a trig problem

[Bryn]

y = psinx + qcosx passes through the points (0,3) and (pi/4,0). Find the values of p and q.

 

Using formula on the coordinates i've got y = 3 + 12x/pi

 

so

 

psinx + qcosx = 3 + 12x/pi

 

but then where? only 2 identities i need for this course is sinx/cosx = tanx and (sinx)^2 + (cosx)^2 = 1 but i cant see how i can use these

[NSX]

hm... well I got y = -3 sin(x) + 3 cos(x)

 

How did you approach the question?

I don't really get where the 12x/π comes from...

[Bryn]

i used

y-y1 = x-x1

y2-y1 x2-x1

 

to get y = 3+12x/pi

 

How did you get -3 and 3?

[NSX]

i used

y-y1 = x-x1

y2-y1 x2-x1

 

to get y = 3+12x/pi

 

How did you get -3 and 3?

 

Why did you use this:

 

y-y1 = x-x1

y2-y1 x2-x1

 

To get q = 3' date=' I substituted the point (0,3) into the equation (1): y = p sin(x) + q cos(x).

 

To get p = -3, I substituted q and the point of (pi/4) back into (1) and voila!

 

[edit']

I don't think you need the identities: sinx/cosx = tanx and sin²(x) + cos²(x) = 1 for this question.

[Bryn]

ah ok cheers