If dy/dx = sinx (cosx)^2 and if y = 0 when x = pi/2, what is the value of y when x = 0?
82524
you're given dy/dx = sinx(cosx)^2.  you can "multiply" both sides by dx.  this "multiplication" works, but don't think of "dx" as a variable or something...ask your teacher - it's too hard to explain.  so anyways, after multiplying by "dx", you should get a nicer equation:\n\ndy/dx = sinx(cosx)^2\ndy = [sinx(cosx)^2]dx\n\nnow integrate:\n\nint(dy) = int{[sinx(cosx)^2]dx}\ny = ( )\n\nto integrate the right side, use u substitution with u=cosx.  if u=cosx, then du=-sinxdx.\n\ny = int{[sinx(cosx)^2]dx}\n   = -int[(-sinxdx)(cosx)^2]\n   = -int[(du)(u)^2]\n   = -int(u^2du)\n   = -(u^3)/3 + C\n\nnow replace u with cosx, since that's what we defined u as earlier.\n\ny = -(u^3)/3 + C\n   = -[(cosx)^3]/3 + C\n\nnow since you know that y=0 when x=pi/2, plug in these values to find "C".\n\n\ny = -[(cosx)^3]/3 + C\n0 = -[(cos(pi/2))^3]/3 + C\n0 = 0 + C since cos(pi/2)=0\nC = 0\n\nthus our final equation is: y = -[(cosx)^3]/3.  now to find y when x=0.\n\ny = -[(cosx)^3]/3\n   = -[(cos(0))^3]/3\n   = -1/3\n\nfinally, y=-1/3.  any mistakes?  hope it helps.
844759
Look at the various values of cosx when x = pi/2, pi, etc. (recall that pi/2 = 90 degrees).\n\nFrom looking at these values in my book, I see two values for x that will satisfy this equation:  x=pi/3, and x=pi\n\ncos pi/3 = 1/2\ncos 2pi/3 = -1/2\n\ncos pi = -1\ncos 2pi = 1\n\n~ ♥ ~
784325
I think the formula for sin(c-d) is sin c cos d - cos c sin d\n\nso sin (x - pi/3) = sin x cos pi/3 - cosx sin pi/3\n\nwhich is 1/2 sin x - sqrt3 / 2 cos x\n\nMultiplied by the -2 in front, this is -1 sin x + sqrt 3 cos x\n\nSo A = -1 and B = sqrt 3\n\nLook up the original formula, I may have messed up the signs
433960
ok. well if y = sin^-1 (x/a), then that means that \nsin y = (x/a). \nso if you do implicit differentiation, you get \ncos y * dy/dx = 1/a.\n\nso thatz not hard. now if you draw a right angle triangle, and label 1 of the angles y. so sin of y is x/a, but it is also equal to opposite/hypotenuse. so then that meanz the opposite is equal to x and the hypotenuse is equal to a. now if you use pythagoras theorem, the other side (the adjacent side) is sqrt(hyp^2 - opp^2). because \nadj^2 + opp^2 = hyp^2\nso adj = sqrt(hyp^2 - opp^2)\nso adj = sqrt(a^2 - x^2).\n\nso now you need cos of y, which is adj/hyp\nso cos y = sqrt(a^2-x^2) / a.\n\nnow back to the first equation\ncos y * dy/dx = 1/a\nsqrt (a^2 - x^2)/a  * dy/dx = 1/a\nsqrt (a^2-x^2) * dy/dx = 1\nso dy/dx = 1/sqrt(a^2-x^2)\n\nyou can do this with any inverse trig derivative stuff. i hope you understood that. thatz the way i remember it but there might be other easier ways (like just memorize the relations), but there you go. hope that helped. let me know if you need more help
806885
dy/dx of xsinx = x(d/dx sinx) + sinx(d/dx x) = xcosx + sinx(1) = xcosx + sinx\n\nHope that helps - it's the first thing multiplied by the differential of the second, plus the second multiplied by the differential of the first. :)
831486
Differentiate both sides, we have \n\ndy =  (1/x)dx + (1 + tan^2(y))/tany dy\n\ntherefore\n(1 - (1 + tan^2(y))/tany)dy=(1/x)dx,\n\ni.e.\n\ndy/dx=tany/(x(tany - tan^2- 1)).
355137
first, this equation is wrong, it should look likle this:\nsin2x=2(sinx (multiplied by) cosx). If yu want, i'll try to solve the true equation for yu.
