The angle 90degrees has only four of six trigonometric functions. What r they??? PLEASE HELP ME!!?
921194
sin(90)=1\ncos(90)=0\ncsc(90)=1\nCot(90)=0
1159205
pair up the  opposite sines, for example sin(10) and sin(350) and add them:\nsin(10) + sin(350) = 0\nsin(20) + sin(340) = 0\n...\nuntil you only have sin(180) and sin(360) which both equal 0\n\nPS. I assumed you were using degrees (sexagesimal). Won't work in radians (RAD) or metric (decimal) angles.
728575
U should find out old class 9 Mathe. It can be helpful to u to understand easily basic of trigonometry i.e., 'Ө' and 'sine' in ur class standard.\nOld class 9 book can be available in old book shop. Find it at ur nearest old book shop.
763822
Divide the right angle into three equal parts, or three 30-degree angles.\n\nSounds like homework to me (which, hey, I'm cool with!), as if you owned this parcel this is NOT the way you'd want to divide it for real -- you wouldn't want three neighbors all bunched up like that at the confluence of the former right angle.\n\n.
1059709
i think ur question meant sinx = cosecx\nsinx = 1/sinx\nsin^2x = 1\nsinx = 1\nsinx is 1 at 90degree\nthus  x = 90 degree
954217
The sum of the angles of a rectangle is always 360 degrees. Four angles of 90 degrees each; you will always get 360 anyway you add them up. \n\nReminds me of 360 degrees in a circle. Full circle.
595637
if you dont have one already, make a nice big copy of the entire circle with everything labeled.  repetition and review is really just the best way to remember anything.  when i try to think of the unit circle values for 30-60-90 triangles, i just think of it in terms of the long side and the short side.  the short side is always 1/2 and the long side is always (sq. root of 3)/2.  if i need to remember the values for, say, a 120 degree angle, i always draw a quick little sketch- just enough to show which are the short and long sides.  the horizonal is short, so the cosine is 1/2.  the vertical is long, so the sine is (root 3)/2.  this method works for all angles that are multiples of 30 degrees.  45, 135, 225, and 315 degrees are easy because you only have to remember one value for all the sine and cosines: (root 2)/2\n\njust dont forget to look at your sketch to see which of the values is negative! (and if the picture method doesnt work for you, theres a trick to knowing those too . . .\n\nhere are the quadrant numbers: (imagine the axis in between)\n\n2 1\n3 4\n\nuse this acronym to remember what trig functions are positive in each quadrant:\n\nALL---------> ALL of them are positive\nStudents----> Sine is positive\nTake--------> Tangent is positive\nCalculus----> Cosine is positive
714757
(sinX + 1) = 0---> sinX = -1--->x = 3pi/2 + (2pi)n, radians, or x = 270 + 360n degrees, where n is any whole number including 0. \n\nThe first step was simple algebra.  In the second step, I just remembered that the sine is how far up on the unit circle that angle points, so -1 means it points straight down, which is 3/4 of a circle.  Adding a whole circle puts it right back where it started from.\n\ncosX + 1/2 = 0---> cosX = -1/2 ---> x = 2pi/3 or 4pi/3 + (2pi)n radians, 120 or 240 + 360n degrees.  \n\nAgain, the first step was simple algebra.  As for the second, my trig teacher made us memorize the sines/cosines of 30, 45, and 60 degrees, so I just know that the cosine of 60 degrees is 1/2.  Go around to the other side of the unit circle and the cosine of 120 degrees is -1/2.  Go down to the bottom of the unit circle, and the cosine of 240 degrees is also -1/2.
1092612
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(–)
1094719
It's quite dificult to explain trigonometry this way, so it;s better for you to have an electronic teacher, this teacher is at your service day or night, sun or rain and  you could find it\n\nhttp://aleph0.clarku.edu/~djoyce/java/trig/\n\nhttp://www.sosmath.com/trig/trig.html\n\nhttp://fergusmurray.members.beeb.net/trig.htm\n\nhttp://mathworld.wolfram.com/Trigonometry.html\n\ngood luck
811153
Hi, \n\nThe answer lies in the pythagorean theorem. \n\nFor any right (one of the angles is 90 degrees) triangle with sides a,b and c where c is the side opposite the 90 degree angle:\n\na^2 + b^2 = c^2\n\nso for your first excercise:\n\na = 4\nb = 4 Sqrt(3)\nc = ?\n\na^2 + b^2 = c^2\n4^2 + {4 Sqrt(3)}^2 = c^2\n16 + 4^2 * 3 = c^2\n16 + 16 * 3 = c^2\n64 = c^2 \nc = +- 8 (since we're dealing with a triangle, -8 doesn't make sense, so just ignore it.)\n\nc = 8. \n\nFollow that same pattern, and you should be all set. \n\nHope that helps, \nMatt
722736
the angles are just unknown if i get you correctly.\nthere are sin, cos, and tan proportions you can work out to find the rest. i think you only need three of the measurements to work out the rest of the triangle, like 2 sides and an angle or 2 angles and a side.  not positive.
