Sine and Cosine of Sum/Proof from Algebraic Definitions

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.

Proof

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:

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

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

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

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

Hence:

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

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.

$\blacksquare$