-i a} e^{-i b} } {4 i}$ expanding $\ds$  $\, \ds + \,$ $\ds \frac {e^{i a} e^{i b} - e^{i a} e^{-i b} + e^{-i a} e^{i b} - e^{-i a} e^{-i b} } {4 i}$ $\ds$ $=$ $\ds \frac {e^{i a} e^{i b} - e^{-i a} e^{-i b} } {2 i}$ simplifying $\ds$ $=$ $\ds \frac {e^{i \paren {a + b} } - e^{-i \paren {a + b} } } {2 i}$ Exponential of Sum $\ds$ $=$ $\ds \map \sin {a + b}$ Sine Exponential Formulation

$\blacksquare$

## Proof 4

 $\ds \sin \left({a + b}\right)$ $=$ $\ds \cos \left({\frac \pi 2 - \left({a + b}\right)}\right)$ Cosine of Complement equals Sine $\ds$ $=$ $\ds \cos \left({\left({\frac \pi 2 - a}\right) - b}\right)$ $\ds$ $=$ $\ds \cos \left({\frac \pi 2 - a}\right) \cos b + \sin \left({\frac \pi 2 - a}\right) \sin b$ Cosine of Difference $\ds$ $=$ $\ds \sin a \cos b + \sin \left({\frac \pi 2 - a}\right) \sin b$ Cosine of Complement equals Sine $\ds$ $=$ $\ds \sin a \cos b + \cos a \sin b$ Sine of Complement equals Cosine

$\blacksquare$

## Proof 5

 $\text {(1)}: \quad$ $\ds 2 \sin a \cos b$ $=$ $\ds \sin \paren {a + b} + \sin \paren {a - b}$ Simpson's Formula for Sine by Cosine: Proof 2 $\text {(2)}: \quad$ $\ds 2 \cos a \sin b$ $=$ $\ds \sin \paren {a + b} - \sin \paren {a - b}$ Simpson's Formula for Cosine by Sine: Proof 2 $\ds \leadsto \ \$ $\ds 2 \sin \paren {a + b}$ $=$ $\ds 2 \sin a \cos b + 2 \cos a \sin b$ $(1) + (2)$ $\ds \leadsto \ \$ $\ds \sin \paren {a + b}$ $=$ $\ds \sin a \cos b + \cos a \sin b$

$\blacksquare$

## Proof 6

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 $\ds AB$ $=$ $\ds \cos a \cos b$ $\ds AE$ $=$ $\ds \cos a \sin b$ $\ds ED$ $=$ $\ds \sin a \cos b$ $\ds DF$ $=$ $\ds \sin a \sin b$ $\ds \map \sin {a + b }$ $=$ $\ds BC$ $\ds$ $=$ $\ds AE + ED$ $\ds$ $=$ $\ds \cos a \sin b + \sin a \cos b$

$\blacksquare$

## Historical Note

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