what is the sin 45 degree & cos 45 degree\n            sin 30 degree & cos 30 degree\n            sin 60 degree & cos 60 degree ?
136921
sin 90 = 1\n\ncos 90 = 0\n\nYou dont need to byheart the table. you can create table on your own.\n\n0 30 45 60 90\n0 1 2 3 4\nHere the numbers in above line show the degree of an angle that you need.\nNow just do one thing. Divide each number written below i.e. 0 1 2 3 4\nby 4\nyou will get 0, 1/4, 1/2, 3/4, 1.\nNow take their square roots. you will get\n\n0, 1/2, 1/root2, root3/2, 1.\nthis values are for sin.\ntable becomes\n\nangle 0 30 45 60 90\nsin 0 1/2 1/root2 root3/2 1\n\ncos 1 root3/2 1/root2 1/2 0\n\nhere value of cos you will get by reversley writting all values of sin. i.e. sin 90 = cos 0.\n\nsimilarly you will get\n\n0 30 45 60 90\ntan 0 1/root3 1 root3 not defined\n\ncot not defined root3 1 1/root3 0\n\nsec 1 2/root3 root2 2 not defined\n\ncosec not defined 2 root 2 2/root3 1\n\nyou can get values of tan and sec by inserting values of sin and cos in the formulae.\ntan = sin/ cos\n\nsec = 1/ cos\n\nthen for cot and cosec write the values of tan and sec respectively in reverse direction.\n\nThis method will help you to use this trigonometric table even when you dont have it. Hope it helps you atleast a bit.\n\n\n\nBy the way do you have any information about RUCHA VAIDYA smallest yet the brightest star? She has acted in movies like Ek Ajnabee and Main Aisa Hi Hoon. She has also done innumerable ads on TV. Please reply if you have any information about her. Absolutely any Kind of good information about her (even anything personal) is acceptable. Please answer my this question from celebrity or movies section. Please Please Please reply
485016
I assume that 30 deg is the included angle between the legs given.\n\nThen c^2 = a^2+b^2 -2abcos 30\n\nplug and chug:\nc^2= 31.518\nc= 5.614\n\nWhy is this called the ambiguous case?  I'd recommend making a sketch to see if this makes sense.  Btw, cos 30 comes from a 1, sqrt(3), 2 right triangle.  cos 30 is sqrt(3)/2 = 0.866
299922
I would convert both vectors to cartesian form using simple trigonometry. 250 degrees is a 3rd quadrant angle so its sine and cosine are -cos(70) and -sin(70) respectively.\n\nA is (-13 cos(70), -13 sin(70)) = (-4.4463, -12.2160)\n\nA quick check here: square these numbers and add them together, and you get 169.0000, which means I've got the right answer so far, by Pythagoras. \n\nSimilarly \nB is (-27 cos(60), +27 sin(60)) = (-13.5, +23.383)\n\nSame check still says I am on the right track. \nBut note that these things are vectors so you write them with the x displacement above the y displacement, not on one line as I have because I'm using a dumb keyboard instead of a pencil like real mathematicians do. \n\nA + B = (-4.4463, -12.2160) + (-13.5, +23.383)\n= (-16.6623,  +11.167)\n= (-16.66, +11.17) to 2 dp\n\nC = A - B = (-4.4463, -12.2160) - (-13.5, +23.383)\n= (+9.0537, -35.599)\n= (+9.05, -35.60) to 2 dp\n\nIf you prefer, you can convert the vectors back into polar form\n\nA + B = (-16.6623,  +11.167)\nx is -ve and y is +ve so this is a 4th quadrant angle\narctan(16.6623/11.167) = arctan(0.670196) = 33.8298 deg\nSo the angle of A + B is 33.83 + 270 = 303.83 deg to 2 dp\nThe length of the vector by Pythagoras again, jolly useful fellow he must have been\n16.6623^2 + 11.167^2 = 402.334 = 20.0583^2\nSo A + B = 20.06 in a direction 303.83 deg\n\nSimilarly\nC = (+9.0537, -35.599)  2nd quadrant angle\narctan(35.599/9.0537) = 75.7308 deg\n90 + 75.7308 = 165.7308\n9.0537^2 + 35.599^2 = 36.7323^2\nHence the vector C is 36.73 in a direction 165.73 to 2 dp
4562
Here are a couple i learned for double angle formulas.\n\nFor sin(a+b)= sin(a)cos(b)+cos(a)sin(b)\nSay out loud: "sine cosine cosine sine"\n\nFor cos(a+b)= cos(a)cos(b)-sin(a)sin(b)\nSay out loud: "cosine cosine SIGN sine sine"\nNotice the word "SIGN". The word SIGN is there to remind you that the SIGN of the rightmost term has the opposite SIGN of the input to the cosine on the left hand side of the equation (if its a+b, then subtract. if its a-b, then add).\n\nJust say it out loud like 3 times. It has a really nice rythm:\n\n"Sine Cosine Cosine Sine"\n"Cosine Cosine SIGN Sine Sine"\n"Sine Cosine Cosine Sine"\n"Cosine Cosine SIGN Sine Sine"\n"Sine Cosine Cosine Sine"\n"Cosine Cosine SIGN Sine Sine"\n\nSee? Easy.
