exactly how did the Pythagoreans find cos, sin, and tan?
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The Pythagorean didn't find these. In fact, trigonometry didn't get started until Hypparchus in the second century BC (Pythagorus was in the 6th century BC). But Hypparchus didn't use the trig functions we use today. The sine function wasn't used until about the early 5th century AD in India.
20199
The basic idea behind trigonometry is that once you know the angles of a triangle, you'll know the ratios of the sides of the triangle.  And so if you know one side, you can use trig to find out the two other sides.\n\nSine, cosine, and tangent deal with right triangles.  They tell you the ratios of different pairs of sides in the right triangle.\n\nHere's a brief intro:\nhttp://opera.answers.com/trigonometry
748672
using Maclaurin series you find the series for sin and cos and then use long division for the Maclaurin series of tan
36505
Great question.\n\nI think a maths history project would be wonderful.\n\nSpecifically, I would like to see a paper which explains how the trigonometric functions were calculated in history.\n\nFor example, what angles did the Greeks know trig functions for?\n\nWhat resources did Isaac Newton have for trig functions?\n\nHow were these tables calculated before electronic computers?\n\nThe combination of history and maths would be very interesting.\n\nCheers!
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http://en.wikipedia.org/wiki/Pythagorean_theorem#Proofs\n\nCheck below for the "proof of converse".  It's the most straightforward there is.\n\nBasically, you try to construct a triangle with x,y, and z, only to discover that it is congruent to a right triangle!
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Brian gave you the answer, but he didn't explain it thoroughly.\n\nPlace your vector at the origin of an x and y plane.  The x and y axes are perpendicular, so you can use the Pythagorean formula to get the magnitude which is the length of the hypotenuse of the right triangle formed by the points (0,0), (10,0), and (0,8). sqrt(10^2+8^2)\n\nNow you need direction which is the angle between the x axis and the vector. Since you have solved for all three lengths of the triangles sides, you can use any inverse trig function to find the angle.  Brian chose tangent.\n\nGood work, Brian!
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I know of two:\n\nTranslation and rotation: \n\n1) translation :\n\n Take a rectangle and cut a diagonal and extended the lines  from all three side of the triangle at right angle to it  from the side that is made from cutting the rectangle until those lines equal the lenght of the side of the triangle drawn from  and it  form three boxes  around all three sides of the triangle. Now let take the big side an calulate the square  of it ,then calulate the squares of the other two sides and add themup,  it should equal the square of the big side and that proves the pythagorean theorem. \n\nRotation: \n\nSee the show : Project Mathematics by Cal tech : Entitled "The Pythagorean Theorem" it explain better  than I can, It worth watching. Very good show on Math & Geometry
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(1) Subtitute the damn values for what you have.\n\n(2) Sin theta=opp/hyps,\n    Tan   "  =opp/adj,\n    Cos   "  =adj/hyps.\n\nNow usually a blind man follows his stick now you might know where you are at the moment - get it???
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sin² x + cos² x = 1\n\nCenter a unit circle on the origin.  Let x be a point on the circle x units of distance around the circle from a starting point of (1,0). [so x is in radians].  Then the horizontal coordinate of x is sin x and the vertical coordinate is cos x, and so by Pythagorean theorem sin² x + cos² x = 1.
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This lesson will allow you to figure out the Pythagorean Theorem all by yourself.  Go ahead and click on the preface link.\n\nhttp://www.arcytech.org/java/pythagoras/index.html
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You can use sin and cosine.
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Pythagoras? As in the Pythagorean theorem. (In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.)
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I am not sure what you mean. I assume you mean what is tan(x+a)=sin(x+a)/cos(x+a)\n=[sin(a)cos(x)+sin(x)cos(a)] / [cos(x)cos(a)-sin(x)sin(a)]\n=[tan(a)+tan(x)] / [1-tan(x)tan(a)]\n\nSo tan(x), if the argument x is translated by a, becomes this expression.
