trigonometric equations

• Apr 19th 2009, 08:35 AM
Tweety
trigonometric equations
The diagram shows part of the curve with equation y=f(x) , where f(x)=1+2sin(px°+q°) , p and q being positive constants and q ≤ 90. The curve cuts the y-axis at the point A and the x-axis at the points C and D. The point B is a maximum point on the curve.

Given that the coordinates of A and C are (0, 2) and (45, 0) respectively:
(a) Calculate the value of q.
(b) Show that p=4.
(c) Find the coordinates of B and D.

I can do question a, but I am really stuck on b and c.
• Apr 19th 2009, 10:44 AM
skeeter
Quote:

Originally Posted by Tweety
The diagram shows part of the curve with equation y=f(x) , where f(x)=1+2sin(px°+q°) , p and q being positive constants and q ≤ 90. The curve cuts the y-axis at the point A and the x-axis at the points C and D. The point B is a maximum point on the curve.

Given that the coordinates of A and C are (0, 2) and (45, 0) respectively:
(a) Calculate the value of q.
(b) Show that p=4.
(c) Find the coordinates of B and D.

I can do question a, but I am really stuck on b and c.

$\displaystyle f(0) = 2 = 1 + 2\sin(0+q)$

$\displaystyle \sin(q) = \frac{1}{2}$

$\displaystyle q = 30$

$\displaystyle f(45) = 0 = 1 + 2\sin(45p + 30)$

$\displaystyle \sin(45p + 30) = -\frac{1}{2}$

$\displaystyle 45p + 30 = 210$

$\displaystyle 45p = 180$

$\displaystyle p = 4$

B is at a maximum ...

$\displaystyle 1+2\sin(4x+30) = 3$

$\displaystyle 4x+30 = 90$

$\displaystyle x = 15$

B is at $\displaystyle (15,3)$

I'll leave it for you to find the coordinates of D
• Apr 19th 2009, 10:46 AM
stapel
The graph looks like a slightly-delays cosine curve, one that isn't quite up to its maximum value at the y-axis. The curve appears to be shifted upward somewhat.

You are given that y = 1 + 2sin(px + q), so this is really a sine curve that was shifted backwards somewhat. The curve starts back before the y-axis, already having nearly reached its maximum by the time it arrives (horizontally) at the y-axis. The upward shift is seen, from the above, to be 1. The amplitude is given as being 2.

You are given that the y-intercept of the curve is at the point A, having coordinates (0, 2). Then:

2 = 1 + 2sin(p*0 + q)

2 = 1 + 2sin(q)

1 = 2sin(q)

1/2 = sin(q)

What then must be the measure of q?

The point B appears to be the maximum point of the curve. The points C and D are the first and second x-intercepts, respectively.

You are given that C has coordinated (45, 0), with "45" being in degrees. Then:

0 = 1 + 2sin(p*45 + q)

-1/2 = sin(p*45 + q)

What then must be the value of 45p + q? You have already found (from above) the value of q, so what then must be the value of p?

If you get stuck, please reply showing how far you have gotten.

Thank you! :D
• Apr 19th 2009, 11:01 AM
Tweety
for part 'b' I dont understand where 210 comes from ?

$\displaystyle sin(45p + 30 ) = -\frac{1}{2}$

$\displaystyle 45p + 30 = -30$
• Apr 19th 2009, 11:08 AM
skeeter
Quote:

Originally Posted by Tweety
for part 'b' I dont understand where 210 comes from ?

$\displaystyle sin(45p + 30 ) = -\frac{1}{2}$

$\displaystyle 45p + 30 = -30$

Quote:

p and q being positive constants ...
your equation above makes p < 0
• Apr 19th 2009, 11:17 AM
Tweety
Quote:

Originally Posted by skeeter
your equation above makes p < 0

Does that not mean that you could add any number to make it positive? And would you not have to add it to both sides?
• Apr 19th 2009, 11:28 AM
skeeter
Quote:

Originally Posted by Tweety
Does that not mean that you could add any number to make it positive? And would you not have to add it to both sides?

adding any number to both sides of the equation will not change the value of p