I need this for my trig. class and have already Dr. Norman Wildberger . Thanks!
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It depends on what you mean by modern.  If you mean "not ancient," then three big names are:\n\nJohn Napier who discovered laws for solving spherical triangles.\n\nIsaac Newton who defined the trigonometric functions in terms of infinite series.\n\nLeonhard Euler who defined trigonometric functions in terms of complex numbers.
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I assume you know the basic trigonometry identity\n\n1) (1 + cscx) / sec x - cotx = (1 + 1/sinx)/(1/cosx) - (1/tanx)\n                                           = (cosx)(1 + 1/sinx) - (1/tanx)\n                                           = cosx + cosx/sinx - 1/tanx\n                                          = cosx + 1/tanx - 1/tanx\n                                          = cosx\n 2) sec^4x - tan^4x = [(secx)^2]^2 - [(tanx)^2]^2\n                               = [(secx)^2 - (tanx)^2][(secx)^2 + (tanx)^2]\n                               = 1[1 + (tanx)^2 + (tanx)^2]\n                               = 1 + 2(tanx)^2\n\n3) sin4x = sin2(2x) = 2sin(2x)cos(2x)\n                               = 2(2sinxcosx)(1 - 2(sinx)^2))\n                               = 4sinxcosx(1 - 2(sinx)^2)\n\n4) (cos3x - cosx)/(sin3x - sinx)\n= (cos(2x + x) - cosx)/(sin(2x + x) - sinx)\n= (cos2xcosx - sin2xsinx - cosx)/(sin2xcosx + cos2xsinx - sinx)\n=[cosx(cos2x - 1) - 2cosx(sinx)^2]/[2sinx(cosx)^2 - sinx + cos2xsinx]\n=[cosx(-(2sinx)^2) - 2cosx(sinx)^2]/[sinx(2(cosx)^2 - 1) + cos2xsinx\n= [-4cosx(sinx)^2]/[2cos2xsinx]\n= [-2cosxsinx]/[cos2x]\n= -sin2x/cos2x\n= - tan2x
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Switch your calculator from degrees (DEG) to radians (RAD). That should do the trick. \nHere are some quick trig rules you need to know:\nIn radians just remember that pi is 180 degrees. Pi/3 is 60 degrees, pi/2 is 90, pi/4 is 45 degrees and so on. \nIn a unit circle, the hypotenuse is always 1.\nSin is always the far side away from the angle (the vertical side). We call that the opposite side.\nCos is always the adjacent side (or horizontal side).\nIn a right triangle with angles 60 and 30, the shortest side is always 1/2 the hypotenuse, so in a unit triangle, it is just 1/2. You can use the pathagorean theorem to find that the longer side is the square root of (3/4), or (sqrt(3))/2.\nIn a right triangle with angles 45 and 45, the opposite and adjacent sides are equal, and we can use the pathagorean theorem to find that the sides are the square root of (1/2), or 1/(sqrt(2)).\n\n1. Sin 1/3 pi. This is the same as sin pi/3 which is a 60 degree angle. If you draw the triangle on a unit circle you would see that the opposite side is the longer side of a 90/60/30 triangle. We have seen above that the length is (sqrt(3))/2.\n\n2. Cos pi. Easy. pi is 180 degrees, so draw the unit circle. You see that it is just a striaght horizontal line going the other way so the answer is -1.\n\n3. cos pi/4 is a 45 degree angle. We know from above that both sides have to be 1/(sqrt(2)).\n\n4. sin pi/4. Same as #3.\n\n27. What is 315 degrees? Well draw your unit circle. Go, 90, then to 180, then to 270, then go 45 more degrees and you will end up at 315 degrees. Draw your triangle. We know that the hypotenuse is 1, so the coordinates will simply be, (cos 315, sin 315). Cos 315 is just a 90/45/45 triangle in the bottom right quadrant so the adjacent side is positive. The opposite is negative, so you end up with coordinates (1/(sqrt(2)), -1/(sqrt(2))). Do the same to solve the last question youeself.
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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!!
