Obviously sin(arcsin(x))=x and sinh(arcsinh(x))=x, but what if there is a coefficient inside the first function? i.e. arctanh(tanh(x)/2) or arcsinh(sqrt(2)sinh(x))
121533
Let sinh(K*arcsinh(x))=y then\n\ny= ((x+Sqrt(1+x^2))^(2K)-1) /(2(x+Sqrt(1+x^2))^K)\n\n\n(x+Sqrt(1+x^2))^(2K)-1\n-------------------------------- = y\n     2(x+Sqrt(1+x^2))^K\n\nExample: for K=1/2 and x=2\ny= ((1+Sqrt(5))/2)/Sqrt(2+Sqrt(5))\nY= 0.78615 = sinh(arcsinh(2)/2)\n\nCuriously in this example\ny= 1/Sqrt(Phi)\n\nwhere Phi= 0.5*(1+Sqrt(5))= 1.618...
166428
Let alpha = arcsin 6x\nThen sin(alpha) = sin(arcsin 6x) = 6x\n\nNow note d(sin(alpha))/dx = d(sin(alpha))/d(alpha) * d(alpha)/dx (Chain Rule)\n\nd(6x)/dx = cos(alpha) * d(alpha)/dx\n6 = cos(alpha) * d(alpha)/dx\nTherefore, d(alpha)/dx = 6/cos(alpha)\n\nNow consider a right triangle with angle alpha < pi/2.  From before, we know sin(alpha) = 6x so we know the following:\n\nlength of leg opposite angle (a)/length of hypotenuse (c) = 6x\n\nso that...\n\na = 6cx\n\nUsing the Pythagorean Theorem, we get the length of the other leg b:\n\nb = sqrt(c^2 - 36c^2x^2) = c*sqrt(1 - 36x^2)\ncos alpha = b/c = sqrt(1 - 36x^2)\n\nTherefore d(alpha)/dx = 6/cos(alpha) = 6/sqrt(1 - 36x^2)\n\nNow you can proceed to solve using L'Hopital OR...\n-----------\nMake this problem more interesting and recall one of the first indeterminate forms you probably solved using the Rule (i.e. lim x-> 0 sin x/x = lim x -> 0 cos x/1 = 1/1 = 1).  This tells us that sin x ~ x for small angles x.  Therefore, arcsin x ~ x for small angles x so that that lim x -> 0 arcsin 6x/5x = lim x->0 6x/5x = lim x->0 6/5 = 6/5.  This can be made more precise with delta-epsilon definitions.
173778
sin(x) - cos(x) - tan(x) = -1\nsin(x) - cos(x) - ((sin(x))/(cos(x))) = -1\n(sin(x)cos(x)) - cos(x)^2 - sin(x) = -cos(x)\n(sin(x)cos(x)) - sin(x) = cos(x)^2 - cos(x)\n((sin(x))(cos(x) - 1) = (cos(x))(cox(x) - 1)\n\ndivide both sides by (cos(x) - 1)\n\nsin(x) = cos(x)\nor\nsin(x) = 1cos(x)\n\ndivide both sides by cos(x)\n\n(sin(x))/(cos(x)) = 1\n\ntan(x) = 1\n\ninverse tan both sides\n\nx = 45° or 225°\n\nAlso i found out that "x" also equals 0\n\nIf you go to www.quickmath.com, it will give you answers of (pi/4) which equals 45°, (-3pi/4) which equals 225°, and 0.
79831
arctan'(x) = 1/(1+x^2), so differentiating using the good old rules, we get the result (using 'sq' for the square root to avoid yahoo cutting the equation short):\n\nsq(a+x^2)-1\n-------------------------------\n(x*sq(a+x^2)(1+(x-sq(a+x^2))^2\n\nMan, that's ugly. :-)
204361
sin(300/A) = B/C\n300/A = arcsin(B/C)\nA = 300/arcsin(B/C)\n\nwhere y=arcsin(x) is function inverse to x=sin(y).\n\nWhat's a problem?
532428
The problem is that thw word 'inverse' is used in two different ways here. It is perhaps easier to think of sec x as the *reciprocal* of cos x. Thus, if, say x=pi/3, cos x=1/2 and sec x=2.\n\nOn the other hand, arccos x is the function that *undoes* cos x. Thus, since cos(pi/3)=1/2, arccos(1/2)=pi/3.
