Can someone answer this for me?  Write a tangent function that has the same graph as y = cot Θ.?
94370
I won't write it for you, but cot is the inverse of tangent, if that helps.....
488283
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.
148719
You just draw a graph yourself.\n\n! ain't gonna graph for you. Rather ! love calculus.
816872
Best to plot this on a piece of graph paper; "+i" is  to the right' , "+j" is upwards, "-i" is to the left, "-j" is downwards, all with respect to the origin.\n\n1. Draw the vectors on the graph paper. "u" is in the second quadrant (i is neg and j is pos). "v" is in the fourth quadrant, with (i is pos, j neg) . There are two answers to this question, since you could go from u to v or from u to v in the opposite direction. I always take the angle from zero to "u" and add it to the angle from 0 to "v".  For "u", the tangent of the angle to the vertical axis is 1/4, then angle = 14%. So the angle from zero degrees to the vert axis = 90, and "u"s angle from zero degrees is 90 + 14 = 104 degrees. For "v", the tangent of the angle to the horizontal axis is 4/3 = 1.333.. and the angle = 53 degrees. Therefore the angle from "u" to "v" is 104 + 53 = 157 degrees. (the other answer is 360 degrees - 157 degrees = 203 degrees if you rotate in the other direction).\n\n2. Again use graph paper. 240 degrees is 30 degrees less that 270 degrees (the vertical axis, down) draw a line 10 graph units long. It will be in the third quadrant where i and j will both be negative. At this point you can use a polar to rectangular converter, or do it by hand. By hand: the i side of the triangle formed with the down axis can be found by saying sin(angle) = i/10, and cos(angle) = j/10. The angle is 30 degrees (from vert downwards). so sin(30) = 0.5, and, cos(30) = 0.866; thus i= 0.5 and j = 0.866 (ignoring pos/neg). Since the vector is in the third quadrant, both i and j are negative.\nAnswer = -0.5i -0.866j\n\nHope this helps!
943973
Tangent 25° 56' 30" = cot (90 - A)\n\n89 59 60\n25 56 30\n- - - - - - -\n\n64 3 30\n\nCotangent 64 degrees 3 minutes 30 seconds\n\nThe answer is. The complemenary angle of tangent 25° 56' 30" is cotangent  64° 3' 30 "\n\n- - - - - - - - - - - -s-
619642
Think about it. tangent is an odd function such that\nf(-x) = -f(x)\ntan(-3pi/4) = -tan(3pi/4)\n\nObviously we have a measure in radians, we want degrees.\nYou should know that:\n1 radian = 180/pi degrees\n1 degree = pi/180 radians\n\nConvert to radians\n3pi/4 radians\n3pi/4 (180/pi degrees)\n45 x 3\n135\n\nThus 3pi/4 radians = 135 degrees\n\nPlug back in \n-tan (135)\n\nWe have a reference angle of 45 degrees with the x axis. Tangent in the second quadrant is negative. What is the tangent of 45? It's 1, but because we're in the second quadrant, tan (135) = -1.\n\nHowever, don't forget we have that -tan. So, -1 x -1 = 1.\n\nAs a result, tan(-3pi/4) = 1.
757786
cot x = cos x / sin x. \nNow, to get the sin x on the bottom, it looks like we've differentiated ln(sin x). When we differentiate that we get 1/sin x * cos x, which is exactly what we want.\nSo its ln(sin x).
706522
It's often handy to draw yourself out a little sketch.  From it you can see the location of the given point.\n\nTo identify the angle the given point makes with the x-axis and the y-axis use trigonometry.  Switch to radians.\n tan Θ = opp./ adj.\n Θ = tan^-1 opp./ adj.\n Θ = tan^-1 (-0∙4425./ -0∙8968)\n Θ = tan^-1 (-0∙493 210 52)\n Θ =  0∙458 370 613 rad.\n    This is the value of the angle below the x-axis.  You need to include the angle above the x-axis, which is 180º or π radians.\n Θ =  0∙458 370 613 rad. + π rad.\n Θ =  (0∙458 370 613 + 3∙14) rad.\n Θ =  3∙598 370 613 rad.\n Θ ≈  3∙6 rad.
885716
cot(-x)cos(-x) + sin(-x) = -csc x\n\nLHS = cot(-x)cos(-x) + sin(-x)\n= cos(-x)/sin(-x) * cos(-x) + sin (-x)\n= cos²(-x)/sin(-x) + sin²(-x)/sin(-x)\n= (cos²(-x) + sin²(-x))/sin(-x)\n= 1/sin(-x)\n= 1/-sin(x)\n= -csc(x)\n= RHS
263848
cot = o/a\ncsc = r/o\n\no = opposite side\na = adjacent side\nr = radius\n\ncalculate csc(2x) in terms of o and r.\nSubstitute o/a for cot(x) and substitue the value you determine for csc(2x) \nmultiply these, then convert to a trig function in terms of x.\n\nIf you know trig identities better than I do, you can reduce this expression using identities, without resorting to o, a, and r.  However, either route will work.