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The word "Trigonometry" is derived from two Greek words meaning measurement or solution of triangles. Trigonometry is a branch of mathematics that deals with the ratio between the sides of a right triangle and its angles. Trigonometry is used in surveying to determine heights and distances, in navigation to determine location and distances, and in fields like nondestructive testing for determining things such as the angle for reflection or refraction of an ultrasound wave.\n\nThere are three principle functions in trigonometry: sine A, cosine A, and tangent A where A is an angle. These are typically abbreviated for use in algebra to: sinA, cosA, and tanA. These terms are defined in terms of a right triangle. SinA is equal to the side opposite of the angle A (side a) divided by the hypotenuse of the triangle (side c). From the right triangle below, it can be seen that the value of angle A is directly linked to the ratio of side a and side c. In other words, if the length of side a is changed (and side c is not changed by the same amount), then angle A must change. When the division of the length of side a by side c is performed, the resulting value is directly related to angle A. The relationship between the decimal value of the ratio of the side and the angular value of angle A can be looked up in trigonometry tables or, as is more common these days, programmed into a scientific calculator. The relationship between angle A and the ratio of any two of the three sides can be determined by using either the Sin, Cos or Tan functions. CosA is equal to the side adjacent to the angle A divided by the hypotenuse and tanA is the sine divided by the cosine and is therefore the side opposite the angle divided by the side adjacent.
726104
Pythagoras (of Samos, born about 569-82 BC - died about 507 BC) was a Greek philosopher who made important developments in mathematics, astronomy, and the theory of music. The theorem now known as Pythagoras's theorem was known to the Babylonians 1000 years earlier but he may have been the first to prove it. \n\nIn any right triangle, the area of the square whose side is the hypotenuse (the long side of a right triangle opposite the right angle) is equal to the sum of areas of the squares whose sides are the two legs (i.e. the two sides other than the hypotenuse, the adjacent and opposite in sine, cosine and tangent workings). \n\nIf we let c be the length of the hypotenuse and a and b be the lengths of the other two sides, the theorem can be expressed as the equation\n\nThis equation provides a simple relation among the three sides of a right triangle so that if the lengths of any two sides are known, the length of the third side can be found.\n\nI was taught this in a long jumpo sand pit, with 25 billard balls.\nmy old teacher, Mr Chadwick, made a line of 5 balls, and this was the hypotenuse. he then drew in the rest of the triangle. he squared the 5 balls, by making a square of 5 by 5 balls, (25 in all).\n\nthen he put balls on the other lines. three fitted onto the short side, and four onto the next side. he then squared the two other sides. 3 x3 and 4 x 4 (9 and 16, when added together = 25. \n\nhe showed me that 27 years ago, It was amazing, and the reason I am now a teacher.
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One. \n\nThis relation is really just Pythagoras' theorem, hiding in a new form. Recall:\n\nH^2 = A^2 + B^2 \n\nThat is, the square of the hypotenuse H is the sum of the squares of the other two sides A and B, for a right triangle with sides A, B and hypotenuse H.  \n\nOf course, A=H*cosx and B=H*sinx for some angle x, so that \n\nH^2= (H^2)*(sinx*sinx + cosx*cosx)\n\nleading to the result:\n\n1=sinx*sinx + cosx*cosx\n\nThis result also is nicely connected to the simple fact that \n\nx - x = 0 \n\nbecause \n\nexp(i*x)=cosx + i*sinx and\nexp(-i*x)=cosx -i*sinx\n\nwhere i=sqrt(-1). \n\nThen multiplying these two, \n\nsinx*sinx + cosx*cosx=exp(i*x)*exp(-i*x)=\nexp(i*[x-x])=exp(0)=1
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its just triangle mathe. remember soh -cah -toa... sine =opposite side/hypotenuse,  cosine=adjacent side/hypotenuse, tangent= opposite side/adjacent side....cosecant(hypotenuse/opposite side) is the opposite of sine, secant(hypotenuse/adjacent side) is the opposite of cosine and cotangent(adjacent side/opposite side) is the opposite of tangent