using the 1st principle differenciate cos2x?
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
1154463
cos(x)*cos(x) = cos^2(x)= 1-sin^2(x)\ntherefore the problem is now down to 1-sin(x)-Sin^2(x)\nthen integrate each one separately. \nfor inter of 1 is just x\nintegral of  -sin(x) is cos(x)\nand for sin^2 use the half angle identity \nsin^2(x) = .5*(1-cos(2X))\nand therefore -sin^2(x) = -.5*(1-cos(2X))\nintergral of which is -.5x+.25sin(2x)\n\nand the final answer add them and you'll get \n.5x + cos(x) + .25sin(2x)\nor \n.5x + cos(x) + .5sin(x)*cos(x)\nbecause sin(2x) = 2sin(x)*cos(x)
950010
replace [Cos(2x)]^2 with 1 - [Sin(2x)]^2.\n\nnow multiply to get...[Sin(2x)]^3 - [Sin(2x)]^5.\n\nthen separate integral into two parts...\n\nnow use power rule and chain rule and don't forget to add a constant.
1048691
(r*x)*(-sin(x))+(cos(x))*(r)+\n(0)*(lnrsin(x))+(r)*(cot(x)) \n\nfor the cot(x) you end up for the derivative of lnrsin(x) is (1/sin(x)) *cos(x)=cot(x)\n\nsorry it cut off on the top. hopefully this works
913797
You can begin by converting sin x^2 to 1-cos2x and since they have the same denominator x^2 just split it in two integrals (1/x^2) - (cos2x/x^2).\nThe first one is easy to solve, do initegration by parts for the second integral.
764433
Finding max's and mins is done thru the first derivitive and setting it to 0, so\n\nf(x)= sinx - cosx\nf1(x)= cosx + sinx\n0=cosx + sinx\n-cosx = sin x\nthis occurs when x= (3pi)/4\nmax is root 2, as was stated.\n\nThe answer that used to be directly above is wrong.
397497
I would use the ideal gas law...more specifically...the combined ideal gas law...\n\nP1*V1/T1 = P2*V2/T2
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.)
562730
cos2t=cos^2t-sin^2t or 2cos^2t-1 or 1-2sin^2t\n\nso you can arrive at something like 2cost-sect (remember 1/cost=sect). that's the simplest answer that you can get.
624769
You can use sin and cosine.
226513
csc(x) is the same as 1/sin(x) and sec(x) is the same as 1/cos(x) so the left side of the equation can be simplified to \n\nsin(x)sin(x) + cos(x)cos(x) \nor sin^2(x) + cos^2(x)    {"sin squared x plus cos squared x"}\n\nUsing the same steps, the right side of the equation becomes sin(x)/sin(x) or 1.\n\nSo the identity has been reduced to  sin^2(x) + cos^2(x) = 1 which is another common trigonometry identity!
829160
y'=2 cos(7x)*(-sin 7x)*7 By the Chain Rule
68963
If it's an identity, then you aren't allowed to move terms from one side to the other. Instead you must manipulate one side (leaving the other untouched) until it is identical to the other.\n\nSince the right side only depends on sinx (it has no cosx's), a good strategy would be to replace the cosx's on the left side.\n\nWe know (sinx)^2 + (cosx)^2 = 1, so\n(cosx)^2 = 1 - (sinx)^2\n\nSubstitute this for (cosx)^2 on the left side, then simplify by combining like terms. Voila - you'll have the right side.
953609
2cos(x) - sin(2x) = 0 \nremembering that sin(2x) = 2sin(x)cos(x), \n2cos(x) -2sin(x)cos(x) = 0\n2cos(x) * [1 - sin(x)] = 0\nor cos(x) = 0 or 1-sin(x) = 0\ncos(x) = 0 or sin(x) = 1\n\nπ/2, 3π/2
704235
I would expand (sin 2x+1)^2, then replace (sin 2x)^2 by\n(1-cos 4x)/2, then integrate term by term.  If you can't do that, I expect another answerer will do the whole thing, or else you can email h_chalker@yahoo.com.au
837061
it can be the original way you had it (c=mx +y) but it is alot simpler in the long run to rearrange it to (y=mx+c) (y being y, x being x, m being gradient, c being a constant). this way it is easyer to solve.
546258
cos2x=cos(x+x)=cos^2x-sin^2x=1-sin^2x-sin^2x=1-2sin^2x\ncos2x=1-2sin^2x  \ncos2x+2sin^2x-sinx=1-2sin^2x+2sin^2x-sinx=1-sinx=0\nsinx=1\nx=2kπ+π/2  ,   k=...,-1,0,1,...