Rigorous Proofs of Important Trig Facts and Formulas
Here we will prove some of the most fundamental facts about sine and cosine from their definitions as power series.
1. Definitions
We begin with the definitions
\[{}\sin x{}\]
\[{}:= \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!},{}\]
\[{}\cos x{}\]
\[{}:= \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!}.{}\]
We can use the Weierstrass \(M\)-test to demonstrate that these functions are well-defined for all \(x \in {\mathbb R}\) and are everywhere continuous: if \(|x| \leq R\),
\[{}\left| \frac{(-1)^n x^{2n+1}}{(2n+1)!}\right|{}\]
\[{}\leq \frac{R^{2n+1}}{(2n+1)!}{}\]
and
\[{}\sum_{n=0}^\infty \frac{R^{2n+1}}{(2n+1)!}{}\]
\[{}\leq \sum_{n=0}^\infty \frac{R^k}{k!} = \exp(R) < \infty{}\]
and likewise for the sum of \(R^{2n}/(2n)!\) (exercise: prove \(\exp(R) < \infty\)). We can further see from the series definitions that \((\sin x)' = \cos x\) because the term-by-term derivative of the series for \(\sin\) is exactly the series for \(\cos\), and since we know the series for cosine converges uniformly, we may conclude that the derivative of the infinite series equals the infinite series of the term-by-term derivatives. The formula \((\cos x)' = - \sin x\) follows similarly after we reindex. Since (up to \(\pm 1\)) the functions sine and cosine are derivatives of one another, this implies in particular that they are infinitely differentiable and hence all derivatives must also be continuous.
2. Most Basic Properties
Theorem (Sine Squared Plus Cosine Squared is One)
For any \(x \in {\mathbb R}\),
\[ \cos^2 x + \sin^2 x = 1. \]
Proof
To prove the famous identity \(\cos^2 x + \sin^2 x = 1\), observe that
\[{}\frac{d}{dx} \left( \cos^2 x + \sin^2 x \right){}\]
\[{}= 2 (-\sin x) (\cos x) + 2 (\cos x) (\sin x){}\]
\[{}= 0,{}\]
which implies that \(\cos^2 x + \sin^2 x\) is a constant function of \(x\); evaluating at \(x = 0\) and using the Taylor series definitions again (i.e., \(\sin 0 = 0\) and \(\cos 0 = 1\)) gives that the value of the constant equals \(1\).
Proposition
There exists \(\pi \in (0,\infty)\) such that \(\cos \frac{\pi}{2} = 0\) and \(\cos x > 0\) for all \(x \in [0,\frac{\pi}{2})\). Furthermore \(\sin x\) is nonnegative and strictly increasing on the interval \([0,\frac{\pi}{2}]\).
Proof
Because \(\cos x\) is continuous and \(\cos 0 = 1\), we know that there must exist some \(\rho >0\) such that \(\cos x \geq 0\) on the interval \([0,\rho]\). Let \(S\) be the set of all real numbers \(\rho\) such that \(\cos x \geq 0\) on the interval \([0,\rho]\). If \(\rho_1 < \rho_2\) are both positive numbers and \(\rho_2 \in S\), then clearly \(\rho_1 \in S\) also. For any \(\rho \in S\), since \(\cos x \geq 0\) on \([0,\rho]\), \(\sin x\) must be increasing on \([0,\rho]\) because \((\sin x)' = \cos x\). In particular, \(\sin x \geq 0\) on \([0,\rho]\) as well.
There is a fundamental dichotomy which must be true of the set \(S\): either it is bounded above or it isn't. We will show that it must be bounded above. To that end, suppose not; we know there is some \(\rho \in S\) (i.e., it's not empty). If \(S\) were not bounded above, then it would necessarily have to contain all positive real numbers. But then for any \(x \geq \rho\), we would have
\[ \cos x - \cos \rho = (- \sin \xi) (x - \rho)\]
for some \(\xi\) between \(x\) and \(\rho\); because \(\sin x\) would have to be an increasing function on \([0,\infty)\) if \(S\) were unbounded, this would mean that \(\sin \xi \geq \sin \rho\), and consequently that
\[ \cos x \leq \cos \rho - (\sin \rho) (x - \rho). \]
Since \(\sin \rho > 0\), taking \(x\) very large (greater than \(\rho + \cos \rho / \sin \rho\)) shows that \(\cos x\) must eventually take on negative values. Thus \(S\) must be a bounded set after all.
