What is the derivative function of sin(x)?

Question

Maybe this will help me grasp the concept of trigonometry :-/

Haha, duh. I should've figured. Well, you guys are smart, can you explain to me how sine & cosine were derived? Like what really is the relationship between an angle to a line dependent on another line's length?

Answer 1

the derivative of sin(x) is cos(x). i don't know if that knowledge would help you grasp trigonometry - basically, you should understand trig before you take calculus, anyway.

EDIT

are you wondering about the definition of sin(x) and cos(x)? If you take a right triangle ABC with hypotenuse BC and right angle A, then consider theta = angle BCA with vertex at C and rats BC and CA. sin(theta) = AB/BC and cos(theta) = AC / BC.

if you're talking about the derivations of the derivatives of sin(x) and cos(x), it's a little complicated. You need to go back to the definition of a derivative, let f(x) = sin(x), and you need to use a product to sum formula and the fact that lim x==>0 sinx/x = 1. The derivation is annoying to type in, but go to this website for details - it shows all the steps and is pretty good:
http://www.maths.abdn.ac.uk/~igc/tch/ma1002/diff/node27.html

good luck!

Answer 2

Draw an x -axis and a y- axis perpendicular to it.
The intersectionn of these two axes is called the origin. Label the origin as point O. Now draw a circle with center at the origin and radius =1. This is called the unit circle.

Now imagine that the x- axis is the initial side of an angle. Now rotate the radius about 30 degrees CCW. the radius is now the terminal side of a 30 degree angle. Call the point where the radius meets the circle B. From b draw a perpendcular to the x=axis meeting the x=axis at point A. OAB is a right triangle with angle OAB A being the right angle. The sine of the angle BOA is defined to be the ratio of the side opposite angle over the radius
which is AB/r or just AB since r=1.

Can you see that as you rotate the radius more and more CCW, AB becomes larger and large until at 90 degrees it becomes 1 sinnce ab/r =1/1=1 at that point . That is why we say sin 90 degrees =1. If you rotate the radius back so that the radius is on top of the x-axis then the angle is 0 degrees and sin 0 = 0 since AB=0 and 0/r =0 So the sine of the angle changes from 0 to 1 as the radius rotates from the x-axis to the y-axis. This ratio is always the same. Use you calcuulator to look at the value of different angles such as 30, 45, 60 ,90, 135, 180, 270, and 360.

Answer 3

It looks like you want to know how the trig functions were derived.
Consider a right triangle ABC, where C is the right angle. Let a, b, & c be the respective sides opposite the angles.
Then the trig functions are defined as follows:
sin(A) ≡ a/c (sometimes said opposite over hypoteneus)
cos(A) ≡ b/c (sometimes said adjacent over hypoteneus)
tan(A) ≡ a/b (sometimes said adjacent over hypoteneus)

You can form three other ratios from the sides, but these are rarely used:
c/a ≡ cosecant(A) = csc(A)
c/b ≡ secant(A) = sec(A)
b/a = cotangent(A) = ctn(A)

Angles A and B are complimentary, so the trig functions are complimentary as well:
sinB ≡ b/c = cosA
cosB ≡ a/c = sinA
tanB ≡ b/a = ctnA

Answer 4

The derivative of sin(x) is cos(x). This can be proven using the definition of a derivative, which is defined to be
f'(x) = limit as h approaches 0 of [f(x+h) - f(x)]/h

The algebra itself is not straightforward, as other limits are involved (such as lim (x -> 0, sinx/x) = 1).

Answer 5

the derivative of this is cos(x)

dont know how that will help with trig though.

Answer 6

cos(x)

Answer 7

cos(x)

Answer 8

-cosx

Answer 9

well duh cosine(x).