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:

$\displaystyle \sin x = \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {x^{2 n + 1}} {\left({2 n + 1}\right)!}$
$\displaystyle \cos x = \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {x^{2 n}} {\left({2 n}\right)!}$

Let:

\(\ds g \left({a}\right)\) \(=\) \(\ds \sin \left({a + b}\right) - \sin a \cos b - \cos a \sin b\)
\(\ds h \left({a}\right)\) \(=\) \(\ds \cos \left({a + b}\right) - \cos a \cos b + \sin a \sin b\)

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

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

\(\ds g' \left({a}\right)\) \(=\) \(\ds \cos \left({a + b}\right) - \cos a \cos b + \sin a \sin b = h \left({a}\right)\)
\(\ds h' \left({a}\right)\) \(=\) \(\ds - \sin \left({a + b}\right) + \sin a \cos b + \cos a \sin b = - g \left({a}\right)\)

Hence:

\(\ds D_a \left({\left({g \left({a}\right)}\right)^2 + \left({h \left({a}\right)}\right)^2}\right)\) \(=\) \(\ds 2 g \left({a}\right) g' \left({a}\right) + 2 h \left({a}\right) h' \left({a}\right)\)
\(\ds \) \(=\) \(\ds 0\)

Thus from Derivative of Constant:

$\forall a \in \R: g \left({a}\right)^2 + h \left({a}\right)^2 = c$

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

$g \left({0}\right)^2 + h \left({0}\right)^2 = 0$

So:

$g \left({a}\right)^2 + h \left({a}\right)^2 = 0$

But from Square of Real Number is Non-Negative:

$g \left({a}\right)^2 \ge 0$ and $h \left({a}\right)^2 \ge 0$

So it follows that:

$g \left({a}\right) = 0$

and:

$h \left({a}\right) = 0$

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
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \frac {e^{i a} e^{i b} - e^{i a} e^{-i b} + e^{-i a} e^{i b} - e^{-i a} e^{-i b} } {4 i}\)
\(\ds \) \(=\) \(\ds \frac {e^{i a} e^{i b} - e^{-i a} e^{-i b} } {2 i}\) sinplifying
\(\ds \) \(=\) \(\ds \frac {e^{i \paren {a + b} } - e^{-i \paren {a + b} } } {2 i}\) 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

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

$\blacksquare$


Historical Note

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


Also see


Sources