Cosine of Sum

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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:

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

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

\(\ds \cos a \cos b - \sin a \sin b\) \(=\) \(\ds \paren {\frac {e^{i a} + e^{-i a} } 2} \paren {\frac {e^{i b} + e^{-i b} } 2} - \sin a \sin b\) Cosine Exponential Formulation
\(\ds \) \(=\) \(\ds \paren {\frac {e^{i a} + e^{-i a} } 2} \paren {\frac {e^{i b} + e^{-i b} } 2} - \paren {\frac {e^{i a} - e^{-i a} } {2 i} } \paren {\frac {e^{i b} - e^{-i b} } {2 i} }\) Sine Exponential Formulation
\(\ds \) \(=\) \(\ds \frac {e^{i a} e^{i b} + e^{i a} e^{-i b} + e^{-i a} e^{i b} + e^{-i a} e^{-i b} } 4\)
\(\ds \) \(\) \(\, \ds - \, \) \(\ds \frac {e^{i a} e^{i b} - e^{i a} e^{-i b} - e^{-i a} e^{i b} + e^{-i a} e^{-i b} } {4 i^2}\)
\(\ds \) \(=\) \(\ds \frac {e^{i a} e^{i b} + e^{i a} e^{-i b} + e^{-i a} e^{i b} + e^{-i a} e^{-i b} } 4\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \frac {e^{i a} e^{i b} - e^{i a} e^{-i b} - e^{-i a} e^{i b} + e^{-i a} e^{-i b} } 4\) 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
\(\ds \) \(=\) \(\ds \map \cos {a + b}\) Cosine Exponential Formulation

$\blacksquare$


Proof 4

Tri1.PNG


$AB$, $AC$, $AE$, 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 Equality, $\triangle DAB \cong \triangle CAE$.

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

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

$\blacksquare$


Historical Note

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


Also see


Sources