Who in the world ever thought of the sinh(x) and cosh(x) functions?
146293
I know of one application in an abstruse area of physics: The sinh( ) of a scalar field occurs in gravitational theories where, for whatever reason, one emplys a metric disformally related to the 'physical' metric tensor.\n\nAlso, the shape of a freely hanging level chain is cosh( ) of displacement along the line joining ends
94920
you mean sinh(ax)?\n\nwhat is this hax
956637
Sorry tried a lot, searched a lot & surfed a lot but found no solution, I think Pascal is right.\n\nP.s- Saket it is ∫e^sinx dx & not ∫(e^x)*sin x dx, if it was that, it would have been trivial.
1058665
|sin(z)|^2 + |cos(z)|^2\n\nWhere |f(z)|^2 is understood to mean f(z)*~f(z), where ~f(z) is the complex congugate of f(z): ~f(x + iy) = f(x - iy)\n\nWriting the real and imaginary parts of the arguments explicitly, we have:\n\n= |sin(x+iy)|^2 + |cos(x+iy)|^2\n\nUsing the sum of angles formulas:\n\n= |(sin(x)cos(iy) + cos(x)sin(iy)|^2 + \n  |(cos(x)cos(iy) - sin(x)sin(iy)|^2\n\nExpanding in terms of complex congugates:\n\n= ((sin(x)cos(iy) + cos(x)sin(iy))*(sin(x)cos(-iy) + cos(x)sin(-iy)) +\n  ((cos(x)cos(iy) - sin(x)sin(iy))*(cos(x)cos(-iy) - sin(x)sin(-iy))\n\nNoting that cos(-x) = cos(x) and sin(-x) = -sin(x):\n\n= ((sin(x)cos(iy) + cos(x)sin(iy))*(sin(x)cos(iy) - cos(x)sin(iy)) +\n  ((cos(x)cos(iy) - sin(x)sin(iy))*(cos(x)cos(iy) + sin(x)sin(iy))\n\n\nMultiplying this out (both terms are the product of a sum and difference of two terms, which yield the difference of squares):\n\n= (sin(x)cos(iy))^2 - (cos(x)sin(iy))^2 + (cos(x)cos(iy))^2 - (sin(x)sin(iy))^2\n\nFactoring:\n\n= ((sin(x))^2 + (cos(x))^2)*cos(iy) - ((sin(x))^2 + (cos(x))^2)*sin(iy)\n\nbut ((sin(x))^2 + (cos(x))^2) = 1, so\n\n= (cos(iy))^2 - (sin(iy))^2\n\nFrom Euler's formula (exp(ix) = cos(x) + i*sin(x)), we can show that:\n\ncos(iy) = (exp(-y) + exp(y))/2 = cosh(y)\nand\nsin(iy) = (exp(-y) - exp(y))/(2*i) = i*sinh(y)\n\nSubstituting these identities in for cos(iy) and sin(iy), we have:\n\n= (cosh(y))^2 - (i*sinh(y))^2 \n\n= (cosh(y))^2 + (sinh(y))^2\nQED
812461
As I mentioned in another post here the other day, there is one technique that *always* works, even if it isn't the fastest.\n\nChange everything to sin and cos. Use sin^2 x + cos^2 x = 1. Simplify. Thats it.\n\nSo, the LHS is \n1/tan x - tan x\n= cos x / sin x  - sin x / cos x\n= (cos^2 x - sin^2 x) / (sin x cos x).\nNow, we have the right denominator, but the numerator isn't right yet. What else can we simplify? We know how to change cos2x into something just involving cos x or sin x:\nYou may already know that cos 2x = cos^2 x - sin^2 x. If not, maybe you know this one:\ncos 2x = 2 cos^2 x - 1\n= 2 cos^2 x - (sin^2 x + cos^2 x)\n= cos^2 x - sin^2 x.\nAnd thus the two sides are equivalent.
