Can you simplify sin^2 (pi/(2-x)) + sin^2 x?
822150
If you're certain that you don't mean sin^2((pi/2)-x) + sin^2 x (= cos^2 x + sin^2 x =1), then no.
567245
simple x=r*cos(theta), y=r*sin(theta), r^2=x^2+y^2\nsubstitude, not hard
109151
It looks like your equation got cut off, but I think you were writing\n\nR = (V^2/2g)(sin (2 theta) +sin (2 theta))\n\nwhich can be immediately simplified, by adding the sin (2 theta) terms, as \n\nR= (V^2/2g )(2)(sin(2 theta) = (V^2/g )sin (2 theta)\n\nwhich is the formula that the other correspondent already gave.  To solve it, simply multiply both sides by g, divide by V^2, and you get \n\n(R g)/v^2 = sin (2 theta)\n\n[Arcsin (Rg/V^2) ]/2 = theta
913334
x^2 + 2x + 7 Remainder 10\n\ni can add the steps if you don't think i did it by hand
339966
The integral of sin(x)/x cannot be written in finite terms with the 'elementary functions', which consist of trig, inverse trig, exponential, logarithms, and algebraic functions. It can be written as a power series or in other ways, depending on what you want to do with it. Good amth programs will compute values for this integral (as approximations).\n\nAs for the integral of sin(x)/x from 0 to infinity, the result is pi/2, but that usually is done with complex analysis. It cannot be done easily with elementary methods.
841196
This is interesting....\n\nTry this.  First, change all tan(x) and cot(x) into sin(x)/cos(x) and cos(x)/sin(x) respectively.  Then combine the fractions using common denominator which is cos(x)sin(x).\n\nEventually, the fraction will reduce to sin^2(x) -cos^2(x)\n\nPlug this back into the original equation and you get:\nsin^2(x) - cos^2(x) + 2cos^2(x) = 1\nsimply and you get:\nsin^2(x) + cos^2(x) = 1\n1 = 1\n\nX can be any real number.....
876211
2 cos^2 x + sin^2 x = 1\n\ncos^2 x + cos^2 x + sin^2 x = 1\n\ncos^2 x + (cos^2 x + sin^2 x) = 1\n\ncos^2 x + 1 = 1\n\ncos^2 x = 0\n\ncos x = 0\n\nx = pi/2 or 3 pi/2
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
542056
Start by factoring:\n\n(cos^2(x))(sin^2(x))-(cos^2(x)) = (cos^2(x))(sin^2(x) - 1)\n= -(cos^2(x))(1 - sin^2(x))\n= -(cos^2(x))(cos^2(x))\n= -cos^4(x)
825318
(a^2 -2ab + b^2) = (a-b)(a-b) = (a-b)^2\nYou cannot cancel anything, so \n(a^2 -2ab + b^2)/(a+b) does not simplify.\n\nYou may wish to write it as (a-b)^2 /(a+b)
876696
1)\nRHS = (1-cot^2 x)/(1+ cot^2 x)\n= (1- cos^2 x/sin^2 x)/(1+ cos^2x/sin^2 x)\n= (sin^2x - cos^2 x)/(sin^2x + cos^2 x) muliypy numerator and denominator by sin^2 x\n= sin^2x -cos^2 x as denominator = 1\n\n2)\nRHS = sin x + cot x (cos x)\n= sin x + cos x/sinx .cos x\n= 1/sinx(sin^2 x+ cos^2x)\n= 1/sin x \n= csc x
729562
1/sin(x) - cos(x) * cos(x)/sin(x)\n1/sin(x) - cos^2(x)/sin(x)\n(1-cos^2(x))/sin(x)\nsin^2(x)/sin(x)\nsin(x)
972642
(A^2+2ab+b^2)/(a+b) + (a^2-b^2)/(a-b) + (a-b)/(a+b)/(a^2-b^2)/a+b)\n(a+b)^2/(a+b)+(a+b)(a-b)/(a-b)+ (a-b)/(a+b)*(a+b)/(a^2-b^2)\na+b+a+b+(a-b)/((a+b)(a-b))\n2a+2b+1/(a-b)\n\nif I got all the ( ) in the right places.
732786
Using Trig identity 2sin(x)cos(x) = sin2x\n\n(2sin3x)(cos3x) = sin6x
938878
x^2 - 2(3-x^2)\nDistribute the -2\nx^2 + (-2)(3) + (-2)(-x^2)\nSimplify\n= x^2 - 6 + 2x^2\nCombine like terms\n= 3x^2 - 6
624769
You can use sin and cosine.
847590
sin (2x ) = 2.sinx.cosx so;\n6- 6(2.sinx.cosx) = 6-12sinxcosx;\n6(1 - 2sin2x) is the simplified form.