More C2 trig

Last 2 questions of chapter = struggling in a couple of the parts - will rep asap.

19bii) The diagram shows part of the curve with equation y=cos(px-q)degrees, where p and q are positive constants and q<180. The curve cuts the x-axis at points A,B and C as shown.

Given that the coordinate of A and B and C are (100,0) , (220,0) and (340,0) respectively, find that value of p and the value of q
Tell me if you need the diagram

20c) The diagram shows part of the curve with equation y = 1 + 2sin(pxtheta + qtheta), p and q being positive constants and q< 90. The curve cuts the y axis at point A and hte x axis and the points C and D. Point B is the maximum point on the curve.

Given that the coordinate of A and C are (0,2) and (45,0) respectively

calculate the value of q - done = 30
show that p = 4 done
Find the coordinates of B and D

Diagram is attatched for that question (really bad though lol)

Thanks all for your help in advance,

Regards, Asad

19)b)ii)
A: 0= cos(100p-q)
B: 0= cos(220p-q)
C: 0= cos(340p-q)
cos X=o
X=90,270,450...
let,
100p-q=90 (a)
220p-q=270 (b)

(b-a) 120p=180
p=1.5 and q=60

Reply 2

danniella

4

Hello!

Here's 19ii)
You know that cos(px-q) = 0
cos^-1(0) gives 90 or 270 or 450
which means that (px-q) = 90, 270 and 450

You also know that cos(px-q) = 0 when x=100, 220 or 340

So you substitute those values in for x in (px-q)=90, 270, 450
which gives you

px-q = 90, but x=100
100p-q=90

Similarly
px-q = 270, but x=220
220p-q = 270

You now have two simultaneous equations.
100p-q = 90
220p-q = 270

Multiply the top one by 2.2
220p-2.2q=198
220p-q =270

this gives 1.2q=72 and so q=60.
Substitute back into one of your equations:
100p-q=90
100p-60=90
100p=150
p = 1.5

love Danniella

Reply 3

danniella

4

Second part...

(Actually, by "theta" you mean "degrees... at least in my copy of C2! :smile:

)
So looking at
f(x) = 1 + 2sin(4x+30)
The point B is at the top of the curve (the maximum)
Max value of sin (whatever) occurs at sin (whatever)=1
So the equation has its max value when sin(4x+30)=1, i.e. at 1+2(1)=3

sin(4x+30)=1
4x+30 = 90 (first value is the only one needed)
4x=60
x=15

co-ordinates of B are (15,3)

Then for the value of D,
1+ 2sin(4x+30) = 0
Sin(4x+30) = -0.5
4x+30=(180+30), (360-30)
4x+30=210, 330
4x=180, 300
x=45, 75

Co-ords (45,0) are those of C
and D is (75,0)

love Danniella

Reply 4

asadtamimi

OP

13

Excellent - thanks a lot daniella - just repped you

Reply 5

r0manticide17

3

Absolutely nothing to do with the topic...but message to the original poster : Regarding your GCSE grades : Don't you think we can count?

Reply 6

username5632290

1

Original post by asadtamimi

Last 2 questions of chapter = struggling in a couple of the parts - will rep asap.

19bii) The diagram shows part of the curve with equation y=cos(px-q)degrees, where p and q are positive constants and q<180. The curve cuts the x-axis at points A,B and C as shown.

Given that the coordinate of A and B and C are (100,0) , (220,0) and (340,0) respectively, find that value of p and the value of q
Tell me if you need the diagram

20c) The diagram shows part of the curve with equation y = 1 + 2sin(pxtheta + qtheta), p and q being positive constants and q< 90. The curve cuts the y axis at point A and hte x axis and the points C and D. Point B is the maximum point on the curve.

Given that the coordinate of A and C are (0,2) and (45,0) respectively

calculate the value of q - done = 30
show that p = 4 done
Find the coordinates of B and D

Diagram is attatched for that question (really bad though lol)

Thanks all for your help in advance,

Regards, Asad

Hi, sorry how did you get q=30?