72848
60+30=90\nand you know that sin 30 = cos 60, also sin 45 would be equal to cos (90-45) or cos 45, sin 75=cos (90-75) or cos 15(you can try this with a calculator or a log book to check)\nthe sin and cos values, for the first quadrant in a graph range from a minimum to a maximum value. (sin 0*=0 and cos 90*=1, then the values begin repeating themselves, decreasing and ten increasing again. (sin 180 = 0)\nso \ncot 65 = tan (90-65) = sin 25\n\nsimilarly\nif sin 2x = cos 8x, then 2x+8x must equal 90\n2x+8x=90\n10x=90\nx=9\nsee that sin 18 = cos 72\n\ncosec 10 = sec 80\n\ncosec 5x = sec x\n5x + x = 90\n6x=90\nx= 90/6\nx=15
114165
cos(315)degrees=cos(360-45degrees)\n               =cos 45degrees\n               =1/root2
72822
The Sec of X is -2sqrt3/3.\n\nThe reference angle of X is 30 degrees. My trig teacher requires that we memorize the values of all 30, 45, and 60 degree angles so I knew that off the top of my head. If you don't understand why, make a simple 30, 60, 90 degree triangle with side values x, xsqrt3 and 2x. You can see the sin (opposite/hypotenuse) of the 30 degree angle is x/2x, or 1/2. The tan (opposite/adjacent) is x/xsqrt3, or 1/sqrt3, which is sqrt3/3 when rationalized.\n\nThe 30 degree reference angle is in the 2nd quadrant. \n\nIn the 1st quadrant, all sin, cos, tan, etc. values are positive. In the 2nd, only sin and it's reciprocal (csc) are positive. In the 3rd tan and cot are positive. In the 4th, cos and sec and positive. You can remember this by the sentence "All Student Take Calculus". Starting in the first quadrant and moving counterclockwise (to 2nd,3rd and 4th) you know the positive values... A,S,T,C.. (All,Sin,Tan,Cos).\n\nKnowing that, we need to use process of elimination to pick a quadrant. Sin the tangent is negative, quadrants 1 and 3 are ruled out because tan is always positive in those quadrants. This leaves 2 or 4. We know the Sin is positive, so quadrant 2 is the right answer. In quadrant 4 only cos and sec and positive.\n\nTo figure out the sec, refer back to your 30, 60, 90 triangle picture. Sec is the reciprocal of cos. Cos is adjacent/hypotenuse, so Sec must be hypotenuse/adjacent. This is 2x/xsqrt3, or 2/sqrt3. When you rationalize the denominator you get 2sqrt3/3.\n\nSince we know the angle is in the 2nd quadrant with a 30 degree reference angle, we know X is 150 degrees. However, we don't need to know the angle measure to answer the problem. All we need to remember is only sin and csc are positive in the second quadrant, so our answer of 2sqrt3/3 must be negative.\n\nTherefore, the sec x is -2sqrt3/3.\n\nSorry this was very wordy. It's difficult to explain math over the internet! I hope it kinda cleared some of your confusion.
404829
I think the formula for sin(c-d) is sin c cos d - cos c sin d\n\nso sin (x - pi/3) = sin x cos pi/3 - cosx sin pi/3\n\nwhich is 1/2 sin x - sqrt3 / 2 cos x\n\nMultiplied by the -2 in front, this is -1 sin x + sqrt 3 cos x\n\nSo A = -1 and B = sqrt 3\n\nLook up the original formula, I may have messed up the signs
539200
(4.7)(9.81)cos 20 degree   ... you do the calculator punching..\n\ncheers
454344
In this case involving trigonometric operation on angles, you will need to carry the error over to your operation results. What I mean by that is:\n\nYou have an angle known only to 1 degree of precision, i.e.\n\nθ = 32° ± 1°\n\nTherefore, when you compute Cos (32°) = 0.848 \n\nYou will also need to perform the trig operation including the error:\n\nCos(31°) = 0.857 and Cos(33°) = 0.839\n\nSo your measurement is 0.848 ± 0.009 or 0.85 ± 0.01 depending on how many significant digits you want to keep.
159608
Given: \n1) a right triangle\n2) hypotenuse = 2\n3) the other two angles are 70 and 20 degrees\n\nFind:\nLengths of the remaining sides\n\nNote: \nWe have to be more specific here.\n\nLet:\na = side opposite the 70 degree angle (side adj. 20 degrees)\nb = side opposite the 20 degree angle (side adj. 70 degrees)\nc = hypotenuse = 2\n\nSolution:\n\n1) For side a:\n\nsin 70 = opp./hyp. = a/c\nTherefore, a = csin70 = 2sin70 = 1.88 ANSWER\nOR\ncos 20 = adj./hyp. = a/c\nTherefore, a = ccos20 = 2cos20 = 1.88 ANSWER\n\n2) For side b:\n\ncos 70 = adj./hyp. = b/c\nTherefore, b = ccos70 = 2cos70 = 0.68 ANSWER\nOR\nsin 20 = opp./hyp. = b/c\nTherefore, b = csin20 = 2sin20 = 0.68 ANSWER\nOR\nPythagorean Theorem: a^2 + b^2 = c^2\nb^2 = c^2 - a^2\nb^2 = 2^2 - 1.88^2\nb^2 = 0.4656\nTherefore:\nb = 0.68 ANSWER
554866
a) W = mg = 30*9.90662 = 294.1986 N\nb) Ft = 294.1986sinθ = 155.0897 N\nc) cosθ = 250/294.1986, θ  = 31.81376°\nd) a = 155.0897/30 = 5.19657 m/sec^2
593903
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!!
