We all know that\nsin^2 x + cos^2 x = 1\nso\nsin^2 x = 1 - cos^2 x\nand\nsin x = sqrt(1 - cos^2 x)\n\nGiven cos 240 = -1/2, the value of sin 240 is given by\nsin 240 = sqrt[1 - (-1/2)^2]\nsin 240 = sqrt(1 - 1/4)\nsin 240 = sqrt(3/4)\nsin 240 = sqrt3/2\n\nbut actually sin 240 = -sqrt3/2\nWhy?\n\n^_^...[easy]...^_^
137613
The final equation you mentioned should be actually:\n\nsin x = +/- sqrt(1 - cos^2 x)\n\nmeaning if you substitute the value of sin x as - sqrt(1 - cos^2 x) in the original equation which is sin^2 x + cos^2 x = 1, it should satisfy it, which it does.
412590
Just put everything over a common denominator, and expand.\nThe numerator becomes:\n(cos x - cos y)(cos x + cos y) + (sin x - sin y)(sin x + sin y)\n= (cos x)^2 - (cos y)^2 + (sin x)^2 - (sin y)^2\n= [(cos x)^2 + (sin x)^2] - [(cos y)^2 + (sin y)^2]\n= 1 - 1 = 0.
607936
(cosx)^2 - 2sinx = 2\ncosx^2 = 1 - sinx^2\n(1 - sinx^2) - 2sinx = 2\n1 - sinx^2 - 2sinx = 2\n-sinx^2 - 2sinx + 1 = 2\n-sinx^2 - 2sinx - 1 = 0\n-(sinx^2 + 2sinx + 1) = 0\n-(sinx + 1)(sinx + 1)\n\n-sinx - 1 = 0\n-sinx = 1\nsinx = -1\nx = sin^-1(-1)\nx = 270°\n\n--------------------------------------\n\n1.5^2 = 1.7^2 + 1.8^2 - 2(1.7)(1.8)cosA\n2.25 = 2.89 + 3.24 - 6.12cosA\n2.25 = 6.13 - 6.12cosA\n-6.12cosA = -3.88\ncosA = (388/612)\ncosA = (97/153)\nA = 50.6551127\n\n1.7^2 = 1.5^2 + 1.8^2 - 2(1.5)(1.8)cosB\n2.89 = 2.25 + 3.24 - 5.4cosB\n2.89 = 5.49 - 5.4cosB\n-2.6 = -5.4cosB\ncosB = (13/27)\nB = 61.2177953\n\n1.8^2 = 1.5^2 + 1.7^2 - 2(1.5)(1.7)cos(C)\n3.24 =  2.25 + 2.89 - 5.1cosC\n3.24 = 5.14 - 5.1cosC\n-1.9 = -5.1cosC\ncosC = (19/51)\nC = 68.1270919\n\nAs you can see, the angle opposite of side A, is the smallest angle.\n\nANS : about 51°\n\n----------------------------------------------\n\ncos(A) = 4/7\nx = 4\nr = 7\n\nr = sqrt(x^2 + y^2)\n7 = sqrt(4^2 + y^2)\n49 = 16 + y^2\ny^2 = 33\ny = sqrt(33)\n\nsin(A) = y/r\nsinA = (-sqrt(33))/7\n\ntanA = y/x\ntanA = (-sqrt(33))/4\n\ncscA = r/y\ncscA = -7/(sqrt(33))\ncscA = (-7sqrt(33))/33\n\nsecA = r/x\nsecA = 7/4\n\ncotA = x/y\ncotA = -4/(sqrt(33))\ncotA = (-4sqrt(33))/33
195106
lets assume b = a/2\ntherefore 2b = a\n\nnow sin 2b = 12/13\n\nsin 2b is also equal to 2sinbcosb\n\nso 2sinbcosb = 12/13\n\nsinb * cosb = 6/13\n\nnow u know that cosb = sqrt ( 1 - sin^2 b)\n\nso sinb * sqrt(1-sin^2 b) = 6/13\n\nor sin^2 b * (1 - sin^2 b) = 36/169\n\nsubstituting sin ^ b = x\n\nyou get x - x^2 - 36/169 = 0\n\nThis is a simple quadratic equation\n\nsolving for x you get\n\nx = 234/338 or x = 104/338\n\nso you get sin b is sqrt (234/338) or sin b is sqrt(104/338)\n\nsince a is in the first quadrant, the value of sin a/2 will be less than that of cos a/2 since a/2 is less than 45 degrees.\n\nSo, sin a/2 = sqrt of (104/338)= 0.554\n\ncos a/2 = sqrt of (234/338) = 0.832
