YOU are better than YOU think. Show yourself how:
|
-/[]\- Logic
Mastery
|
-/[]\- |
Derivatives of sin(x) and cos(x) at x = 0
3 Preliminary Limit Calculations.
Let h be a small positive real number. < (¼)p. Then h drawn in standard position gives an angle in the first quadrant.
The area A of the sector is proportional to its radian measure. Here A = k h for some constant h. But when h = 2p, the Area = p = k 2p. So k = ½. Therefore the area of the sector determine by h is A = (½)h.
From the diagram (impure mathematics), we see
| small triangle area |
< | sector area |
< | big triangle area |
Therefore from the (½)(base)(height) triangle area formula we find
| (½)cos(h)sin(h) | < | (½)h. | < | (½)(1) tan(h) |
This leads to two in-equalities
| cos(h)sin(h) | < | h. | or | sin(h) h |
< |
1 cos(h) |
|
| h | < | tan(h) . | = | sin(h) cos(h) |
and hence | ||
| cos(h) | < |
sin(h) |
|||||
Therefore, we have the inequality sandwich for h > 0.
| cos(h) | < |
sin(h) |
< | 1 cos(h) |
Here cos(h) is an even function and so is the ratio sin(h)/h. Therefore this inequality sandwich holds for h < 0 as well.
4 Derivative of sin(x) at x = 0
Now both
| cos(h) |
and its reciprocal |
1 cos(h) |
approach the value 1 as h approaches zero. Therefore, for every whole n, we can make find an m > 0 such that
0< |h| < (½)10-m
implies |cos(h) - 1| < (½)10-n
and | < (½)10-n and
| |1 - | 1 cos(h) |
| | < | (½)10-n |
Therefore the inequality sandwich implies
| |1 - |
sin(h) |
| | < | (½)10-n |
Therefore our decimal view of limits implies
| lim h® 0 |
sin(h) |
= 1 |
since sin(x) = 0 when x = 0, we conclude sin'(x) = 1 when x = 0. That is,
| sin'(0) = | lim h® 0 |
sin(h) - sin(0) |
= | lim h® 0 |
sin(h) |
= 1 |
So the foregoing limit is useful and will be more useful in calculating the derivatives of cos(x) and sin(x) below.
Conclusion:
Remark: The value of this limit follow because of the use of radian measure in place of degrees. |
5. Derivative of cos(x) at x = 0.
For the calculation of derivatives of trig functions, we need to evaluate one more limit, that of
|
cos(h) - 1 |
as h approaches zero. Since cos(0) = 1, we are seeking here the derivative of cos(x) at x = 0.
Now
|
cos(h) - 1 |
= |
cos(h) - 1 |
cos(h) + 1 cos(h) + 1 |
|
|
= |
cos2(h) - 1 |
1 cos(h) + 1 |
|
|
= |
(-1)sin2(h) |
1 cos(h) + 1 |
|
|
= |
sin(h) |
(-1) sin(h) cos(h) + 1 |
| = | |||
Therefore
|
lim |
cos(h) - 1 |
= |
lim |
sin(h) |
(-1) sin(h) cos(h) + 1 |
|
|
|
= |
1 |
(-1) 0 0 + 1 |
|||
| = |
0 |
Conclusion:
|
|
|
www.whyslopes.com
More Calculus
Volumes
To Learn More, visit Volumes 2 and 3.
Advanced Topics
To Learn More: Visit Real Analysis.