Inverse trig functions?
931057
f(x) = sin½x + a \nSo x = sin(½f^(-1)( x)) + a\nx - a = sin(½f^(-1)( x))\nso ½f^(-1)( x) = arcsin(x - a)\nSo f^(-1)( x) = 2 arcsin(x - a) \nDomain: a -1 ≤ x ≤ a + 1\nRange: - π ≤ f^(-1)( x) ≤ π
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They don't have values, they are operations, which means each one tells you to do something different (and shame on your teacher for not explaining this to the class BEFORE having you do them on your calculator). \n\nYou always have to have right triangles when dealing with these three trig functions (that's what sine, cosine and tangent are, trig functions). They are each ratios, which means a comparison of two numbers. Sine means to take the length of the opposite side (the side opposite the angle you're measuring) and put that number in a fraction as the numerator, then in the denominator put the length of the hypotenuse of the triangle. So for short, sine means "opposite over hypotenuse." Similarly, cosine means "adjacent over hypotenuse" and tangent means "opposite over adjacent." The adjacent side is the side forming the angle that is NOT the hypotenuse (this is really difficult to explain without being able to draw you a picture). My students called it SOHCAHTOA for short to remember them all.\n\nI'll try and walk you through an example as best I can...Draw an isoceles right triangle (which means the triangle has one right angle and two 45 degree angles). Since it's isoceles, the two sides that are NOT the hypotenuse must be the same length. So let's say those two sides each have a length of 1. If you do the pythagorean theorem (a squared plus b squared equals c squared) you'll get that the length of the hypotenuse is equal to the square root of 2, or approximately 1.414. Did you draw the pictures and label all the parts? \n\nNow pick one of those 45 degree angles and circle it. Let's calculate the sine of 45 degrees, WITHOUT using a calculator. Using that angle you chose, find the length of the opposite side (the side of the triangle that isn't touching the angle at all). It's 1 right? So put that in a fraction as the numerator, the number on top. Sine means opposite over hypotenuse, so in the denominator we need the length of the hypotenuse, which is the square root of 2 or 1.414. So your equation now says....\n\nsin 45=1/1.414\n\nOn your calculator, punch in 1 divided by 1.414 and you'll get the number 0.70721....\n\nMake sure your calculator is in degree mode, if it's not you'll get the wrong answer, it's a very common mistake!\n\nNow punch in sin 45 and hit enter. What number do you get? 0.70710....\n\nThe reason the two numbers are different is because we rounded the length of the hypotenuse. If we'd actually used the square root function on our calculators and typed in the square root of 2 instead of the decimal, they would have been exactly the same number. \n\nAnd it doesn't matter what the lengths of the sides of the triangles are, as long as they're right triangles. IF YOU DON'T HAVE RIGHT TRIANGLES THOUGH, YOU CAN'T USE THESE FUNCTIONS. So be careful. But that's all the sine, cosine and tangent represent. They are the ratios of the side lengths of right triangles. \n\nHere's a final example. The tangent of 51 degrees (tan 51) equals 1.23489..... That means that in a right triangle with a 51 degree angle, if you take the length of the side opposite the 51 degree angle and divide it by the length of the side adjacent to that angle (remember, tangent means "opposite over adjacent"), then you'd get 1.23489....\n\nI hope that helped. Good luck! I'm sorry I couldn't draw you a picture.
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sec(72.3pi)=1.701301618\n\nfirst change the mode of ur calculator to radians and then press the sequence of keys.\n\nand sec =cos-1 not sin-1\nso if u want sec , find cos first and then its inverse.\ncosec=sin-1\nsec=cos-1 \ncot=tan-1
247689
Use analytical trig (hint: sin function).  That should give you the answer you are looking for.
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Grab a caluclator.  Read what it says when you enter the square root of 4.  It gives you 2... 2 only, and not -2.\n\nA square root is a function, meaning it can only have one value.  A bunch of fellow math geeks a long time ago decided the non-negative one was better.\n\nIt's kind of like inverse trig functions.  The sine of a 0° angle is 0.  The sine of a 180° angle is 0.  So are the sines of 360°, 540°, -180°, and so on.  When it comes to the arcsine, though, arcsin(0) could equal -180°, 0°, 180°, 360°, or any of a host of others.  To make the Arcsin a function, the math geeks picked one, Arcsin(0) = 0°.\n\nYes, (2)(2) = 4.  Yes, (-2)(-2) = 4.  But 4 has only one square root, and it's 2.  People answering otherwise are thinking of solving the equation:\nx² = 4, which does have two solutions... x = 2 or -2.\n\nx² = 4\nSqRt(x²) = SqRt(4)  [Take the square root of both sides.]\n|x| = 2  [Simplifying.  Note on the right side, the square root of 4 is 2.  It's a function, and can only have one value.  On the left, taking a square root of a variable that would leave an odd power (x^1) means the expression must be an absolute value.]\nx = 2 or -2.  [It's solving the absolute value part of this problem that yields plus or minus 2.]\n\nAn equation [like x² = 4, or sin(a)=0] can have many solutions.\nA function [like the square root of 4, or Arcsin(0)] can have only one value.