388197
http://www.amazon.com/Cosmos-Carl-Sagan-Jarom%C3%ADr-Hanzl%C3%ADk/dp/B000055ZOB
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ok so i assume you know the basic rules of trigonomerty for a triangle.\ni.e sinR = opposite/hypotenuse and cosR= adjacent/hypotenuse.\n\ndraw the triangle on a piece of paper.\nYou will see that sin R = ST/SR\n\n(where ST is the edge running from S to T and similarly SR is the edge running from Sto R)\n\nalso cos R= RT/SR\n\nso sinR^2 + cosR^2 = (ST)^2/(SR^2) + (RT)^2/(SR^2)\n\n= [(ST)^2 + (RT)^2]/(SR^2)\n\nNow we use pythagoras' theorem. I assume u know this? The sum of the squares of the two shorter sides of the triangle equals the square of the longer side (the hypotenuse)\n\nso (ST^2) + (RT)^2 = SR^2\n\nSO WE NOW HAVE\nsinR^2 + cosR^2 = SR^2/SR^2 = 1\n\nthere you go\n\n\nCANT E-MAIL YOU BACK WESTILE BECAUSE IT SAYS YOUR E-MAIL IS NOT VERIFIED
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http://www.wtamu.edu/academic/anns/mps/math/mathlab/beg_algebra/\n\nhttp://www.math.armstrong.edu/MathTutorial/
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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
155241
Please visit http://mathforum.org/library/drmath/view/52240.html
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You should be able to email him at:\n\ngreene@math.columbia.edu\ngreene@physics.columbia.edu\n\nHowever, he's a busy guy so don't be surprised if you get an auto-reply or no reply.\n\nI like this site to ask physicists questions:\n\nhttp://imagine.gsfc.nasa.gov/docs/ask_astro/ask_an_astronomer.html
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Try to find a book on analytic geometry (which should also cover trigonometry), and one that has lots of exercises, which will help you retain all that information.  There used to be a book called Geometry & Trigonometry for Calculus, but it's out of print.  As per the algebra, try to find a cheap Dover Classics book through amazon, they are usually only $10-20.  Also, the following site may help you:\nhttp://ltcconline.net/greenl/courses/152a/152a.htm\nhttp://ltcconline.net/greenl/courses/152b/152b.htm\nhttp://ltcconline.net/greenl/courses/154/154.htm\nhttp://tutorial.math.lamar.edu/AllBrowsers/2413/TrigEquations.asp\nhttp://tutorial.math.lamar.edu/AllBrowsers/2413/TrigFcns.asp\n\nGood luck!\n\nSteve
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When I was first trying to do trigonometry problems like this, I found it helpful to draw a picture.\n\nDraw a horizontal line to represent the ground, and then put a vertical line to represent the radio tower.  We want to know the height of the tower, so label it "x".  The angle between the tower and the ground is 90 degrees, so label it too.\n\nNow draw a dot (or a little stick figure with a protracter if you feel like it) to the side of the tower, on the ground.  Label the line between the tower base and the dot "50", as it's 50 meters from the tower.\n\nNow draw a line between the dot and the top of the tower, representing the person sighting to the top and measuring the angle.  Lable the angle between this sight line and the tower top "67 degrees".\n\nYou now have a drawing of the triangle with two angles and one adjacent side known.  This should be enough to get you started picking the right trig function to apply here.\n\nIf that's not enough to get going, here's another hint: Remember the unit circle with a triangle and the sides labeled with "sin(theta)", "tan(theta)" etc?  Are there any similarities to your tower drawing?  I always found being able to draw that unit circle on my scratch paper at the beginning of a problem set or test helped immensely.\n\nGood luck.
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here's a link, page down and see if you can follow the steps. :\ngood luck.\n\nhttp://cstl.syr.edu/fipse/Algebra/Unit2/expo.htm
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So you know the ones for sin(2a) and cos(2a). Cool.\n\nNow, tan(2a) is the same as sin(2a)/cos(2a), since tan = sin/cos (definition of tangent).\n\nSub in the formulae for sin(2a) and cos(2a) :\n\n2sin(a)cos(a)\n----------------\ncos(a)^2-sin(a)^2\n\nDivide both numerator and denominator by (cos(a))^2\n\n2tan(a)\n----------\n1-tan(a)^2
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Try "Calculus" - By Gilbert Strang\nThis is available along with Student study guide &\nInstructor's manual!\n\nhttp://ocw.mit.edu/ans7870/textbooks/Strang/strangtext.htm
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In quadrant one, sine, cosine, and tangent are positive. In quadrant two, sine is positive, and cosine and tangent are negative. In quadrant three, tangent is positive, and sine and cosine are negative. And in quadrant four, cosine is positive, and sine and tangent are negative.\n\nQuadrant 1 – Sin(+) Cos(+) Tan(+)\nQuadrant 2 – Sin(+) Cos(–) Tan(–)\nQuadrant 3 – Sin(–) Cos(–) Tan(+)\nQuadrant 4 – Sin(–) Cos(+) Tan(–)
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The Edifier has it right, but darned if I can follow the notation. You really need to see this written out like a textbook to understand what's going on.\n\nClick the link below in "sources" to see the answer hosted at my website www.calculus-help.com. Right now, Yahoo! answers, as far as I can tell, doesn't allow graphics, which makes it tricky to answer ANY math questions.\n\nPlease don't post math answers, guys, if the poster knows the answer but can't get there, and you get something completely different.