479974
Yes you are correct.\n\nx=tan t = (sin t)/(cos t)\nso cos t = (sin t)/x\nSo y^2 = (sin^2 t)/x^2 = (1 - cos^2 t)/x^2 = (1 - y)/x^2\n\nRearranging this you get y = 1/(x^2 +1)
347644
Wait a minute, did you want to know when the function is zero or when its derivative is zero?  \n\nFor derivative:\ny' = 2x + sin x, right?\n\nIt looks like there's one place where it's 0.  Requires and numerical solution.\n\nWell, it's 0 at x=0, right?\nI assume that x is in radians.  1 radian is about 57 deg.  As x decreases from 0 both parts become more negative.\n\nIt looks like x = 0 is the only zero of the dirivative.
489270
cos2x = 1 - 2 (sin x) ^2 = sin x\n\nthen 2 ( sin x)^2 + sin x  - 1 = 0\n or (2sin x - 1 )(sin x + 1) = 0\nsin x = 1/2 , - 1\nx = 30 deg, or - 90 deg
421590
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. :)
57941
well, you don't have to worry about sinx and cosx as they are always between -1 and +1. so the first part of the function (1/2)* sinx is bounded.\ncos(x) is bounded.\nSo, you can rewrite the function as,\ny(x) = f(x) - x*g(x).\nwhere both f(x) and g(x) are bounded.\nso, as x goes to infinity, y(x) tends to minus infinity because of -x in the right hand side. \nSimilarly when x goes to minus infinity, y(x) goes to plus infinity\n----------------------\nResponse to the counter-example - \nin case y(x) = 2+x*(-3),\nas x -> infinity y -> -infinity.\nso y(x) is unbounded\nwhat's wrong with that??????
559787
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.)
449074
If sin A = x and sinB = -x  then \nA = .... , -4π - B, -3π + B, -2π - B, -π + B,.-B, π + B, 2π - B, 3π + B, 4π - B, .... \n= nπ + (-1)^nB (n is any integer)\n\nSo arcsin(x) - arcsin(-x) = nπ + (-1)^nB - B\n= nπ + ((-1)^(n) -1)B\n\nNo this looks too clumsy ....new strategy!!\n\nsinA + sinB = 0\nSo 2 sin(½(A + B))cos(½(A - B)) = 0\n\nThus sin(½(A + B)) = 0 or cos(½(A - B)) = 0\nie ½(A + B) = 0, ±π, ±2π, ±3π, .... , ±nπ, \nie A + B = ±2nπ\nie arcsinx + arcsin (-x) = ±2nπ\n\nOr ½(A - B) =  ±π/2, ±3π/2, ±5π/2, ... , ±(2n - 1)π/2\nSo A - B = ±(2n - 1)π\nSo arcsinx - arcsin(-x) = ±(2n - 1)π
384205
tan² x - 2 tan x = 0\n\nFactor out tan x\ntan x(tan x - 2) = 0\n\nTherefore,\ntan x = 0, or tan x - 2 = 0\n\nThus,\nx = arctan 0 or arctan 2\n\n^_^
143804
2. (2sin^2 x - 1)^2 / sin^4 x - cos^4 x = 1 - 2cos^2 x\n\nAssuming you meant\n(2sin^2 x - 1)^2 / (sin^4 x - cos^4 x)\nthen\n\n(2sin^2 x - 1)^2 / (sin^4 x - cos^4 x)\n = (2sin^2 x - 1)^2 / ((sin^2 x + cos^2 x)(sin^2 x - cos^2 x))\n = (2sin^2 x - 1)^2 / (sin^2 x - cos^2 x)\n = (2(1 - cos^2 x) - 1)^2 / ((1 - cos^2 x) - cos^2)\n = (1 - 2cos^2x)^2 / (1 - 2cos^2 x)\n = 1 - 2cos^2 x\n\n\n\nUpdate (added after wiildmunkii's answer)\n= = = = = = = = = = = = = = = = = = = = =\n\n1. sin^2 x + sin^4 x / cos^2 x + cos^4 x = tan 3x\n\nIf you meant\n(sin(2x) + sin(4x)) / (cos(2x) + cos(4x)) = tan3x\nthen yes.