Find the exact value of:  sin (arctan 1/2).   How does one know to draw the triangle in Quadrant I?
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The trig functions are not one-to-one hence they don't have an inverse.  The only way for them to have an inverse is to make them 1-1.  You can only do this by restricting the domain.  In the case of the tangent, this is from -pi/2 <x<pi/2.  The inverse trig functions are all positive in the first quadrant only.  Look at the graphs of the trig functions to find the restrictions on the domain.  As for your problem, draw a triangle in QI.  arctan1/2 means "the angle whose tangent is 1/2."  Draw a triangle with tan = 1/2.  Since tan = y/x, y = 1 and x = 2.  Solve for r and you get r = sqrt5.  Since sin = y/r, you get 1/sqrt5 or sqrt5/5.  Since you have all three sides, you can define the other 5 functions of arctan1/2.
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You're looking for the angle that has a sine of 1/2.\nDrawing a triangle representing this is very helpful.  You will see that it is a 30-60-90 triangle.  \nIf you can find it in degrees, then you can convert it to radians by multiplying it by pi/180.\nsine(30) = 1/2\n30 degrees = 30(pi/180) = pi/6
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just draw a triangle has right angle which dimensions 1,sqrt(3),and hypothenus 2. if you want it, i can send to you
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The general formula is\narctan(x)-arctan(y)=\narctan[(x-y)/(1+xy)].\nIn your case, x=1+sqrt(2), y=1-sqrt(2), so\nx-y=2sqrt(2) and xy=-1. Thus, you get \narctan(2sqrt(2)/0)=pi/2.
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Because of (one of) the definition(s) of sin and cos.\nWhich is: take a unit circle, and draw in any radius at some angle theta. Then sin theta = the y value, and cos theta = the x value. Those are obviously less than 1.\n\nOr you could just take a typical right angled triangle - the length of the opposite/adjacent sides are always less than the length of the hypotenuse.
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I think that what you really want here is the tangent ratio. You know the slope of the line, right (hint, it is right there in the equation)? You know that the tangent ratio is a over b (where a and b are the non-hypotenuse sides of the triangle), right? You should be able to use that to figure out the tangent of the angle made by the line. Now all you have to do is figure out the angle that has that number as a tangent. That would be the "tan -1" (inverse tangent) function on your calculator.\n\nIf you are working from trig tables rather than a calculator, find the number in the second column of the tangent table that is closest to your tangent number then look in the first column for the angle.\n\nIf you are required to use sin and cosine remember your Pythagorean theorem and use it to find the hypotenuse.\n\nASIDE TO Drakestone: Your solution only finds the sine of the angle, not the angle itself. You need to apply "sin -1" to your answer to get the final answer.\n\n\nASIDE TO eternal...: You are wrong too. graph y=5x+3 and you will see that it is way too steep to be an 11 degree angle. You mixed up your opposite and adjacent values in calculating your tangent.
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First, you need a protractor. Create a right triangle ABC with C as the right triangle. Let angle B be your angle. Make rough measurements of the three sides of the triangle. now:\nsin B = AC/AB\n\nOnce you know its sine, use these formulae for the other trigonometric functions:\ncos B = sqrt(1 - sin² B)\ntan B = sin B/cos B\n\nSpecial cases are:\nsin 0 = 0\nsin 90 = 1\nsin 30 = 1/2\nsin 60 = sqrt3/2\nsin 45 = sqrt2/2\nFrom these, you can compute all angles.\nremember,\nsin B = sin (B - 360n) where n is an integer\nsin B = cos (90 - B)\ncos B = sin (90 - B)\ncos (-B) = cos B\nsin (-B) = -sin B\n\nexample:\ncalculate sin, cos and tan of -150.\nfirst, sin -150 = -sin 150\n-sin 150 = -cos(90 - 150)\n= -cos (-60)\n= - cos 60 = -sin(90 - 60) = - sin 30\n= -1/2\nsin -150 = -1/2\ncos -150 = cos 150 = sin(90 - 150) \n= sin (-60) = -sin 60\n= -sqrt3/2\ntan -150 = sin -150/cos -150\n= (-1/2)/(-sqrt3/2)\n= sqrt3/3\n\nJust use these sign rules:\nif angle is in Q1 = sin cos and tan are +\nif angle is in Q2 = sin is +,cos is -,tan is -\nif angle is in Q3 = sin is -,cos is -,tan is +\nis angle is in Q4 = sin is -,cos is +,tan is -\n\n^_^
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Depends\n\nlim θ→π/2- (ie from below) tanθ = +∞\n\nlim θ→π/2+ (ie from above) tanθ = -∞ \n\nThis is because in the 1st quadrant tanθ >0 and in the second quadrant tanθ <0
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It really depends upon your problem.\n\nIf you are asked to find the sine, cosine, and tangent then for any right triangle:\n\nsine = opposite side/hypotenuse  = y/r\n\ncosine = adjacent side/hypotenuse = x/r\n\ntangent = opposite side/adjacent side = y/x\n\nwhere opposite side is the side opposite the angle in consideration.\n\nwhere adjacent is the side closest to the angle.\n\nwhere hypotenuse is the longest segment in the triangle.