299283
first draw a right triangle and label the angles and sides. now angle B cut in half is 15 degrees and angle A is 90 degrees.  lets assign AM a value of x.  so the tangent function is equal to the opposite leg over ajacent leg=> tan(15degrees)=x/(2*sqrt(3)).  take the tan of 15 and then multiply that by 2*sqrt of 3 and that will be your answer for the length of AM which is approx=.9282  to find MC you must first find the length of AC with the same tangent function using 30degrees as your angle so lets assign AC a value of y. so tan(30degrees)=y/(2*sqrt(3)) which is = 2 so MC=AC-AM= 1.0718
893232
f(g(x)) = f(√5 * x)\n= cot(π/10 * (√5 * x)\n= cot(πx/(2√5))\nWhen x = 1\nf(g(x)) = cot(π/(2√5))\n≈ 1.1813
333191
Consider the function f as mapping the points x of a set X to the points y of a set Y. Draw a horizontal line (the x axis) so that each point on the line represents an element of the set X. Draw a perpendicular line (the y axis) to intersect the x axis at the null point or zero. Each point on the y axis represents an element of the set Y. \n\nFor each x in X calculate the value of y= f(x) in Y. This gives a pair of co-ordinates (x,y). (Depending on the function, you may have more than on value of y for each x eg the function f(x)=square root of x will give + and - values.)\n\nFor any x on the x axis draw a line parallel to the y axis to intersect the x axis at x. Draw another line to intersect the y axis at the point y=f(x) and parallel to the x axis. Where these lines intersect is a point on your graph. The complete graph consists of all such points.\n\nWhere the set X is infinite (eg the set of real numbers) you cannot calculate the value of f(x) individually for every x but by calculating the values for a series of points at suitable intervals you will see a pattern emerging to the graph and you will be able to fill in the gaps by joining the (x,y) points, hopefully with a smooth and continuous line.
973710
"Main Idea: If f(x) is a function that has a derivative at a point x = a (i.e., f'(a) exists), then the line tangent to the graph of f at x = a, given by:\n\ny - f(a) = f'(a)(x - a);\nor, equivalently y = f(a)+f'(a)(x - a) can be an exceptionally good approximation of f(x) as long as x is close to a.\n\nAs a definition, we call the tangent line\nL(x) = f(a)+f'(a)(x - a)\nthe linearization (or linear approximation) of f at a. The\npoint x = a is the center of the approximation." (Ref 1)\n\nSo for f(x) = lnx at x = 8\n\nL(x) = ln8 + 1/8 (x - 8) (as f(8) = ln 8 and f'(x) = 1/x so f'(8) = 1/8)\n= 1/8 x + ln 8 - 1\n\nThis is useful to say estimate ln 8.1 if you know ln 8 (≈ 2.07944)\n\nSo ln (8.1) ≈ (8.1)/8 + 2.07944 - 1\n= 1.0125 + 2.07944 - 1\n= 2.09194\n\nBy calculator ln 8.1 = 2.09186..... so its pretty close (8^O)
225717
Yes. Think of the trignometric circle (radius = 1 and centered at the orgin (x=0, y=0). The so called tangent, is the line x=1 (that *is* tangent to this circle). The numeric value that we call tangent of an angle (alfa) is the y value of the intercept of a line passing thru the origin and forming an angle = alfa with the x-axis.\nIn other words, draw a trignometric circle and a verticle line tangent given by x=1. Draw another line forming an angle w/ the horizontal axis and passing thru the origin an stretch it so that it crosses that line. The height of the point of intercept of these two lines is the value of the tangent.\n \nThe co-tangent is the same but w/ y=1.\n \nEdit:\nThe secant is the distance from the origin to the intercept in the explanation above. Since it is a segment of the diameter, it is over a line that separates the trig circle, and is thus a secant.\n\nCheck out this link (skip the beginning of the page):\nhttp://www.npcts.edu/math/PreCalculus/Transcendental/Trigonometric/Definition/index.html
443025
(tanA + tanB + tanC - tanAtanBtanC) / (1 - tabAtabB - tabBtanC - tanCtanA)\ngets nasty fast, so remember these rules\ntanX = sinX / cosX\nsin(X+Y) = sinXcosY + cosXsinY\ncos(X+Y) = cosXcosY - sinXsinY\n\n(tanA + tanB + tanC - tanAtanBtanC) / (1 - tabAtabB - tabBtanC - tanCtanA)\n=[cosAcosBcosC * (tanA + tanB + tanC - tanAtanBtanC)] / [(1 - tabAtabB - tabBtanC - tanCtanA) * cosAcosBcosC]\n=[sinAcosBcosC + sinBcosAcosC + sinCcosAcosB - sinAsinBsinC] / [cosAcosBcosC - sinAsinBcosC - sinBsinCcosA - sinCsinAcosB]\n=[(sinAcosC + cosAsinC)cosB + sinB(cosAcosC - sinAsinC)] / \n[cosC(cosAcosB - sinAsinB) - sinC(sinAcosB + cosAsinB)]\n=[sin(A+C)cosB + sinBcos(A+C)] / [cosCcos(A+B) - sinCsin(A+B)]\n=sin(A+B+C) / cos(A+B+C)\n=tan(A+B+C)
730760
θ = α + β  where α = tan^-1 (1/3) and β = tan^-1 (1/2)\nie tanα = 1/3 and tanβ = 1/2\n\ntanθ = tan(α + β)\n= (tanα + tanβ)/(1 - tanαtanβ)\n= (1/3 + 1/2)/(1 - 1/3 * 1/2)\n= (5/6)/(5/6)\n= 1\nSo θ = ....., π/4, 5π/4, .....\nSo cosθ = ±1/√2 = ±√2/2