Since \(S\) is bounded and nonempty, it has a finite supremum. Let \(\tau\) be the infimum of all such \(\rho\) for which \(\cos x \geq 0\) on the interval \([0,\rho]\). Because \(\tau\) is a least upper bound, \(\cos x\) must be greater than or equal to zero on \([0,\tau-\epsilon]\) for every \(\epsilon > 0\), and by continuity of \(\cos x\), it follows that \(\cos \tau \geq 0\). In fact, we must have that \(\cos \tau = 0\), since if it were true that \(\cos \tau > 0\), then cosine would be nonnegative on some interval \([0,\tau + \epsilon]\) for some positive \(\epsilon\) (contradicting the minimality of \(\tau\)).
Finally, observe that if \(\cos x_0 = 0\) for any \(x_0 \in [0,\tau)\), then as we showed above, it would follow that
\[ \cos x \leq - (\sin x_0) (x - x_0) \]
for all \(x \geq x_0\), which would force \(\cos x\) to be strictly negative for \(x \in (x_0,\tau)\). Thus it cannot happen that \(\cos x_0 = 0\). If we define \(\pi := 2 \tau\), then the proposition is complete.
We note that the value of \(\pi\) identified by the proposition must be unique, since if \(\pi < \pi'\) were two numbers satisfying the conclusion of the proposition, then it would follow that \(\cos \frac{\pi}{2} = 0\) but also that \(\cos \frac{\pi}{2} > 0\) because \(\frac{\pi}{2} \in (0,\frac{\pi'}{2})\). Because \(\sin \frac{\pi}{2}\) is positive and because \(\cos^2 \frac{\pi}{2} + \sin^2 \frac{\pi}{2} = 1\), it must be true that \(\sin \frac{\pi}{2} = 1\) as well.
Proposition
Let \(u,v\) be positive real numbers such that \(u^2 + v^2 = 1\). Then there exists a unique \(\theta \in [0,\frac{\pi}{2}]\) such that \(\cos \theta = u\) and \(\sin \theta = v\).
Proof
Because \(\cos 0 = 1\) and \(\cos \frac{\pi}{2} = 0\), by the Intermediate Value Theorem, there must be some \(\theta \in [0,\frac{\pi}{2}]\) at which \(\cos \theta = u\). Furthermore, since the derivative of \(\cos x\) is strictly positive for \(x \in (0,\frac{\pi}{2})\), cosine must be strictly monotone on \([0,\frac{\pi}{2}]\), so this \(\theta\) must be unique. Lastly, since \(1 = u^2 + v^2 = \cos^2 \theta + \sin^2 \theta\), the equality \(u = \cos \theta\) implies \(v^2 - \sin^2 \theta = 0\), i.e., \((v - \sin \theta)(v + \sin \theta) = 0\). Since both \(v\) and \(\sin \theta\) are nonnegative, they are either both zero (and hence \(v = \sin \theta\)) or \(v + \sin \theta > 0\), which then implies \(v - \sin \theta = 0\) (hence \(v = \sin \theta\)).
Theorem (Angle Addition Formulas I)
For any \(x, y \in {\mathbb R}\),
\[ \cos (x+y) = \cos x \cos y - \sin x \sin y \]
and
\[ \sin (x + y) = \sin x \cos y + \cos x \sin y. \]
Proof
It suffices to show the modified formulas
\[{}\cos y{}\]
\[{}= \cos x \cos (y-x){}\]
\[{}- \sin x \sin (y-x) {}\]
and
\[{}\sin y{}\]
\[{}= \sin x \cos (y-x){}\]
\[{}+ \cos x \sin (y-x).{}\]
To prove it, we differentiate the right-hand sides with respect to \(x\):
\[{}\frac{d}{dx} \left[ \cos x \cos (y-x) - \sin x \sin (y-x)\right]{}\]
\[{}= - \sin x \cos (y-x){}\]
\[{}+ \cos x \sin (y-x) {}\]
\[{}- \cos x \sin (y-x){}\]
\[{}+ \sin x \cos (y-x) = 0,{}\]
\[{}\frac{d}{dx} \left[ \sin x \cos (y-x) + \cos x \sin (y-x) \right]{}\]
\[{}= \cos x \cos (y-x){}\]
\[{}+ \sin x \sin (y-x){}\]
\[{}- \sin x \sin(y-x){}\]
\[{}- \cos x \cos (y-x) = 0.{}\]
As was argued above, this means that the functions \(\cos x \cos (y-x) - \sin x \sin (y-x)\) and \(\sin x \cos (y-x) + \cos x \sin (y-x)\) are constant functions of \(x\) for each \(y \in {\mathbb R}\), and to conclude, the constant can be computed by fixing \(x = 0\).
