Cosine of Sum

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

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

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

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

Proof from Euler's Formula
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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 $$D_a \left({g \left({a}\right)^2 + h \left({a}\right)^2}\right)$$

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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.

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:

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Similarly:

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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

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

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

As above, we have:


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

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

Therefore:

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