170888
i am assuming we are differentiating with respect to x...\n\nd sin((k-x)^2) / dx = cos ((k-x)^2) * d (k-x)^2 / dx = cos((k-x)^2) * 2(k-x) * (-1)\nby Chain rule
477866
Let y=cos2x, then, \ndy/dx = limit (as h tends to zero) [(cos2(x+h) - cos2x)/h]\n      = lim. [(cos2x cos2h  - sin2x sin2h - cos2x)/h]\n      = lim. [(cos2x(cos2h-1) -sin2x sin2h)/h]\n      = lim. [(cos2x (-2(sin2h)^2)/h] - sin2x* lim.(sin2h/h)\n      = -2 cos2x * (lim. sin2h/h)*(lim. (sin2h))- 2 sin2x\n      = -2 cos2x * 2 * 0 - 2 sin2x\n      = -2 sin2x
932480
sin(A+B)sin(A-B)/(cos²Acos²B)\n= (sinAcosB + cosAsinB)(sinAcosB - cosAsinB)/(cos²Acos²B)\n= (sin²Acos²B - cos²Asin²B)/(cos²Acos²B) (since (x + y)(x - y) = x² - y²)\n= sin²A/cos²A - sin²B/cos²B\n= tan²A - tan²B ← Option c)\n\nPlease note: sin(A + B) = sinAcosB + cosAsinB\nand sin(A - B) = sinAcosB - cosAsinB
948239
xy = y^½\n y = y^½ /x\n y = y^½*x^-½\n\ndy/dx = -½ y^½ x^ -3/2\ndy/dx = - [½ y^½ ]/x^+3/2\ndy/dx = - (y^½) / (2x^3/2)
730722
put sinx=t\ncosxdx=dt\nI=fcos(pi t)dt with limits -1,+1\nI=(sin(pi t))/pi with limits +1,-1\nI=(sin(pi)-sin(-pi))/pi=0\nthanks & good luck
901670
(sin x +cos x)^2 =1    [0=<x=<360]\nsin^2x + 2sinx * cosx +cos^2x =1 or 2sinxcosx=0      Because sin^2x+cos2^x =1 \n2sinxcosx is zero when x =0, 90, 180, or 270 degrees\n\n1-sinx = cosx     [0=<x=<360]\nsinx + cosx = 1\nThis occurs when x= 0, 360
873429
cos5x = sinx\ncos5x = cos(pi/2 - x)\n5x = 2npi +- (pi/2 - x)\n6x = (4n + 1)pi/2   ,   4x= (4n-1)pi/2\nx= (4n+1)pi/12    ,  x = (4n-1)pi/8
70526
=((-4y)^2)/(10y-5)= (16y^2)/(10y-5)\nand if the equation is equal to zero then y=4i
699547
sin[x-(pi/2)]=kcosx\n-sin[(pi/2)-x]=kcosx\n-cosx=kcosx\nk=-1\ncos[x+{pi/2)]=ksinx\n=cos[(pi/2)+x]=ksinx\n=-sinx=ksinx\nk=-1
994111
First, factor the left hand side of the equation:\n\n2sin(x)*cos(x) + cos(x) = 0\n\ncos(x)*[2sin(x) + 1] = 0\n\nNow, this equation is satisfied whenever either cos(x) = 0 or (2sin(x)+1) = 0\n\ncos(x) = 0 at x = pi/2 and 3*pi/2  (You should memorize when the cosine and sine functions are equal to zero!)\n\nThese are the first two solutions to the original problem.\nNow deal with the factor containing the sine term.  We need to find the values of x when:\n\n2sin(x) + 1 = 0\n\nRearrange this equation to read:\n\n2sin(x) = -1\n\nthen\n\nsin(x) = -1/2\n\nx = 11pi/6 and 7pi/6 .  These are the other two solutions to the original problem.\n\nYou should also memorize when the sine and cosine functions are equal to 1/2.  For the sine function, this happens at x = pi/6 = 60 degrees and 5pi/6 = 150 degrees.  If the sine or cosine function has a certain value at x, then it will have -1 times that value at x + 180 degrees (x + pi radians), so in this case if sin(pi/6) = 1/2, then sin(pi/6 + pi) = sin(7pi/6) = -1/2.  Similarly, if sin(5pi/6) = 1/2, then sin(5pi/6 + pi) = sin(pi/6+pi) = sin(7pi/6) = -1/2.\n\nNote that despite what the previous answerer said, the sin(2pi/3) and the sin(4pi/3) do not equal -1/2.  [It's the cosine that's equal to -1/2 at these points.)