can someone explain sin, cos, and tan?
469918
In order to define the trigonometric functions for the angle A, start with an arbitrary right triangle that contains the angle A:\n\nWe use the following names for the sides of the triangle:\n\nThe hypotenuse is the side opposite the right angle, or defined as the longest side of a right-angled triangle, in this case h. \nThe opposite side is the side opposite to the angle we are interested in, in this case a. \nThe adjacent side is the side that is in contact with the angle we are interested in and the right angle, hence its name. In this case the adjacent side is b. \nAll triangles are taken to exist in the Euclidean plane so that the inside angles of each triangle sum to π radians (or 180°); therefore, for a right triangle the two non-right angles are between zero and π/2 radians. The reader should note that the following definitions, strictly speaking, only define the trigonometric functions for angles in this range. We extend them to the full set of real arguments by requiring that they be periodic functions.\n\n1) The sine of an angle is the ratio of the length of the opposite side to the length of the hypotenuse. In our case\n\n. \nNote that this ratio does not depend on the particular right triangle chosen, as long as it contains the angle A, since all those triangles are similar.\n\nThe set of zeroes of sine is .\n\n2) The cosine of an angle is the ratio of the length of the adjacent side to the length of the hypotenuse. In our case\n\n. \nThe set of zeroes of cosine is .\n\n3) The tangent of an angle is the ratio of the length of the opposite side to the length of the adjacent side. In our case\n\n. \nThe set of zeroes of tangent is .\n\nThe remaining three functions are best defined using the above three functions.\n\n4) The cosecant csc(A) is the multiplicative inverse of sin(A), i.e. the ratio of the length of the hypotenuse to the length of the opposite side:\n\n. \n5) The secant sec(A) is the multiplicative inverse of cos(A), i.e. the ratio of the length of the hypotenuse to the length of the adjacent side:\n\n. \n6) The cotangent cot(A) is the multiplicative inverse of tan(A), i.e. the ratio of the length of the adjacent side to the length of the opposite side:\n\n. \n\nMnemonics\nThere are a number of mnemonics for the above definitions, for example SOHCAHTOA (sounds like "soak a toe-a" or "sock-a toe-a" depending upon the use of American English or British English. Can also be read as "soccer tour"). It means:\n\nSOH ... sin = opposite/hypotenuse \nCAH ... cos = adjacent/hypotenuse \nTOA ... tan = opposite/adjacent. \nMany other such words and phrases have been contrived.
890202
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)
324104
Uncomputable. It's is either plus infinity or minus infinity. The plot of tan from 0 to 180 degrees looks a bit like a flat line becoming ever steeper until you reach 90 degrees, when it's vertical (IE at infinity), also at 90 degree it's a line 'coming from' minus infinity.\n\nThe source below has a good plot of tan(x) but it's in radians - same plot, just replace angle(radians) with angle(the radians number * 360 degrees)/(2 * pi) to get the angle in degrees.
106150
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.
1090403
x = r .cos ө\ny = r. sin ө\nsince x=y\nr.cos ө = r.sin ө\nsin ө/cos ө =1\nө = tan-1 ( 1)\nө = ∏/4\nr^2 = x^2 + y^2\nr^2 = 2.x^2\nr = x√2\n\npolar form would be ( r.CiS ө)  { i.e. r.cos ө + i.sin ө }\n\nso the polar form is x√2.CiS (∏/4)
1050391
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!!
388496
There are six ratios of the three sides of a right triangle. These ratios have been identified by name throughout the centuries. In the USA they are known as sine, cosine, tangent, cotangent, secant, and cosecant. Your calculator has buttons for three of these: [SIN], [COS], and [TAN]. Remember them with the non-word SohCahToa. \n\nThe tangent ratio of an angle is simply the ratio of the opposite side to the adjacent side of a right triangle. \n\nIf you know the angle, type \n\n[TAN] > angle > [ENTER].\n \nThe answer will be the tangent (ratio of the opposite side to the adjacent side).\n\nIf you know the opposite and adjacent sides and need to find the angle, type \n\n[2nd] > [TAN] > opposite > [/] > adjacent > [ENTER]. \n\nThe answer given will be the desired angle.\n\nCaution: \n(1) Different calculators require these buttons to be pushed in a different order.\n(2) If your angle is in degrees, then make sure your calculator's mode is in degrees.\n(3) Don't get the numbers backwards or you'll find the cotangent of the angle and not the tangent.