543045
sin(y-270) \n\nIs that sin(y - 270°)?\nIf so sin(y - 270°) = siny cos270° - cosy sin270°\n= siny * (0) - cosy * (-1)\n= cosy
588506
I got  the first part. just get the cosx by itself. then you wind up with the cosx = -(the square root of 2/2)\nwhich is 225+360n and 315 +360n(in degrees)\nSorry i can't help you on the other one. Good luck, just keep trying new things with it, would be my advice.
350677
C - 3 possible answers\n\nIn any given right triangles with the two other angles being a and b,\n\nsin a = cos b AND sin b = cos a,\n\n\nas such, the first and fourth combination will give one answer, the other two will either give square trigonometric values of each angle i.e. sin^2 a and cos^2 a  OR sin^2 b and cos^2 b.\n\nThe only condition that will not satisfy this is when the angle is 45 degrees, which means a = b and contradicting the given statement of a being less than b.\n\nThus the answer is C.
560385
The first point is not at exactly 210 degrees.  It is a little over 212 degrees.\n\nThe second point (3,3) is obviously at 45 degrees, not 30 degrees.\n\nThe distance between (-8, -5) and (3, 3) is\nsqrt((-8 - 3)^2 + (-5 - 3)^2) = sqrt((-11)^2 + (-8)^2) = sqrt(121 + 64)\n = sqrt(185)\n\nThe original angle from (3,3) to (-8, -5) is\natan2(-8 - 3, -5 - 3) = atan2(-11, -8) = 180 degrees + arctan(8/11)\ncounterclockwise.\n\nThe clockwise rotation subtracts thirty degrees from that, so in the final coordinate system the original (-8, -5) is sqrt(185) magnitude in the 150 degrees + arctan(8/11) direction from the new origin at the original (3,3).\n\nLuckily, sin( arctan(8/11) ) = 8 / sqrt(8^2 + 11^2) = 8 / sqrt(185)\nand cos( arctan(8/11) ) = 11 / sqrt(185).\n\nWe have the "new" x is\n\nsqrt(185) cos(150deg. + arctan(8/11)) =\n = sqrt(185) ( cos( 150 deg) cos ( arctan(8/11) ) - sin( 150 deg ) sin( arctan(8/11) ) )\n = 11 cos(150 deg) - 8 sin(150 deg)\n = 11 (-sqrt(3)/2) - 8 (1/2)\n = -(1/2)(8 + 11 sqrt(3)) or approx. - 13.526\n\nand the "new" y is:\n\nsqrt(185) sin(150deg. + arctan(8/11)) =\n = sqrt(185) ( sin( 150 deg ) cos ( arctan(8/11) ) + cos( 150 deg ) sin( arctan(8/11) ) )\n = 11 sin( 150 deg ) + 8 cos( 150 deg )\n = 11 (1/2) + 8 (- sqrt(3) / 2)\n = (1/2)(11 - 8 sqrt(3)) or approx. -1.428\n\nwhich makes sense, as the original slightly over 212 degrees after the rotation is slightly over 212 - 30 = 182 degrees, so both quantities are negative with the magnitude of x large and the magnitude of y small.\n\nCheck: "new x"^2 + "new y"^2 =\n = ( -(1/2)(8 + 11 sqrt(3)) )^2 + ( (1/2)(11 - 8sqrt(3)) )^2\n = (1/4)( 8^2 + 2*8*11 sqrt(3) + 3*11^2 + 11^2 - 2*11*8 sqrt(3) + 3*8^2)\n = (1/4)( 4 (8^2 + 11^2) )\n = 185 as it should be.\n\nCheck: atan2("new x", "new y") = 180 degrees + arctan("new y" / "new x")\n\n"new y" / "new x" =\n = (11 - 8sqrt(3)) / (- 8 - 11sqrt(3))\n = (11 - 8sqrt(3)) (- 8 + 11sqrt(3)) / ( (-8)^2 - (11sqrt(3))^2 )\n = (-88 + 121sqrt(3) + 64sqrt(3) - 88*3) / (64 - 121*3)\n = (- 352 + 185sqrt(3)) / -299 or approx. 0.105587\nwith arctan 6.02737\nand expected value arctan(8/11) - 30 or approx. 6.02737\nso this checks out, too.\n\nChanged coordinates:\n\n( -(1/2)(8 + 11 sqrt(3)), (1/2)(11 - 8 sqrt(3)) ) or approx ( - 13.526, -1.428 )