\n\nCombining\n\nsin(a+b) = sin(a) cos(b) + cos(a) sin(b)\ncos(a+b) = cos(a) cos(b) - sin(a) sin(b)\ntan(a + b) = sin(a + b) / cos(a + b)\n = (sin(a) cos(b) + cos(a) sin(b)) / (cos(a) cos(b) - sin(a) sin(b))\n = (tan(a) + tan(b)) / (1 - tan(a) tan(b))\n\nyou get\n\nsin(2x) = sin(x) cos(x) + cos(x) sin(x)\n = 2 sin(x) cos(x)\ncos(2x) = cos(x) cos(x) - sin(x) sin(x)\n = cos^2(x) - sin^2(x)\n = cos^2(x) - (1 - cos^2(x))\n = 2 cos^2(x) - 1\n [similarly = 1 - 2 sin^2(x)]\ntan(2x) = (tan(x) + tan(x)) / (1 - tan(x) tan(x))\n = 2 tan(x) / (1 - tan^2(x))\n\n\ne^it = (cos(t) + i sin(t))\n\nSo (e^it)^n = (cos(t) + i sin(t))^n\n\nBut (e^it)^n = (e^int) = cos(nt) + i sin(nt)\n\nSo cos(nt) + i sin(nt) = (cos(t) + i sin(t))^n\n\nFor any given n, expand the right hand side using the binomial theorem and equate real and imaginary parts to get:\n\ncos(2t) + i sin(2t) = cos^2(t) - sin^2(t) + 2 i sin(t) cos(t)\ncos(2t) = cos^2(t) - sin^2(t)\n = 2 cos^2(t) - 1\n = 1 - 2 sin^2(t)\nsin(2t) = 2 sin(t) cos(t)\n\ncos(3t) + i sin(3t) =\n cos^3(t) + i (3 cos^2(t) sin(t)) - (3 cos(t) sin^2(t)) - i sin^3(t)\ncos(3t) = cos^3(t) - 3 cos(t) sin^2(t)\n = 4 cos^3(t) - 3 cos(t)\nsin(3t) = 3 cos^2(t) sin(t) - sin^3(t)\n = 3 sin(t) - 4 sin^3(t)\n\ncos(4t) + i sin(4t) =\n cos^4(t) + i (4 cos^3(t) sin(t)) - (6 cos^2(t) sin^2(t)) - i (4 cos(t) sin^3(t)) + sin^4(t)\ncos(4t) = cos^4(t) - 6 cos^2(t) sin^2(t) + sin^4(t)\n = cos^4(t) - 6 cos^2(t) (1 - cos^2(t) + (1 - cos^2(t))^2\n = 8 cos^4(t) - 8 cos^2(t) + 1\nsin(4t) = 4 cos^3(t) sin(t) - 4 cos(t) sin^3(t)\n = 4 cos(t) sin(t) (cos^2(t) - sin^2(t))\n = 4 cos(t) sin(t) (2cos^2(t) - 1)\n = 4 cos(t) sin(t) (1 - 2 sin^2(t))\n\nFinally...\n\n(sin(2x) + sin(4x)) / (cos(2x) + cos(4x))\n = (2 sin(x) cos(x) + 4 cos(x) sin(x) (1 - 2 sin^2(x))) / (2 cos^2(x) - 1 + 8 cos^4(x) - 8 cos^2(x) + 1)\n = (6 sin(x) cos(x) - 8 sin^3(x) cos(x)) / (8 cos^4(x) - 6 cos^2(x))\n # divide by 2 cos(x)\n = (3 sin(x) - 4 sin^3(x)) / (4 cos^3(x) - 3 cos^(x))\n = sin(3x) / cos(3x)\n = tan(3x)
353956
(sinX + 1) = 0---> sinX = -1--->x = 3pi/2 + (2pi)n, radians, or x = 270 + 360n degrees, where n is any whole number including 0. \n\nThe first step was simple algebra.  In the second step, I just remembered that the sine is how far up on the unit circle that angle points, so -1 means it points straight down, which is 3/4 of a circle.  Adding a whole circle puts it right back where it started from.\n\ncosX + 1/2 = 0---> cosX = -1/2 ---> x = 2pi/3 or 4pi/3 + (2pi)n radians, 120 or 240 + 360n degrees.  \n\nAgain, the first step was simple algebra.  As for the second, my trig teacher made us memorize the sines/cosines of 30, 45, and 60 degrees, so I just know that the cosine of 60 degrees is 1/2.  Go around to the other side of the unit circle and the cosine of 120 degrees is -1/2.  Go down to the bottom of the unit circle, and the cosine of 240 degrees is also -1/2.
362067
Use sin(arcsinx) = x...\n\ncos(arcsinB) = sqrt(1-sin^2(arcsinB))\n= sqrt(1-B^2)\n\nSimilarly, sin(arccosA) = sqrt(1-A^2).\n\nUse the cos(x+y) equality to compute the last one...