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Sine Graph

The sine function is a periodic function meaning that it repeats itself every so many (in the case of sine 2pi radians or 360o). It has a range of -1 to 1 and has a domain for -∞ to ∞. Starting at the origin it increase to 1 at 90<sup>o</sup> or pi/2 radians and then decrease to -1 at 270<sup>0</sup> or 3pi/2 radians and then returns to 0 and 360<sup>o</sup> or 2pi radians.

On the graph below the angle, in radians, is along the x axis and the value of the sine function for that angle is on the y axis.

Graph of y=sin(x)

Graph of y=sin(x)

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Categories: maths, trigonometry Tags: , ,

Sine and Cos Graphs Differentiating sin and cos

June 23, 2008 7 comments

This is the basics of the sine cos and tan graphs and how sine and cos relate to give you tan. It also shows how to differentiate sin and cos.

The output or range of both sine and cos is from -1 to 1 when given any angle. They can be shown on a graph where y = sin(x) and y = cos(x). In these graphs all the angles go along the x axis and you can see a wave type shape is formed

Sine Graph
graph of y=sin(x)

Cosine Graph
cosine graph

As you can see both the sin and cos graphs move periodically between -1 and 1 as the angles change, this pattern continues indefinitely because once you pass 360 degrees or 2 pi radians you will return back to the beginning. If you try to perform sin-1 of a value out side the range -1 to 1 you will get an error.

Differentiate Sin and Cos
also notice that the gradient of the sin graph is the value of the cos graph for the same angle and that the gradient of the cos graph is the -value of the sin graph for that angle. This means that we can differentiate the sin and cos graphs:
if f(x) = sin(x) then f ‘ (x)=cos(x)
and
if f(x) = cos(x) then f ‘ (x) = -sin(x)

however if we use ax instead of x we must differentiate it by bringing the a out, when its just x this doesn’t matter as the differential of x is 1.
ie)
let y = sin(f(x))
now let u = f(x)
du/dx = f ‘ (x)
also
y=sin(u) as u = f(x)
dy/du = cos(u)

from the chain rule

dy/dx = du/dx * dy/du
therefore
if y = sin(f(x))
dy/dx = f ‘ (x)cos(f(x))

and similarly for cos
if y = cos(f(x))
dy/dx = -f ‘ (x)sin(f(x))