13. Continuity of Trigonometric Functions

Dated: 30-10-2024

Let us define a constant \(h = x - c\) such that \(x = h + c\).
The benefit of doing so is to give us the ability to replace \(x \rightarrow c\) with \(h \rightarrow 0\).

\[\because \lim_{x \rightarrow 0} \sin(x) = 0\]

\[\because \lim_{x \rightarrow 0} \cos(x) = 1\]

So to prove the continuity of \(\cos\) and \(\sin\) functions, we will modify our definition of continuity.¹ 1. \(f(c)\) is defined. 2. \(\lim_{h \rightarrow 0} f(h + c)\) exists. 3. \(\lim_{h \rightarrow 0} f(h + c) = f(c)\)

Proofs

\[\lim_{h \rightarrow 0} \sin(h + c) = \sin(c)\]

\[\lim_{h \rightarrow 0} \sin(h + c) = \lim_{h \rightarrow 0}\left( \sin(h) \cdot \cos(c) + \sin(c) \cdot \cos(h) \right)\]

\[ = \cos(c) \cdot \lim_{h \rightarrow 0} \sin(h) + \sin(c) \cdot \lim_{h \rightarrow 0} \cos(h)\]

\[= \cos(c) \cdot (0) + \sin(c) \cdot (1)\]

\[= \sin(c)\]

Squeeze Theorem

Let us have 3 functions² such that

\[g(x) \le f(x) \le h(x)\]

If \(\lim_{x \rightarrow a} g(x) = \lim_{x \rightarrow a} h(x) = L\) then \(\lim_{x \rightarrow a} f(x) = L\).

Use case

Let us try to prove why the area of a sector is defined to be:

\[Area = \frac 1 2 r^2 \cdot \theta\]

Where \(r\) is the radius of the circle.

Let us say, the area of this triangle is \(A_1 = \frac 1 2 \cdot 1 \cdot \sin(x)\) that is \(\frac {\sin(x)}{2}\) .
Recall,

\[Area = \frac 1 2 \times base \times perp\]

\(\(\because base = 1\)\)

\(\(\because perp = \sin(\theta)\)\)

Then there is area of the sector that is \(A_2 = \frac 1 2 \cdot 1^2 \cdot x\) that is \(\frac x 2\).

Then there is another triangle.

The area of this triangle is \(A_3 = \frac 1 2 \cdot 1 \cdot \tan(x)\) that is \(\frac{\tan(x)}{2}\).
The base is \(\tan(x)\) because

\[\because \tan(x) = \frac {perp}{base}\]

\[\because base = 1\]

\(\(\therefore \tan(x) = perp\)\)

From visual aids, it is clear that \(A_1 \le A_2 \le A_3\).

\[\frac{sin(x)}{2} \le \frac x 2 \le \frac{\tan(x)}{2}\]

Multiplying all sides by \(\frac{2}{\sin(x)}\).

\[1 \le \frac{x}{\sin(x)} \le \frac{1}{\cos(x)}\]

Taking reciprocals and applying the theorem of inequalities.³

\[1 \ge \frac{x}{\sin(x)} \ge \cos(x)\]

Applying our squeezing theorem, we get

\[\lim_{x \rightarrow 0} 1 \ge \lim_{x \rightarrow 0} \frac{x}{\sin(x)} \ge \lim_{x \rightarrow 0} \cos(x)\]

\[1 \ge 1 \ge 1\]

Hence, the formula for area of the sector of a circle holds true.

13. Continuity of Trigonometric Functions

Proofs

Squeeze Theorem

Use case

References