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Just as some of the other answerers have said,\nfrom 4cos^2(x) = 3, you get cos^2(x) = 3 / 4 and\nso cos(x) = ± sqrt(3) / 2.\n\nI'm imagining that this is where your trouble starts.\nOnce you have the basics, it's useful to remember\nnumbers such as sqrt(3) / 2, sqrt(2) / 2 and 1 / 2,\nwhen it comes to working out angles, as they come\nin handy lots of times.\n\nDraw a right-angled triangle with legs equal to 1 unit,\nthen by Pythagoras, the hypotenuse = sqrt(2) units.\nThis triangle is half of a unit square, so the 2 equal\nangles in the triangle must each be 45º.\nKnowing that sin = opp / hyp and cos = adj / hyp,\nwe have : sin 45º = 1 / sqrt(2) = sqrt(2) / 2, and\nthis is also true for cos 45º. Also, tan 45º = 1.\n\nNow draw an equilateral triangle with each side equal\nto 2 units. Then halve it by constructing a line from\none vertex to the mid-point of the opposite side.\nThis presents us with 2 right-angled triangles.\nUsing Pythagoras again on either of the 2 triangles\nformed, we find that the hypotenuse = 2 units, while\none leg = 1 unit and the other leg = sqrt(3) units.\nThe original equilateral triangle has all angles equal\nto 60º, so each half-triangle will have the angles 90º,\n60º and 30º.\nNow we have sin 30º = cos 60º = 1 / 2 and\nsin 60º = cos 30º = sqrt(3) / 2.\nAlso, tan 30º = sqrt(3) / 3 and tan 60º = sqrt(3).\n\nThese commonly used angles have relatively simple\nratios. Other angles have more complex exact values.\n\nBack to your question,  we have cos(x) = ± sqrt(3) / 2,\nor what is the same, ± cos(x) = sqrt(3) / 2.\nAs we have seen, the principal value of x is 30º. The\nprincipal value is the least positive angle for which the\nequation is true.\n\nTo find more angles, there are a couple of relations you\nprobably know. One is that cos(A) = cos(-A). This says\nthat cos(30º) = cos(-30º) = cos(360º - 30º) = cos(330º).\nHere, 30º is in Quadrant I and 330º is in Quadrant 4.\n\nThe other relation is : -cos(A) = cos(180º ± A).\nThus, -cos(30º) = cos(180º ± 30º) = cos(150º) or cos(210º).\nAs we've seen, it doesn't matter if cos(30º) is positive or\nnegative. In either case, the other 2 angles are 150º, which\nis in Quadrant 2, and 210º, which is in Quadrant 3.\n\nThus, all the angles x, are 30º, 150º, 210º and 330º.\n\nIn terms of π, knowing that π radians =180º, the\nangles in radians are π/6, 5π/6, 7π/6 and 11π/6.\n\nThese are the only angles between 0º and 360º,\nor, between 0 radians and 2π radians.
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in trigonomertic form\n\nz = r cos t + ir sin t = 1 + i\n\nso r cos t = 1\n r sin t = 1\n\nsquare and add r^2 = 2\nso r = sqrt(2)\n\nnow cos t = 1/sqrt(2) and sint = 1/sqrt(2)\n\nso t = pi/4 (as cos pi/4 = 1/sqrt(2) and sin is also positive so 1st quadrant)\n\nso t = pi/4 and r = sqrt(2)
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Unit circle is a circle centred in the origin (0,0) with radius 1.\nYou can use it to find trig. functions value as follows:\nDraw a line from the center to some point in the circle, and another vertical line from that point to Y=0. There you'll get a triangle. The Hypontenuse of that traingle measures ONE, then the the sinus and cosinus of the angle between Y=0 and first line you draw are represented by triangle sides.\nExamples: \n1. Draw an angle of 60° [using Y=0 as one side of angle], and complete the rect triangle, then measure the adjacent side and this is the value of cos(60), the oposite side is Sin(60)\n\n2. You want to calculate arcsin(√2/2), then draw a vertical line from Y=0 with length √2/2 that touch the circle, then draw the line between that point in the circle and the centre (0,0), measure the angle and that's it. Now measure the Adjacent side (over Y=0) of this triangle and 'voila' this is Cos(Arcsin(√2/2)).\nHope it helps.\n\nP.S. In the figure you'll find that Adjacent side is [ Using Pytagoras: H^=A^2+B^2]\nA=√(H^2-B^2)\n√[1-(√2/2)^2)]= √2/2\nCos( ArcSin(√2/2))=√2/2
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sin=opposite/hypotenuse\ncos=adjacent/hypotenuse\ntan=opposite/adjacent
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In rt triangle ABC with B= 90degrees (draw figure)      \nIf C is theta, cosC=7/8(ignoring sign)\ni.e., BC = 7, CA = 8 (in simplest terms)\nusing pythagoras theorem, AB= sqrt(15)\n \nsince C=theta is in the second quadrant tan C will be negative\n\nso tan C = AB / BC \n\n  tan C= - { sqrt(15) } / 7\n\ntan(theta) = -0.5533
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tan(theta)=x/x^3=1/x^2. So theta=arctan(1/x^2)
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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!!
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The first quadrant angle x whose sine is 1/2 is 60 degrees.\n\nThis angle has cosine of sqrt(3)/2.\n\nAnother way that we can do this is to remember that:\n[sin(x)]^2 + [cos(x)]^2 = 1 for all x.\n\n(1/2)^2 + [cos(x)]^2 = 1\n[cos(x)]^2 = 3/4\n[cos(x)] = sqrt(3/4) = sqrt(3)/2