Math 13200
(Don’t freak out you don’t have to know this picture)
Today, we investigate the derivatives of trigonometric functions.
Just like on Wednesday, we hope that the symmetry of the circle/trigonometric identities will allow us to compute these derivatives by first looking at one θ -value. Let f ( θ ) = sin θ . Want to find f
0
( θ ). Again, we investigate θ = 0 first.
sin
cos
0
If we let L be the tangent line of the unit circle C at A = (1 , 0), we see that when we use a microscope at the point A , the distinction between C and L decreases rapidly as we zoom in.
So let Q
θ be the point on the line L obtained by moving up L a distance of θ , and let g ( θ ) be the
1

y -value.
y
L
O sin θ
θ cos θ
P
θ
Q
θ
A x
P
θ
Q
θ
A
P
θ
Q
θ
A
When θ is very small (i.e. close to 0), P
θ and Q
θ are “indistinguishable under a microscope”.
Since the derivative is the essentially the linear approximation that is “indistinguishable under a microscope” we hope f
0
(0) = g
0
(0)
Indeed this is the case. The book has a rigorous proof using areas of circular sectors and triangles and the squeeze theorem.
Now what is g ( θ )? Its just θ !
L is just a vertical line, so moving up θ takes you to the point (1 , θ ), whose y -value is θ . Thus, g
0
( θ ) = 1 from which we conclude f
0
(0) = 1.
A similar argument gives us that the derivative of cos at 0 is 0.
In terms of limits, what do the relations sin
0
(0) = 1 cos
0
(0) = 0 say? Recall the limit definition of derivative: f
0
( c ) = lim h → 0 f ( c + h ) − f ( c ) h
Thus, plugging in for sin, we see that sin
0
(0) ⇔ 1 = lim h → 0 sin h h
2

Similarly, cos
0
(0) ⇔ 0 = lim h → 0 cos h − 1 h
Main Trigonometric Limits lim h → 0 sin h h
= 1 lim h → 0 cos h − 1 h
= 0
In standard calculus courses, these limits are often stated or proved before introducing the derivative.
But the best intuition for these limits is as a derivative, so their value is hard to understand at first.
Let’s take a closer look at the function f ( x ) = sin x x
First rewrite it this way: f ( x ) = sin x − 0 x − 0
= slope of line through O and ( x, sin x )
Thus, f ( x ) is the slope of a secant line for the graph of sin x ! Let’s analyze the secants of the graph of sin x through the origin:
3

At x = 0, the function f ( x ) is undefined. But the graph suggests (and our knowledge of the value of the limit) that when x = 0 .
0000001, x is just smaller than 1. As x increases, f ( x ) oscillates above and below the x -axis. But the oscillations get smaller as x gets bigger, because the secant lines wobble less and less.
Conversely, as x gets close to 0, this picture let’s us remember how we view derivatives geometrically.
The tangent line is the “limit of the secant lines” at a point, and the derivative is that slope. Saying that these slopes approach 1 is equivalent to the derivative of sin at 0 being 1.
sin
cos
We now use these limits and the angle addition formula to derive the derivative of trigonometric functions: sin
0
( θ ) = lim h → 0 sin( θ + h ) − sin θ h sin θ cos h + cos θ sin h − sin θ
= lim h → 0
= lim h → 0
= lim h → 0
= sin θ h sin θ (cos h − 1) + cos θ sin h h sin θ (cos h − 1) lim h → 0 h cos h h
− 1
+ lim h → 0 cos θ sin h h sin h
+ cos θ lim h → 0 h
= cos θ
A similar argument (which you show in your homework) will show that cos
0
( θ ) = − sin θ
Using the trig identities and derivative rules, one can then deduce the formulas for tan , cot , sec , csc.
4

D x sin x = cos x
D x cos x = − sin x
D x tan x = sec
2 x
D x cot x = − csc
2 x
D x sec x = sec x · tan x
D x csc x = − csc x · cot x
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