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

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


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

where $$\tan$$ is tangent.

Proof from Euler's Formula
$$ $$ $$ $$

By equating real and imaginary parts, we have:


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

Proof from Algebraic Definitions
We have:
 * From the definition of sine: $$\sin x = \sum_{n=0}^\infty \left({-1}\right)^n \frac {x^{2n+1}}{\left({2n+1}\right)!} = x - \frac {x^3} {3!} + \frac {x^5} {5!} - \cdots$$;
 * From the definition of cosine:$$\cos x = \sum_{n=0}^\infty \left({-1}\right)^n \frac {x^{2n}}{\left({2n}\right)!} = 1 - \frac {x^2} {2!} + \frac {x^4} {4!} - \cdots$$.

Let:
 * $$g \left({a}\right) = \sin \left({a + b}\right) - \sin a \cos b - \cos a \sin b$$;
 * $$h \left({a}\right) = \cos \left({a + b}\right) - \cos a \cos b + \sin a \sin b$$;

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

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


 * $$g^{\prime} \left({a}\right) = \cos \left({a + b}\right) - \cos a \cos b + \sin a \sin b = h \left({a}\right)$$;
 * $$h^{\prime} \left({a}\right) = - \sin \left({a + b}\right) + \sin a \cos b + \cos a \sin b = - g \left({a}\right)$$.

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 Basic Properties of Cosine Function we have $$\cos \left({- b}\right) = \cos b$$;


 * From Basic Properties of Sine Function we have $$\sin \left({- b}\right) = - \sin b$$.

Thus:

$$ $$ $$ $$

Similarly:

$$ $$ $$ $$

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

Proof of Corollary 2

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

$$ $$ $$ $$


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

As above, we have:


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

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

Therefore:

$$ $$ $$ $$

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