Cosine of Sum

Theorem

 * $\cos \left({a + b}\right) = \cos a \cos b - \sin a \sin b$


 * $\sin \left({a + b}\right) = \sin a \cos b + \cos a \sin b$

where $\sin$ and $\cos$ are sine and cosine.

Corollary 1

 * $\cos \left({a - b}\right) = \cos a \cos b + \sin a \sin b$


 * $\sin \left({a - b}\right) = \sin a \cos b - \cos a \sin b$

Corollary 2

 * $\displaystyle \tan \left({a + b}\right) = \frac {\tan a + \tan b} {1 - \tan a \tan b}$


 * $\displaystyle \tan \left({a - b}\right) = \frac {\tan a - \tan b} {1 + \tan a \tan b}$

where $\tan$ is tangent.

Proof from Euler's Formula
By equating real and imaginary parts, we have:

Proof from Algebraic Definitions
We have:
 * From the definition of sine:
 * $\displaystyle \sin x = \sum_{n=0}^\infty \left({-1}\right)^n \frac {x^{2n+1}}{\left({2n+1}\right)!} = x - \frac {x^3} {3!} + \frac {x^5} {5!} - \cdots$


 * From the definition of cosine:
 * $\displaystyle \cos x = \sum_{n=0}^\infty \left({-1}\right)^n \frac {x^{2n}}{\left({2n}\right)!} = 1 - \frac {x^2} {2!} + \frac {x^4} {4!} - \cdots$.

Let:

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

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

Hence:
 * $D_a \left({g \left({a}\right)^2 + h \left({a}\right)^2}\right)$

Thus from Differentiation of a Constant, $\forall a: g \left({a}\right)^2 + h \left({a}\right)^2 = c$.

But this is true for $a = 0$, and $g \left({0}\right)^2 + h \left({0}\right)^2 = 0$.

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

But $g \left({a}\right)^2 \ge 0$ and $g \left({a}\right)^2 \ge 0$ from Even Powers are Positive.

So it follows that $g \left({a}\right) = 0$ and $h \left({a}\right) = 0$.

Hence the result.

Geometric Proof

 * [[File: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 $E$:

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

$\forall x, y \in \R^2: d \left({x, y}\right) = \sqrt {\left({x_1 - y_1}\right)^2 + \left({x_2 - y_2}\right)^2}$

where $x = \left({x_1, y_1}\right), y = \left({x_2, y_2}\right)$.

Thus $DB \cong CE \iff d \left({D, B}\right) = d \left({C, E}\right)$.

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

In the above, repeated use is made of the identity $\cos^2 \theta + \sin^2 \theta \equiv 1$ from Sum of Squares of Sine and Cosine.

Now, using the identity $\cos \left({\frac \pi 2 - a}\right) = \sin a$ from Sine equals Cosine of Complement, we have:

Proof of Corollary 1

 * From Cosine Function is Even we have $\cos \left({- b}\right) = \cos b$


 * From Sine Function is Odd we have $\sin \left({- b}\right) = - \sin b$.

Thus:

Similarly:

So:

Proof of Corollary 2

 * $\displaystyle \tan \left({a + b}\right) = \frac {\tan a + \tan b} {1 - \tan a \tan b}$:


 * $\displaystyle \tan \left({a - b}\right) = \frac {\tan a - \tan b} {1 + \tan a \tan b}$:

As above, we have:

Thus:
 * $\tan \left({- b}\right) = - \tan b$

Therefore:

Historical Note
These formulas were proved by François Viète in about 1579.