345601
Brackets are your friend.\n\nIf you mean cos(x/2) = cos(x), then substitute y=x/2 and use the equality cos(2y) = 2cos^2(y) -1\n\nYou have\ncos(y) = cos(2y)\n= 2cos^2(y) - 1\nor 2 cos^2(y) - cos(y) - 1 = 0.\n\nQuadratic equation, solutions are cos(y) = 1 and cos(y) = -1/2.\nThe first solution yields y = 0, and the second yields y = 2pi/3 or y = 4pi/3.\n\nSince x=2y, we have x=0, x=4pi/3 and x=8pi/3.  The latter is greater than 2pi so it won't work.  The answer is:\nx=0 ----> cos0 = cos0 = 1\nx=4pi/3 ---> cos(2pi/3) = cos(4pi/3) = -1/2
166823
Yes. I assume to want to find sin 1/2x, cos 1/2x, and so on.\nIt's quite simple:\n1. Write the expanded form of cos 2x\ncos^2x - sin^2x\n2. Write that twice, in once in a column (remember it equals cos 2x)\ncos^2x - sin^2x  = cos 2x =  cos^2x - sin^2x\n3. Rewrite cos^2x in the first column as 1-sin^2x, and in the second column rewrite sin^2x as 1-cos^2x\n(1-sin^2x) - sin^2x = cos 2x = cos^2x - (1 - cos^2x)\n4. Combine like terms\n1-2sin^2x   and -1+2cos^2x\n5. In order to find cosx and sinx, we must isolate each:\n-2sin^2x = cos 2x -1  and 2cos^2x = cos 2x + 1\n6. Divide by the coefficent of each\nsin^2x = 1/2 - 1/2(cos 2x) and cos^2x = 1/2 + 1/2(cos 2x)\n7. Find the square root of both sides\nsin x = sqrt(1/2 - 1/2(cos 2x)) and \ncos x = sqrt(1/2 + 1/2(cos 2x))\n8. We found what sin x and cos x equal, we want what 1/2x equals. Let x = 1/2x\nsin 1/2x = sqrt(1/2 - 1/2(cos 2(1/2x)) and\ncos 1/2x = sqrt(1/2 + 1/2(cos 2(1/2x))\n9. Simplify\nsin 1/2x = sqrt(1/2 - 1/2(cos x)) and\ncos 1/2x = sqrt(1/2 + 1/2(cos x))\n10. You're done. You can find tan 1/2x, csc 1/2x, sec 1/2x, and cot 1/2x by using the trigometric identities with sin 1/2x and cos 1/2x:\n\ntan 1/2x = (sin 1/2x)/(cos 1/2x)\ncsc 1/2x = 1/(sin 1/2x)\nsec 1/2x = 1/(cos 1/2x)\ncot 1/2x = (cos 1/2x)/(sin 1/2x)\n\nObviously there are other expansions for the half angle formulae. These are just the easiest.
25819
Here's a head start.  The identity you wrote is wrong.  Maybe you mistyped it?\n\ncos^(-1/2)xsinx-cos^(3/2)xsinx\n= [sinx-cos^(2)xsinx]/[cos^(1/2)x]\n= sinx[1-cos^(2)x]/[cos^(1/2)x]\n= sinx[sin^(2)x]/[cos^(1/2)x]\n= [sin^(3)x]/[cos^(1/2)x]\n= [sin^(3)x][cos^(1/2)x]/[cosx]\nThis is your square root of cosx/cosx, but times sine cubed.
399131
It won't reduce to a fraction.\n\nThe fundamental relation is: sin^2 x + cos^2 x = 1.\n\nSo \nsin x = square root [ 1 - (-4/7)^2 ]\nsin x = sqrt [ 1 - (16/49) ]\nsin x = sqrt [33/49]\nor\nsin x = sqrt[33] / 7\nwhich is as close as we can come to a fraction, which is about .82.\nAnd in the second quadrant, x = 145.15 degrees.