Corollary (Special Angle Addition Formulas)
For each \(x \in {\mathbb R}\),
\[{}\cos x{}\]
\[{}= \sin\left(x + \frac{\pi}{2}\right){}\]
\[{}= - \cos (x + \pi ){}\]
\[{}= - \sin\left(x + \frac{3\pi}{2}\right){}\]
\[{}= \cos \left(x + 2 \pi \right),{}\]
\[{}\sin x {}\]
\[{}= - \cos \left(x + \frac{\pi}{2}\right){}\]
\[{}= - \sin(x + \pi){}\]
\[{}= \cos \left(x + \frac{3\pi}{2} \right){}\]
\[{}= \sin(x + 2 \pi).{}\]
Proof
These are direct applications of the angle addition formulas just proved. We have already seen that \(\cos 0 = 1\) and \(\sin 0 = 0\) directly from the Taylor series definitions. We have subsequently shown that \(\cos \frac{\pi}{2} = 0\) and \(\sin \frac{\pi}{2} = 1\), so using the angle addition formulas with \(y = \frac{\pi}{2}\) gives the first set of formulas. Evaluating these formulas at \(x = \frac{\pi}{2}\) gives \(\cos \pi = -1\) and \(\sin \pi = 0\). Repeating the process similarly gives \(\cos \frac{3 \pi}{2} = 0\), \(\sin \frac{3\pi}{2} = -1\), and \(\cos 2 \pi = 1\), \(\sin 2 \pi = 0\).
3. Relationship to the Unit Circle
Corollary (Parametrizing the Unit Circle)
For any real numbers \(u,v\) such that \(u^2 + v^2 = 1\), there exists a unique \(\theta \in [0,2 \pi)\) such that \(\cos \theta = u\) and \(\sin \theta = v\).
Proof
We argue systematically based on the known signs of \(\sin x\) and \(\cos x\) on the intervals \([0,\frac{\pi}{2}], [\frac{\pi}{2}, \pi], [\pi , \frac{3 \pi}{2}],\) and \([\frac{3 \pi}{2},2 \pi]\). In particular, we know that cosine and sign are (1) both positive, (2) negative and positive, (3) negative and negative, and (4) positive and negative, respectively, at all points except for endpoints (where \(\sin\) and \(\cos\) may be \(0\) in some cases as we have already indicated). So if neither \(u\) nor \(v\) equals zero, then the desired \(\theta\) can belong to only one of these four intervals, at which point we use the shift formulas just proved together with the unique \(\theta \in [0,\frac{\pi}{2}]\) already shown to exist when \(u\) and \(v\) are nonnegative. If one or more of \(u\) or \(v\) is zero, we know that \(\theta\) must equal an integer multiple of \(\frac{\pi}{2}\), which we can then solve by hand since all values of sine and cosine at these points are known.
Theorem (Minimal Periods of Sine and Cosine)
There does not exist any positive real number \(\rho < 2 \pi\) such that \(\cos (x + \rho) = \cos x\) for all \(x \in {\mathbb R}\). Likewise there is no \(\rho \in (0,2 \pi)\) such that \(\sin (x + \rho) = \sin x\) for all \(x \in {\mathbb R}\). Thus \(2 \pi\) is the unique smallest positive number which is a period for the functions sine and cosine.
Proof
If there were a \(\rho\) such that \(\cos (x + \rho) = \cos x\) for all \(x\), then it would follow that \(1 = \cos 0 = \cos \rho\), and consequently that \(\sin \rho = 0\). But there is only one \(\theta \in [0,2 \pi)\) at which \(\cos \theta = 1\) and \(\sin \theta = 0\). Similarly, if \(\sin (x + \rho) = \sin x\), then \(\sin \rho = 0\), which means that \(\cos \rho = \pm 1\), and this then forces either \(\rho = 0\) or \(\rho = \pi\); however, in this latter case, we would have \(\sin (x + \pi) = - \sin x\), which is not identically equal to \(\sin x\) because \(\sin x\) is not itself identically zero.