Where is Sin and Cos Negative Continuous

Limit and Continuity of Trigonometric Functions

Continuity of Sine and Cosine function

Sine and Cosine are ratios defined in terms of the acute angle of a right-angled triangle and the sides of the triangle.

Here is the graph of Sinx and Cosx
-We consider angles in radians
-Insted of θ we will use x

f(x) = sin(x)

g(x) = cos(x)

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It is evident that as h approaches 0, the coordinate of P approach the corresponding coordinate of B.

But by definition we know
sin(0) = 0 and cos(0) = 1
The values of the functions matche with those of the limits as x goes to 0 (Remind the definition of continuity we have).
lim _{x → 0} sin(x) = sin(0) = 0 lim _{x → 0} cos(x) = cos(0) = 1

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Hence we have the following theorem
      DEFINITION 2.7.1
A function f(x) is said to be continuous at a point c if the following conditions are sarisfied
-f(c) is defined
-lim _{x → c} f(x) exists
-lim _{x → c} f(x) = f(c)

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      THEOREM 2.8.1
The functions sin(x) and cos(x) are continuous

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      Proof
Let h = x - c. So x = h + c. Then x → c is equivalent to the requirement that h → 0
A function f(x) is continuous at c if the following conditions are true:
-f(c) is defined
-lim _{h → 0} f(h + c) exists
-lim _{h → 0} f(h + c) = f(c)

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      Assume
lim _{x → 0} sin(x) = 0 and lim _{x → 0} cos(x) = 1
The first two conditions of continuity definition are met. We have to show that
lim _{h → 0} sin(c + h) = sin(c)

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      Now
lim _{h → 0} sin(c + h) = lim _{h → 0} [sin(c)cos(h) + cos(c)sin(h)] = lim _{h → 0} sin(c)cos(h) + lim _{h → 0} cos(c)sin(h) = sin(c)lim _{h → 0} cos(h) + cos(c)lim _{h → 0} sin(h) = sin(c)(1) + cos(c)(0) = sin(c)

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      Continuity of Other Trigonometric Functions
tan(x) = sin(x)/cos(x)
tan(x) is continuous everywhere except at cos(x) = 0 which gives
x = ± φ/2, ± 3φ/2, ± 5φ/2, ... = ± kφ/2 (k = 1, 3, 5, ...)

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Similarly, since
cot(x) = cos(x)/sin(x)
sec(x) = 1/cos(x)
cosec(x) = 1/sin(x)
They are all continuous on appropriate ontervals using the continuity of sin(x) and cos(x).

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      Obtaining Limits by Squeezing
We will use Squeeze Theorem for finding limits
lim _{x → 0} sin(x)/x = 1
lim _{x → 0} [1 - cos(x)]/x = 0 Consider the graph of

And the graph of

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Here the problem is:
- As x goes to 0, both thr top and the botton functions go to 0.
- sin(x) goes to 0 means that the fraction as a whole goes to 0.
- x goes to 0 means that the function as a whole goes to +∞.

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Here we cannot write these functions in some other form by using algebraic Manipulation to solve this problem. So here we will use some other method. One such method is to obtain by following theorem:

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      Squeezing theorem
Let f, g and h be fucntions satisfying g(x)≤f(x)≤h(x) for all x in some open interval containg the point a, whit the possible exception that the inequalities need not hold at a.
If g and h have the same limits as x approaches a, say
lim _{x → a} g(x) = lim _{x → a} h(x) = L
then f also has this limit as x approaches a, that is
lim _{x → a} f(x) = L

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      Example:
Use the squeezing theorem to evaluate
lim _{x → 0} x ² sin ² (1/x)
Solution
Remember that 0 ≤ sin(x) ≤ 1, so 0 ≤ sin ² (x) ≤ 1 and so 0 ≤ sin ² (1/x) ≤ 1
Multiply throughtout this last inequality by x ²
0 ≤ x ² sin ² (1/x) ≤ x ²
But lim _{x → 0} 0 = lim _{x → 0} x ² = 0
So by the Squeezing Theorem
lim _{x → 0} x ² sin ² (1/x) = 0

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Before proving next theorem, let us see the following formula.
The proof will use basic facts about circles and areas of sectors with angle θ radians and raius r

The area of a sector is given by
A = (1/2).r ² θ

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      THEOREM 2.8.3
lim _{x → 0} sin(x)/x = 1
Let x be such that 0 < x < φ/2 or -φ/2 < x < 0
We made the assumption that
0 < x < φ/2
Also works when -φ/2 < x < 0

From the figure
0 < area of ΔOBP = (1/2) base.height = (1/2) (1).sin(x) = (1/2) sin(x)
area of sector OBP = (1/2)(1) ² .x = (1/2)x
area of ΔOBQ = (1/2) base.height = (1/2) (1).tan(x) = (1/2)tan(x)

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So, the above inequality becomes
0 < (1/2)sin(x) < (1/2)x < (1/2)tan(x)
Multiplying throughout by 2/sin(x)
1 < x/sin(x) < 1/cos(x)
Taking resiprocal gives
cos(x) < sin(x)/x < 1

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Taking limit now and using Squeezing Theorem gives
lim _{x → 0} cos(x) < lim _{x → 0} sin(x)/x < lim _{x → 0} 1 = 1 < lim _{x → 0} sin(x)/x < 1
Since the middles term is between 1 and 1, it must be 1

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Source: https://www.math10.com/en/algebra/functions/continuity-sine-cosine-function/continuity-sin-cos-function.html

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