# Could anyone help me with drawing a  Y=cosx and y=sin2x For 0<=x<=180 (using the graph)  Please!

Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator

Posted on

These functions are very pleasant to draw because we know anything about them:)

1. y(x) = cos(x), 0<=x<=180°.

cos(x) has one root at x=90°, is >0 for 0<=x<90° and is<0 for 90°<x<=180°.
Also, y(0)=1 (the only maximum) and y(180°)=-1 (the only minimum).

And y(x) is monotone decreasing. Also it is convex up for 0<=x<90 and convex down for 90<x<=180, so having a bend at x=90°.

Please look at the pictures cosx.png (created by hand) and cosxw (by a graphing utility, x marked in radians).

2. y(x) = sin2x, 0°<=x<=180°.

y has three roots, at x=0, x=90° and x=180°. y>0 for 0°<x<90° and <0 for 90<x<180. It has one maximum at x=45° (y=1) and one minimum at x=135° (y=-1).

y is increasing from 0 to 45°, decreasing from 45° to 135° and again increasing from 135° to 180°.

It is convex up from 0 to 90 and convex down from 90 to 180°.

Please look at the picture sin2xw.png (by graphing utility).

Images:
This image has been Flagged as inappropriate Click to unflag
Image (1 of 3)
This image has been Flagged as inappropriate Click to unflag
Image (2 of 3)
This image has been Flagged as inappropriate Click to unflag
Image (3 of 3)
Sources:

kseddy123 | College Teacher | (Level 1) Assistant Educator

Posted on

This is simple ....it is as follows

a)

Y=cosx  For 0<=x<=180

sol:-

given ,

`0<=x<=pi `

Now by applying cos ,we get

`cos(0)<=cos(x)<=cos(pi)`

`1<=cos(x)<= (-1)`

so the cos(x) function lies in between -1,1

b)

y= sin 2x for 0<=x<=180

`0<=x<=pi`

multiplying with 2 , we get

` 0<=2x<=2pi`

now applying sin function , we get

` sin(0)<=sin(2x)<=sin(2pi)`

` 0<=sin(2x)<=1`

now let see  the graphs in the attachments given below

Images:
This image has been Flagged as inappropriate Click to unflag
Image (1 of 2)
This image has been Flagged as inappropriate Click to unflag
Image (2 of 2)