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(–)
700676
tan sq x -1 = 2\nadd 1 to both sides\ntan sq x = 3\nsquare root both sides\ntan x = the square root of 3\nrefer to the values of trigonometric funtions\nx = pi/3
857149
we know\n\ntan A/2 = sin A/(1+cos A) = sinA (1- cos A)/(1- cos^2 A)\n= (1- cos A)/ sin A = cosec A - cot A\n\nusing this we get cot A = cosec A - tan A/2\n\n2) let( A+B)/ 2 = x\n(A-B)/2 = y\n\nA = (x+Y)\nB= x-y\nsin A + sin B = sin (x+y) + sin (x-y) = 2 sin x cos y\nsin A - sin B = sin (x+y) - sin (x-y) = 2 sin y cos x\n(sin A+ sin B)/(sin A - SIn B) = tan x/tan y = tan (A+B)/2/tan(A-B)/2\nQED
787158
y = xtan(x)\n\nUse product rule:\ndy/dx = x/cos²(x) + tan(x)\n= x/cos²(x) + sin(x)/cos(x) \n= [x + sin(x)·cos(x)] / cos²(x)\n\nUse product and chain rules:\nd²y/dx² = 1/cos²(x) + (-2)·(x/[cos(x)]^3)·(-sin(x)) + 1/cos²(x)\n= 2/cos²(x) + [2·x·sin(x)] / [cos(x)]^3\n= [2·cos(x) + 2·x·sin(x)] / [cos(x)]^3
932480
sin(A+B)sin(A-B)/(cos²Acos²B)\n= (sinAcosB + cosAsinB)(sinAcosB - cosAsinB)/(cos²Acos²B)\n= (sin²Acos²B - cos²Asin²B)/(cos²Acos²B) (since (x + y)(x - y) = x² - y²)\n= sin²A/cos²A - sin²B/cos²B\n= tan²A - tan²B ← Option c)\n\nPlease note: sin(A + B) = sinAcosB + cosAsinB\nand sin(A - B) = sinAcosB - cosAsinB
1050478
tan(x) = y  implies arctan(y) = x\n\narctan ( x )  = tan-1 ( x )\n\ndoes this help u
983394
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
1021739
Since tanx = sinx/cosx,\n\ntan(A-B) = sin(A-B)/cos(A-B)\n\nNow, write out the expression for each term on the right side of this equation.\n\nWhen you finish that, divide each term by cosAcosB and you will get the result you are seeking.
677828
Yes use the trig identities:\n\ncsc x = 1/sin x\n\nSo for your first question\n\ncsc x = 3 sin x becomes 1 = 3 sin(x)^2\n\nso sin(x)^2 = 1/3  sin(x) = 1/rt(3) = 0.57735 and 35.26 degrees using the arcsin function.\n\nNote 1/cot x = tan x = sinx/cos x\n\nSo your second question becomes:\n\n2 sin x/cosx + 3sinx= 0  or 2sinx(1/cosx +3)=0\n\nThis is true if sinx = 0 or x= 0 or 180 degrees or k2pi, kpi where k=(0,1,2,3... \n\nor 1/cosx + 3 =0 which happens 1/cosx = - 3, means cosx = -1/3 means x = 109.5 degrees using arc cos function etc.
282780
1.sin(2x)=2sin(x)cos(x)\n2.tan(x)=sin(x)/cos(x)\n3.cos(x)=(1-sin^{2}(x))^(1/2)\n\nwe can use (3) in (2) and (1) to eliminate cos(x). Then substitute sin(x)=y:\n\n1.sin(2x)=2y.(1-y^{2})^(1/2)\n2.tan(x)= y/(1-y^{2})^(1/2)\n\ntherefore,\n\nf(y)=2y(1-y^{2})^{1/2}\n     +y/(1-y^{2})^{1/2}\n\nI hope that's what you wanted.