why is eix = cos(x) + i sin(x) helpful?
150881
There are several reasons. One of the most basic ones is that this representation provides a "polar" representation of complex numbers. Any complex number z=a+ib can be written as\n\nz= r exp(i theta)\nr= sqrt(a^2+b^2)\ntheta = atan(b/a)\n\nThis helps because, for example, a product of z1 and z2 would give\n\nz1 z2 = R exp(i phi)= r1*r2 exp (theta1 + theta2)\nR=r1*r2\nPhi = theta1 + theta2\n\nSo multiplying by z2 "stretches" z1 by amount r2 and "rotates" z1 by theta2. Such interpretations are convenient in many instances. Also, exponential functions are easier to deal with than regular a+ib type expressions in other cases.
962866
Anami's answer was correct and very simple.  You should give him credit for it.\n\nHere it is:\n\nID: cos(A+B)-cos(A-B)= -2sinAsinB\nso RHS: A=3x and B=x\n=> -2sin3xsinx = cos4x -cos 2x\nproved.\n\nWhat he did is this (only working with the left side):\n\ncos4x - cos2x\n------------------\n2sin3x\n\ncos(3x + x) - cos(3x - x)\n------------------------------\n2sin3x\n\ncos3x cosx - sin3x sinx - (cos3x cosx - sin3x sinx)\n-----------------------------------------------------------------\n2sin3x\n\n-2sin3x sinx\n---------------\n2sin3x\n\n-sinx\n\nis that more clear?
949562
Use csx =1/sin and  1-cos^2=sin^2  and cot=cos/sin\n Ready...Go\ncscx - (cosx)(cotx) =  1/sin - cos^2/sin = (1-cos^2)/sin=sin^2/sin =sin\n       ole, ole , infree
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.
463304
(sinx)^(-1) - sqrt(1-x^2) + C\nStart by multiplying (1+x)/(1-x) by (1+x) on the top and bottom\n[(1+x)(1+x)]/[(1-x)(1+x)]\nMultiply the denominator out but not the numerator.\n[(1+x)(1+x)]/[(1-x)(1+x)] = [(1+x)(1+x)]/(1-x^2)\nNow\nsqrt[(1+x)/(1-x)] \n= sqrt[[(1+x)(1+x)]/(1-x^2)]\n= (1+x)/sqrt(1-x^2)\nNow use trig substitution again letting x = sin(u)\nthen dx = cos(u) du\n(1+x)/sqrt(1-x^2) dx \n= [(1+sin(u))/sqrt(1-(sinu)^2)]cos(u) du\n= [(1+sin(u))/sqrt((cosu)^2)]cos(u) du\n= [(1+sin(u))/cos(u)]cos(u) du \nthe cosines cancel\n= [1 + sin(u)]du\n= u - cos(u) + C\nNow x= sin(u) so u = (sin x)^(-1)\nand using a triangle you can see that cos(u) = sqrt(1-x^2)
831941
You're supposed to print a table, right? The question is poorly written, so it's not clear, but I'll assume that's the goal.\n\nWell, it sounds like you can write your loop, so shouldn't you just print out the sine, cosine, and tangent of the angle in the body of the loop?  It's not clear what the stumbling block is.  You realize that there are built in functions that compute sine, cosine, etc.?
1050391
EDIT: I was looking at the answer above mine, and the cos(0) and sin(0) values are reversed, so it is not x=0.\n\nOK!! No problem. First of all, let me just say I teach math, but there is no way I would want to memorize all of the trig identities, so open up your precalc book, and if it is a good book, the major identities will be located on the front or back cover. If not, then go to the applicable page.\n\nStart with what is given:\n(sq root:3)sinx + cosx = 1\n\nAll I did was go through the identity lists and see what seemed to match closest to what you have there, and I noticed that:\n\ntan(x/2) = (1 - cosx) / sinx\n\nThen I visualized doing a little bit of algebra with your problem, and saw that if you subtracted cosx from both sides you would have:\n\nsqrt(3) sinx = 1 - cosx\n\nWhich is kind of similar to tan(x/2), except we need a sinx in the denominator. And if you divide both sides by sinx, that is what you get! So now we have:\n\nsqrt(3) = (1 - cosx) / sinx\n\nBut we already know that this is the same as tan(x/2)\n\nSo, we have:\n\nsqrt(3) = tan(x/2)\n\nFrom here, since you want to find the actual value for x which will be an angle measure, you go to inverse tangent:\n\ntan(-1) sqrt(3) = (x/2)\n\nI also have a handy table of all the major trig values open (it is in the back cover of my precalc book) and notice that tan(60) = sqrt(3)\n\nWe are almost done:\n\n60 degrees = x / 2\n\nMultiply by 2,\n\nx = 120 degrees.\n\nOf course, you should check that answer and make sure it is correct!!
