Theorem: lim _{t -> 0} sin (t) = 0.

Note: since sin(0) = 0, this says that sin(x) is continuous at the origin.

This result follows quite easily from the inequality

0 < sin(t) < t, for t>0 and

t < sin(t) <0, for t<0.

By the squeeze theorem , lim _{t -> 0} sin (t) = 0.

Note: this follows since sin²t + cos² t = 1.

Corollary:

lim _t->0 cos(t) = 1 Theorem: lim _t->0 sin(t)/t = 1.

This follows immediately from the inequalities -- t < tan(t), that is cos(t) < sin(t)/t.

Since we know sin(t)/t < 1, we have

cos(t) < sin(t)/t < 1

and by the squeeze theorem, as t->0, we have lim _t->0 sin(t)/t = 1.

Corollary:

lim _t->0 (cos(t)-1)/t = 0 With these preliminaries under our belt, we can show that

d/dx sin(x) = cos(x) which follows by trigonometric identities and the fundamental inequalities. Once we know the derivative of the sin() function, we have

• d/dx sin(x) = cos(x)
• d/dx cos(x) = -sin(x)
• d/dx tan(x) = sec ² x
• d/dx csc(x) = -csc(x)cot(x)
• d/dx sec(x) = sec(x)tan(x)
• d/dx cot(x) = -csc²x

Because the above formulae hold for all real x, it follows that all trigonometric functions are continuous on their domains (that is excluding obvious divisions by zero...)

Last modified Mon Sep 30 22:27:40 CDT 1996