828746
6 cosh x = 6/5 x^5 + 3x^4 + 8x³ + 24x² + 96x + C\n\nIf C is not given, then x can be any real number.  That's all.\n\n^_^
869966
http://www.math.psu.edu/tseng/class/Math140H/Sqrti.html\n\nAnd, BTW, i (or j) is not a 'complex' operator.  It is the 'imaginary unit' ☺\n\n\nDoug
816208
RHS  (simplify it into sin and cos terms)\ncsc^2(x)/cotx\nsinx/cosx*1/sin^2x\nsinx/sinx*1/cosxsinx\n2/2sinxcosx\n2/sin2x\n2csc2x\nhence LHS=RHS
418030
Take a unit circle and a point p on the circle. Place the circle so that it is centered at the origin in a Cartesian xy-plane. The cosine of p, cos(p), is defined to be the x-coordinate, or abscissa, of p. The sine of p is the y-coordinate. These functions are the just the projection functions for the unit circle by the standard embedding in the Cartesian plane as described above.\nIt is common to give the points p on the 1-dimensional circle the coordinate referring to the arclength between p and the positive x-axis (as measured clockwise). Thus, the point (1, 0) on the unit circle is given the native coordinate (0), or 0 for short. Inversely, the point 0 in the unit circle has Cartesian coordinates (cos(0), sin(0)). Since the circumference of a circle is 2Pi, a point that is a quarter of the way around the circle from the positive x-axis is given the coordinate 2Pi/4 = Pi/2 and so forth. This system of coordinates is called radians and is the native way of measuring angles, as each angle corresponds to a unique point on the unit circle.\nThis relationship can be seen by drawing a radius from the origin to mark the angle from the positive x-axis and assigning that angle to the radian measure of wherever it intersects the unit circle. There are more analytical reasons as well, as you go on to study calculus.\nThe radius drawn above also gives the relationship between these circular projection functions, and the way one first encounters trigonometric functions, as ratios of the sides of right triangles. Now you should be able to visualize the triangles inside the unit circle (or any larger circle, dependent on the triangle's hypotenuse), and see why the trigonometric functions are cyclic in 2Pi, and derive some properties of the trigonometric functions that are not easily apparent from just looking at right triangles.\nIt is now however easy to see why all the basic trigonometric functions have a neighbor with prefix co-. The cosine is short for complementary sine. This is because the cosine of a point is obviously the same as the sine of the complementary angle.\nThe tangent function is so named because it is the length of a certain tangent to the circle; a tangent drawn from the positive x-axis to where it meets an extended radius of the circle. You may be able to work out the equation tan(p) = sin(p)/cos(p) from the geometry involved.\nThe arc trigonometric functions are so named because they measure arc length; they give you back the radian measure of the sine or cosine when you restrict these functions to one hemicircle (otherwise you get two possible answers).\nThe secant function does not give you an arclength; it is another function of points on the unit circle (and thus by our unit circle coordinate system, it is a function of arc length). In the right triangle formed by the tangent to the circle, the secant is the length of the hypotenuse. The arcsecant function returns the arc length on the unit circle that the secant intersects. See the sources below for illustrations.
524303
So X is dependent on TSOUT (the source of universal truth)\n\nAnd the TSOUT is dependent on X.\n\nThat, my friend, is circular logic, and as such, is a false proof.
950350
LHS=sin^2x-cos^2x=(sinx+cosx)(sinx-cosx)\nRHS=1-2sinxcosx\n=(sin^2x+cos^2x)-2sinxcosx as sin^2x+cos^2x=1\n=(sinx-cosx)^2\nLHS and RHS can be equal only ifsinx+cosx=sinx-cosx\nso it is not an identity\nan identity must hole for all values
1013129
Please explain what is ~f.  I have curiosity, because neither inverse multiplicative, complex conjugate or inverse function could do that result...\n\nIt should be interesting!! Please post what is ~f !
741912
cos ^1 t = x such that cos x = t\n\nsin ^1 t = y such that sin y = t\n\nThese are called inverse trigonometric functions.\n\nnow cos^-1 t = sin^-1  t = x \nso  cos x = sin x = t\n\ntan x = 1\n\nx =  pi/4\n\nthe above is solution
1047641
since 1= |sin z|^2+|cos z|^2\nthen |cos Z|=sqrt(1-(sin x cosh y +i cos x sinh y)^2)\nAnyway if you were looking for a more elaborated one using the "e" definitions for the hyperbolics function I could do it but naah
433823
Revolving about the x-axis yields a cone whose tip is at the origin and opens in the positive x direction.  Taking discrete x slices (dx) yields a disk whose volume is given by pi*(kx)^2 dx.  Integrating from 0 to h gives us (where "S" is used as the integration symbol):\n\nV = S pi*(kx)^2 dx\n= pi * k^2 S x^2 dx\n= (pi * k^2) * (x^3)/3\n\nEvaluating from 0 to h gives us\n= (pi * k^2) * (h^3)/3 - (pi * k^2) * (0^3)/3\n= (pi * k^2 * h^3)/3
160524
This is not a calculus problem, this is merely a trigonometric identity.  No derivatives, integrals, or numerical approximation were used.\n\nDefinition:\n\ntanh(2x) = sinh(2x)/cosh(2x)\n\nYou know the double angle formulas for sinh and cosh:\n\nsinh(x + x) = sinh(x)cosh(x)+cosh(x)sinh(x) = 2cosh(x)sin(h)\n\ncosh(x + x) = cosh(x)cosh(x)+sinh(x)sinh(x) = cosh(x)^2 + sinh(x)^2\n\nDivide:\n\n(2cosh(x)sinh(x))/(cosh(x)^2 + sinh(x)^2)\n\nDivide numerator and denominator by cosh(x)^2:\n\n((2cosh(x)sinh(x))/cosh(x)^2) / ((cosh(x)^2 + sinh(x)^2)/cosh(x)^2)\n\nSimplify:\n\n(2 sinh(x)/cosh(x))/(1 + sinh(x)/cosh(x)) \n\nSimplify more to solution:\n\n2tanh(x)/(1 + tanh(x)
822981
i)Just substitute in the values to get F\nx = 1.1.  Other values are given.  Hence F.\n\nii) Rearrrange the equation\nx = [cosh^-1(QCK/2F-1)]/k\nF = 2.9.  Other values are given, hence x.