946513
Subject: Re: Proving trig identiies...small error..thisis the right one\nFrom: \nDate: Monday June 14, 2004 at 08:35:54\n\nRHS= 2csc^2x=2(cot^2x+1) \n\nnote :-(csc^2x=cot^2x+1) \n\n=2cot^2x+2 \n\n=2cos^2x/sin^2x +2 \n\nnote :-((cot^2x=cos^2x/sin^2x) \n\n=2cos^2x/(1-cos^2x) +2 \n\nnote :-((sin^2x=1-cos^2x) \n\n=2cos^2x +2(1-cos^2x) /(1-cos^2x) \n\n=2cos^2x +2 -2cos^2x) /(1-cos^2x) \n\n=2/(1-cos^2x) \n\n=2/(1-cosx)(1+cosx) \n\nnote :-((1-cos^2x=(1-cosx)(1+cosx)) \n\n\nTHEREFORE lHS =2/(1-cosx)(1+cosx) (1-cos^2x=(1-cosx)(1+cosx)) \nnow for the RHS we get \n\nRHS= 1/1+COSX + 1/1-COSX \n\n=(1-cosx) + ( 1 + cosx)/(1-cosx)(1-cosx) \n\nnote :-(( through expansion) \n\n=2/(1-cosx)(1+cosx) \n\nnote :-(cosx on the numerator cancel out) \n\ntherefore RHS= LHS \n\nTHEREFORE 2csc^2x= 1/1+cosx + 1/1-cosx \n\nas required\n\n\n\nCurrent Thread\n   Proving trig identiies   a girl - Monday June 14, 2004 at 08:06:52 \n   Re: Proving trig identiies   - Monday June 14, 2004 at 08:31:28 \n--> Re: Proving trig identiies...small error..thisis the right one   - Monday June 14, 2004 at 08:35:54 <--
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. :)
714757
(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.
411302
cos^2 (2x) is equal to 1-sin^2(2x), remember, sin^2(x)+cos^2(x)=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.
1185389
There's ageneral formula for solving questions of this type\n \nRemeber \ncosx+cos2x+...+cosnx\n=cos(nx/2) sin((x+nx)/2)/(sin(x/2))\n\nNow just plug the values.\nhere x=1\nn=100\nSum= cos(100/2)sin((1+100)/2) / (sin1/2)\n=cos50sin(50.5)/sin0.5\n\nFormula is valid for all such series if angles increase in arithmetic series.\n\nTo see how the formula is derived\nJust form the series sinx+sin2x+..+sin nx =S\nAlso let cos x +cos 2x+...+cos nx=C   \nlet i=sqrt(-1)\nForm the sum C+iS=exp(ix)+exp(2ix)+...+exp(nix)\nThis is obtained by Euler's formula cos a+i sin a =exp(ia)\nSum the C+iS as geometric series and separate real and imaginary parts and you can see formula for both the series
908924
Because of (one of) the definition(s) of sin and cos.\nWhich is: take a unit circle, and draw in any radius at some angle theta. Then sin theta = the y value, and cos theta = the x value. Those are obviously less than 1.\n\nOr you could just take a typical right angled triangle - the length of the opposite/adjacent sides are always less than the length of the hypotenuse.
206636
for (i=1,t=1,cosx=1;i<=n;++i)\n{\n  t=(-1)*(t*x*x)/((2*i)*(2*i-1);\n  cosx += t;\n}
1159612
1) You meant to have a minus sign not "over". I've used x in place of "theta"\n    secx = 1/cosx and tanx = sinx/cosx \nSo 1 becomes (1/cosx)*sinx - tanx = 0 \nor tanx - tanx = 0, which is true.\n\n2) tanx = 1/cotx, so tanx*cotx = 1 \nSo 2 becomes 1- tanx*cosx = 1-sinx\nBut tanx = sinx/cosx so we have:\n1- (sinx/cosx)*cosx = 1-sinx, so\n1-sinx =1-sinx, which is true.\n\nI'll do one more:\n\n3) tan^2 x - 3sinx*tanx*secx = - 2tan^2 x\nAgain secx = 1/cosx and tanx = sinx/cosx\nSo 3. becomes:\ntan^2 x - 3(sinx/cosx)*tanx = - 2tan^2 x, so\ntan^2 x - 3tan^2 x = -2tan^2 x, so\ntan^2 x = tan^2 x, which is true.\n\nYou need to learn to do these yourself: Here is a link which gives you all the trig identities to solve the rest. They are not difficult.\n \nhttp://www.sosmath.com/trig/Trig5/trig5/trig5.html
276083
Yes, it can be rewritten as 2sin2xcos2x.  Or, you can use the addition formula for sineto prove this.  sin(x + y) = sinxcosy + cosxsiny.  In your case, sin(2x + 2x) = sin2xcos2x + cos2xsin2x = 2sin2xcos2x.