546529
1. Make sure the angle of interest is in radians.  Radians are the "natural" (non)-units of angles, because they relate the angle measurement to the length of arc the angle cuts off on a unit circle.  So 2*Pi radians are equivalent to 360 degrees.  Basically, if you have that sucker in degrees, multiply it by 2 * Pi / 360.  Unit conversion is done. \n \n2. Make sure the angle is between -Pi and Pi.  Because sine is a periodic function, with period 2*Pi, just subtract or add 2*Pi enough times to get the angle between those two points.\n\nnow for the meat of the problem:\n\nsin(x) = x - x^3/3! + x^5/5! - x^7/7! + x^9/9! - ... + \n+ (-1)^(n-1)*x^(2n-1)/(2n-1)! + ...\n\nExample: sin(15)\n\nStep 1: 15 degrees = 15*(2*Pi/360) radians = Pi/12 radians \n \nStep 2: I need Pi/12 in decimal (let's see, Pi=3.14159 (memorize that much, it's pretty easy... "Pi when calculating sine, 3 point 1 4 1 5 9!")) \nPi/12 approx = .261799 (we only had 6 sig figs, so we can't add any more than that) \n \nNow to the series: \n         first term:  x=.261799 \n         1st estimate: .261799 \n         second term: -x^3/3! = -.0029905  \n         2nd estimate: .258809 \n         third term: x^5/5! = .0000102     \n         3rd estimate: .258819 \n         fourth term: -x^7/7! = -.167*10^-7  \n         stop here - past precision\n\nThe answer is positive \nsin(15) = 0.258819\nsin(18) = 0.309017
98672
use the tangent ratio ( trig function ) or inverse tangent of (6/ (43/8)) OR iNV TANGENT (1.116) = 48.14 degrees and to get the other angle, just subtract it from 90 degrees or 41.86 degrees. hope it helps how I got it. go over the trig function or cohsahtoa rule
762124
I wouldn't use the sine function to solve this problem.\n\nYou know the angle of elevation, and something about both legs of two right triangles. This looks like a job for the tangent function.\n\nCall the initial distance to the kite x, and the elevation of the kite y.\n\ntan 47 = y / x\n\nWhen you increase x by 30, the angle of elevation becomes 39 degrees.\n\ntan 39 = y / (x + 30)\n\nSolve both relations for y and equate the two:\n\ny = x tan 47\ny = (x + 30) tan 39\n\nx tan 47 = (x + 30) tan 39\n\nx tan 47 = x tan 30 + 30 tan 39\nx tan 47 - x tan 30 = 30 tan 39\nx = 30 tan 39 / (tan 47 - tan 39)\nx = 92.516 feet.\n\nNow you can find y:\n\ny = x tan 47\ny = 92.516 tan 47\ny = 99.21 feet.\n\nSince we didn't work with the hypotenuse of either right triangle, we didn't use the sine function, but since you did ask about it:\n\nh^2 = y^2 + x^2\nh = sqrt(x^2 + y^2)\n\nWhere x and y are defined as before.\n\nsin 47 = y / h\nsin 47 = y / sqrt(x^2 + y^2).\n\nSimilarly:\n\nsin 39 = y / sqrt[(x+30)^2 + y^2]\n\nI would take issue with the fact that we don't know x. As it turns out, we can figure it out without using the sine function, as shown previously.
969737
sine 30° 15'\n\nCos ( 90 - A)\n\n89 60\n30 15\n- - - - -\n\n59 45\n\nThe answer is cos 59° 45'\n\n- - - - - - - -s-
215098
You can't have 3 corners at 90 degrees without the fourth also at 90 degrees.\nYou can have two at 90 and the remaining add up to 90 degrees.\nA penta (5sided) gram, with even sides and angles, where the angle is 360/5 = 72 degrees.