Sine of Sum

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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:

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

But from Square of Real Number is Non-Negative:

$\map g a^2 \ge 0$ 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

\(\ds \sin a \cos b + \cos a \sin b\) \(=\) \(\ds \paren {\frac {e^{i a} - e^{-i a} }{2 i} } \cos b + \cos a \paren {\frac {e^{i b} - e^{-i b} }{2 i} }\) Euler's Sine Identity
\(\ds \) \(=\) \(\ds \paren {\frac {e^{i a} - e^{-i a} } {2 i} } \paren {\frac {e^{i b} + e^{-i b} } 2} 
         + \paren {\frac {e^{i a} + e^{-i a} } 2} \paren {\frac {e^{i b} - e^{-i b} } {2 i} }\)
Euler's Cosine Identity
\(\ds \) \(=\) \(\ds \frac {e^{i a} e^{i b} + e^{i a} e^{-i b} - e^{-i a} e^{i b} - e^{-i a} e^{-i b} } {4 i}\) 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
\(\ds \) \(=\) \(\ds \map \sin {a + b}\) Euler's Sine Identity

$\blacksquare$


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

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

Angle-sum.png


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

$\blacksquare$


Proof 7

Sum of Sines using Ptolemy.png

Let two triangles $\triangle ABC$ and $\triangle ADC$ be inscribed in a circle on opposite sides of diameter $AC$.

By Thales' Theorem, they are both right triangles and $\angle ADC$ and $\angle ABC$ are right angles.

Let the diameter $AC = 1$.

Let $\angle DAC = \alpha$ and $\angle CAB = \beta$.

From the construction above, we have the following:

$\cos \alpha = AD$
$\cos \beta = AB$
$\sin \alpha = DC$
$\sin \beta = BC$

By Length of Chord of Circle:

$DB \mathop = 2 r \map \sin {\alpha + \beta}$

Since $2r \mathop = 1$:

$DB \mathop = \map \sin {\alpha + \beta}$

By Quadrilateral is Cyclic iff Opposite Angles sum to Two Right Angles:

$\Box ABCD$ is a cyclic quadrilateral.

By Ptolemy's Theorem:

$DB \times AC = AB \times CD + BC \times AD$

Substituting:

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


By Equivalence of Definitions of Sine of Angle, the definition of sine 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:

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

The result follows.

$\blacksquare$


Historical Note

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


Also see


Sources

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