463304
(sinx)^(-1) - sqrt(1-x^2) + C\nStart by multiplying (1+x)/(1-x) by (1+x) on the top and bottom\n[(1+x)(1+x)]/[(1-x)(1+x)]\nMultiply the denominator out but not the numerator.\n[(1+x)(1+x)]/[(1-x)(1+x)] = [(1+x)(1+x)]/(1-x^2)\nNow\nsqrt[(1+x)/(1-x)] \n= sqrt[[(1+x)(1+x)]/(1-x^2)]\n= (1+x)/sqrt(1-x^2)\nNow use trig substitution again letting x = sin(u)\nthen dx = cos(u) du\n(1+x)/sqrt(1-x^2) dx \n= [(1+sin(u))/sqrt(1-(sinu)^2)]cos(u) du\n= [(1+sin(u))/sqrt((cosu)^2)]cos(u) du\n= [(1+sin(u))/cos(u)]cos(u) du \nthe cosines cancel\n= [1 + sin(u)]du\n= u - cos(u) + C\nNow x= sin(u) so u = (sin x)^(-1)\nand using a triangle you can see that cos(u) = sqrt(1-x^2)
728902
This can be done easily using a scientific calculator or there are many angular / decimal tables in Trig text books or many machinists handbooks or machinists guides.\n\nwww.industrialpress.com/en/tabid/63/Default.aspx
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You're supposed to print a table, right? The question is poorly written, so it's not clear, but I'll assume that's the goal.\n\nWell, it sounds like you can write your loop, so shouldn't you just print out the sine, cosine, and tangent of the angle in the body of the loop?  It's not clear what the stumbling block is.  You realize that there are built in functions that compute sine, cosine, etc.?
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The problem is that thw word 'inverse' is used in two different ways here. It is perhaps easier to think of sec x as the *reciprocal* of cos x. Thus, if, say x=pi/3, cos x=1/2 and sec x=2.\n\nOn the other hand, arccos x is the function that *undoes* cos x. Thus, since cos(pi/3)=1/2, arccos(1/2)=pi/3.
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Switch your calculator from degrees (DEG) to radians (RAD). That should do the trick. \nHere are some quick trig rules you need to know:\nIn radians just remember that pi is 180 degrees. Pi/3 is 60 degrees, pi/2 is 90, pi/4 is 45 degrees and so on. \nIn a unit circle, the hypotenuse is always 1.\nSin is always the far side away from the angle (the vertical side). We call that the opposite side.\nCos is always the adjacent side (or horizontal side).\nIn a right triangle with angles 60 and 30, the shortest side is always 1/2 the hypotenuse, so in a unit triangle, it is just 1/2. You can use the pathagorean theorem to find that the longer side is the square root of (3/4), or (sqrt(3))/2.\nIn a right triangle with angles 45 and 45, the opposite and adjacent sides are equal, and we can use the pathagorean theorem to find that the sides are the square root of (1/2), or 1/(sqrt(2)).\n\n1. Sin 1/3 pi. This is the same as sin pi/3 which is a 60 degree angle. If you draw the triangle on a unit circle you would see that the opposite side is the longer side of a 90/60/30 triangle. We have seen above that the length is (sqrt(3))/2.\n\n2. Cos pi. Easy. pi is 180 degrees, so draw the unit circle. You see that it is just a striaght horizontal line going the other way so the answer is -1.\n\n3. cos pi/4 is a 45 degree angle. We know from above that both sides have to be 1/(sqrt(2)).\n\n4. sin pi/4. Same as #3.\n\n27. What is 315 degrees? Well draw your unit circle. Go, 90, then to 180, then to 270, then go 45 more degrees and you will end up at 315 degrees. Draw your triangle. We know that the hypotenuse is 1, so the coordinates will simply be, (cos 315, sin 315). Cos 315 is just a 90/45/45 triangle in the bottom right quadrant so the adjacent side is positive. The opposite is negative, so you end up with coordinates (1/(sqrt(2)), -1/(sqrt(2))). Do the same to solve the last question youeself.