598140
Please check the statement of the problem, then apply this.\n\nWe know that\nsin (x)=(e^(ix)- e^(-ix))/2i\n\ncos (x)=(e^(ix)+ e^(-ix))/2\n\nalso\n\nsin(ix)=i(sinh(x))=(e^(-x)- e^(ix))/2i\n\ncos(ix)=(cosh(x))=(e^(-x)- e^(ix))/2\n\nThe only thing you need is if z=x+iy then |z|=sqrt(x^2 + y^2)\n\nGood luck
405147
x^2+x+1=0\nx^2+2(1/2)x+1=0\nX^2+2(1/2)x+(1/2)^2+1-(1/2)^2=0\n(x+1/2)^2+3/4=0\n(x+1/2)^2=-3/4\nx=-1/2+/-√(3/4) i=-1/2+/-√3 * i/2
488943
cos²x - 1 - cosx = 0\nie cos²x - cosx - 1 = 0\ncosx = ½(1 ± √5) by quadratic formula\nNow as |cosx| ≤ 1 for real x\ncosx = ½(1 - √5) = -0.61803......\nSo x = 180° - 51°49'38", 180° + 51°49'38"\n= 128°10'22", 231°49'38" (for 0° ≤ x ≤ 360°)
182452
cos^2 (2x) is equal to 1-sin^2(2x), remember, sin^2(x)+cos^2(x)=1
133424
sin(2x) = sin(3x)\nsin(2x) = 2sin(x)cos(x)\n\nsin(3x) = -sin(x)^3 + 3cos(x)^2*sin(x)\nsin(3x) = (sin(x)) (-1 + 4cos(x)^2)\n\nsin(2x0 = sin(3x)\n2sin(x)cos(x) = (sin(x)) (-1 + 4cos(x)^2)\ndivide both sides by sin(x)\n2cos(x) = -1 + 4cos(x)^2\nsubtract 2cos(x) from both sides\n4cos(x)^2 - 2cos(x) - 1 = 0\n\nNow just think of this problem like this\n\n4x^2 - 2x - 1 = 0\n\nUsing the quadratic formula\n\nx = (-b ± sqrt(b^2 - 4ac))/2a\nx = (2 ± sqrt(4 + 16))/8\nx = (2 ± sqrt(20))/8\nx = (2 ± sqrt(4 * 5))/8\nx = (2 ± 2sqrt(5))/8\nx = (1/4)(1 ± sqrt(5))\n\nnow just plug in cos(x) for x\n\ncos(x) = (1/4)(1 ± sqrt(5))\ninverse cosine both sides\n\nx = 36° and 108°
495907
2cos^2x-1+1/2sinxcosx\n=2cos^2x/2sinxcosx\n=cosx/sinx\n=cotx
208067
sin2x = 2sinxcosx\nsin2xsinx = cosx\n(2sinxcosx)sinx = cosx\n2(sinx)^2cosx - cosx = 0\n2(1-(cosx)^2)cosx - cosx = 0  Factor out cosx\ncosx[2(1-(cosx)^2)-1] = 0  Distribute the 2\ncosx[2 - 2(cosx)^2 - 1] = 0  Combine 2 and -1\ncosx(1-2(cosx)^2) = 0\nso cosx = 0 or 1-2(cosx)^2 = 0\ncosx = 0 when x = 90 degrees and 270 degrees\n1-2(cosx)^2 = 0 when\n2(cosx)^2 = 1  Divide both sides by 2\n(cosx)^2 = 1/2 Take the square root of each side\ncosx = sqrt(1/2) or -sqrt(1/2)\ncosx = sqrt(1/2) when x = 45 or 315 degrees\ncosx = -sqrt(1/2) when x = 135 or 225 degrees.\nThus, the answers are\n45, 90, 135, 225, 270, and 315 degrees.
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.)
414846
Have you learned trig identities yet?  I'd check to see if there was a trig identity you could substitute here.\n\nBTW, I am assuming you meant cos²x rather than cosx².\n\nTry this:  cos²x + sin²x = 1, so substitute 1-sin²x for cos²x:\n\n6cos²x + sinx - 4 = 0\n\n6(1-sin²x) + sinx - 4 = 0\n\n-sin²x + sinx + 2 = 0\n\n(multiply by -1 to make it a little easier)\n\nsin²x - sinx - 2 = 0\n\n(sinx-2)(sinx+1) = 0\n\nSo, sinx = 2 or sinx = -1\n\nSolving for x, this means x = sin^-1(2) or sin^-1(-1)