Bryn Posted May 28, 2004 Share Posted May 28, 2004 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 Link to comment Share on other sites More sharing options...
NSX Posted May 28, 2004 Share Posted May 28, 2004 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... Link to comment Share on other sites More sharing options...
Bryn Posted May 28, 2004 Author Share Posted May 28, 2004 i used y-y1 = x-x1 y2-y1 x2-x1 to get y = 3+12x/pi How did you get -3 and 3? Link to comment Share on other sites More sharing options...
NSX Posted May 28, 2004 Share Posted May 28, 2004 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. Link to comment Share on other sites More sharing options...
Bryn Posted May 28, 2004 Author Share Posted May 28, 2004 ah ok cheers Link to comment Share on other sites More sharing options...
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