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QUESTION 1:\n\nAsk yourself: WHICH TRIG FUNCTION IS OPPOSITE OVER ADJACENT?  IT IS THE TANGENT FUNCTION, RIGHT?\n\nThe bottom of the tree is 25 feet away from a point that measures 30 degrees from ground level to the top of the tree.\n\nAnswer: tan30degrees = x/25feet...choice 1.\n\nNotice that they are NOT asking you to solve for x.  Get it?  \n\nQUESTION 2:\n\nYou need to write the given equation 3x - 2y = 12 in the y =mx + b\nform.\n\nLet's do that right now.\n\nWe have:\n\n3x - 2y = 12 \n\nI will subtract -3x from both sides of the equation.\n\nWe get:\n\n-2y = -3x + 12...See what I did?  \n\nNext:\n\nDivide BOTH sides of the equation by -2 to find the value of y.\n\nWe get this:\n\ny = (-3x + 12)/-2\n\nWe can reduce the equation to this:\n\ny = 3/2 (x) - 6\n\nSlope is 3/2 and the y-intercept is -6.\n\nFinal answer: choice 1\n\nGuido
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-------------- edit ---------------\n\nThe function\n\n     sin(πx/180) - x/60\n\nis 0 when x = 0 and x = 30, and can be shown, using derivatives, to have a maximum between 0 and 30\n\n∴     sin(πx/180) > x/60 if 0 < x < 30\n\n∴     sin(xº) > x/60 if 0 < x < 30\n\nPutting x = 6 and 10 get\n\nsin 6º > 1/10\nsin 10º > 1/6\n\nand similarly, also get\n\nsin 1º > 1/60\nsin 2º > 1/30\nsin 3º > 1/20\nsin 4º > 1/15\nsin 5º > 1/12\nsin 12º > 1/5\nsin 15º > 1/4\nsin 20º > 1/3\n\n---------------- end of edit ------------------------------\n\nAlternatively, prove the special cases as follows.\n\nProof that sin 6º > 1/10\n-----------------------------\n\nSee Source 1 to calculate:\n\n     sin 36º = ¼√( 2 ( 5 - √5 ) ) \n     cos 36º = ¼√5\n\n     sin 30º = ½  and cos 30º = ½√3\n\nUse difference formula to get\n\n     sin 6º = sin( 36º - 30º ) = sin 36º cos 30º - sin 30º cos 36º\n\nSimplify to get (and check with Source 2):\n\n     sin 6º  =  [√6 √( 5 - √5) - (√5 + 1 ) ] / 8\n\nTo prove this > 1/10 rearrange, square, etc to get\n\n     30 - 6√5 > 206 / 25 + 18 / 5 √5\n\n     √5 < 34 / 15\n\n     1125 < 1156 which is true.\n\nA simpler method is similar to the following for 10º. Use the multiple angle formula for sin (5 × 6º) and show that when s = 1/10\n\n     32 s^5 - 40 s³ + 10 s - 1 < 0\n\nProof that sin 10º > 1/6\n-----------------------------\n\nUse multiple angle formula (Source 3) to get\n\n     sin 30º = 3 cos² 10º sin 10º - sin³ 10º\n\nLet s = sin 10º and simplify to\n\n     f(s) = 8s³ - 6s + 1 = 0\n\nUse calculus to show that this function decreases from 0 to ½ then increases. So sin 10º < 1/6 if f(1/6) > 0 which is true.
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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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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.
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Job one, convert everything to trig functions of a single angle, namely x.\n\nsin2x=2sinxcosx\ncos2x=cos^2(x)-sin^2(x)=2cos^2(x)-1=1-2sin^2(x)\n\nThen substitute in and simplify:\n2sinxcosx+2(2cos^2(x)-1)=1+sinx-4cosx\n2sinxcosx+4cos^2(x)-2-1-sinx+4cosx=0\n4cos^2(x)+2sinxcosx-sinx+4cosx-3=0\n\nFactor this into:\n(2cosx-1)(2cosx+sinx+3)=0\n\nset each parantheses equal to zero\n2cosx-1=0   \ncosx=1/2\nx=arccos(1/2)\nx={pi/3, 5pi/3} + 2pi*n  where n=0,1,2....\n\nAnd\n\n2cosx+sinx+3=0\n\nSolved numerically gives imaginary solutions of\n{-i*ln((-2/5)-(i/5),-i*ln(-2-i)}\n\nSolved....
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The inverse function is computed as below. In general, if you have a function of the form,\n\nf(x) = y, the inverse function is defined as \nx = f^(-1) (y), basically rearrange the equation so that you have the x on the left hand side of the equation.  This may not be always possible.\n\nIn the present case, when you solve for x,\ny = sqrt(x+5)\ny^2 = x + 5\nx = y^2 - 5\n\nhence the inverse function is y^2 - 5.
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Go to http://mathworld.wolfram.com/Cosine.html