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

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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).

Please ask me if anything is unclear or missing.

**Images:**

**Sources:**

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:**