I know how to draw the two sine curves, but not the curves of the reciprocal functions...
132163
Let y = arccosine x\nthen x = cos y\nfirst put a value for y and get the corresponding value of x\nDraw the graph
75360
I think I get it.  You're given the two circles (which I assume do not intersect and one is not inside the other) and then you have to come up with the equations for two tangential lines to both of these circles.  \n\nOnce you have the two lines this becomes a linear inequality system.  What you're asking is if the point is in the area bounded by one line on top and the other line on the bottom, so write these two lines as two inequalities.   Example: \n\ny1 < mx1 + b\n\ny2 > mx2 + b\n\nI think this is the format, it's been a while since I did stuff like this.  You can graph it by shading the area between the lines.\n\nAnyway, once you have this it's pretty easy: if a point satisfies both inequalities then it's inside the area.  If not, then it's outside.\n\nOkay, how to draw the tangent lines?  Well, this is a little trick to explain, it would be easier if I could graph it.  But the thing to remember is that by definition a tangent line is perpendicular to the circle at that point.  Thus if you draw a line connecting the centers of the two circles, and then draw one tangent line to the circles, then draw a line connecting the center of each circle to the tangent point on each circle, you will see that you have two similar right triangles that share the same angle.\n\nAt this point I don't think you need to use trignometry to find the equations of the tangent line ... but my mind is slipping a bit at this point and I need a moment to think it out.\n\nupdate:  I found the solution to finding the tangent to 2 circles knowing only the center points and the radii, but I'd have to draw you a picture to show you how.  It's complicated, as it involves a bunch of similar right triangles.
201089
Aha, geometry!\n\nDraw a diagram and follow the steps below.\n\nThe trick is to show that the triangles DBE and \ntriangle DAF are similar.\n\nProof That triangles DBE and triangle DAF are similar.\n\nLet <ABC> donote angle ABC.\nWe need to show any two corresponding angles of these traingles are equal.\n\nFirst angle,\n<FDA> = <BDE>  because D,B,F collinear and D,E,A collinear.\n\nSecond angle,\n<DAF> = <DAC> because A,F,C collinear\n      = (pi) - <ACD> - <ADC> angles in triangle ACD\n      = (pi) - <ACD> - <ADC>  ...(1)\n\nNow, separately  manipulate,<ACD> and <ADC>.\n\n<ACD> = <ACB> + <BCD> splitting <ACD> with line CB\n      = <ACB> + <CAB>, tangent property <BCD> = <CAB> \n      = (pi) - <CBA>  angles in triangle ABC\n      = <ABE>  because C,B,E are all colinear\n\nNow, similarly\n\n<ADC> = <ADB> + <BDC> splitting <ADC> with line DB\n      = <ADB> + <BAD>, tangent property <BDC> = <BAD> \n      = (pi) - <ABD>  angles in triangle ABD\n         \nSubstituiting these two info into (1), we get,  \n\n<DAF> = (pi) - <ACD> - <ADC>\n      = (pi) - <ABE> - [ (pi) - <ABD> ]\n      = <ABD> - <ABE>\n      = <BDE> because BE splits <ABD> \n\nDone. In showing the triangle DAF and triangle DBE are similar.\n\nHence, by ratio for similar triangles,\n\n        DA/DF = DB/DE\n\nOr Rewriting we get\n \n      DA * DE = DB * DF\n\nProved.\n\nNot so difficult, was it. hehehe.\n\nRegards
648074
Yeah where's angle 1 and 2.\nBut remember the angle opposite the longest side has the greastest measurement.
448008
With these data a lot of angles fulfill that requirement; If  you have alpha first look at cos alpha, after divide by1.33  and look for theta.\n\nDid you not make a mistake. I suppose you mean sin and not cos\n\nSnell law uses sinus
11526
The Edifier has it right, but darned if I can follow the notation. You really need to see this written out like a textbook to understand what's going on.\n\nClick the link below in "sources" to see the answer hosted at my website www.calculus-help.com. Right now, Yahoo! answers, as far as I can tell, doesn't allow graphics, which makes it tricky to answer ANY math questions.\n\nPlease don't post math answers, guys, if the poster knows the answer but can't get there, and you get something completely different.
440237
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.?
391480
b isnot true, reciprocal of 3 should be 1/3
554546
They taught us SOHCAHTOA back in OL trig, but I never really felt the need for it, I mean just picture a right angled triangle and the longest side is the hypotenuse,h. Take an angle and the side opposite it is h.sin x , where x is the angle. The other side is h.cos x
100562
The derivative of sinx is just cosx, but the derivative of cosx is -sinx.  Then the derivative of -sinx is -cosx, and the derivative of that is sinx.  So it actually takes four applications of the derivative to get back where you started rather than two.  \n\nIf you're comfortable with the definition of the a derivative you can work it out and see that that's true.  Otherwise, just draw the graphs and look at where the slopes are negative and where they're positive.\n\nHope that helps!\n\nPS.  When you're looking at the page your question is on, you can click "additional information," and add more there rather than making a new question.  Saves you five points. :-)
552711
Take the first few derivatives of sin -- it is cyclical and you will see a pattern.  After you see the pattern, you will be able to say exactly what the 53rd derivative is, without having to actually take the derivative 53 times!\n\n~ ♥ ~
15370
(1) Subtitute the damn values for what you have.\n\n(2) Sin theta=opp/hyps,\n    Tan   "  =opp/adj,\n    Cos   "  =adj/hyps.\n\nNow usually a blind man follows his stick now you might know where you are at the moment - get it???
19211
There is no closed form for the nth derivative of this function.  I symbolically calculated and factored the first 50 derivatives on MathCAD, and there was not a pattern.  Sorry, but you're out of luck....
570307
You have two ways to do it...\nOne... if you have to do it on a graph paper... \nthen.. put some values of x and solve for y and make a table... and then plot those points on graph\n\nOtherwise.. use graph plotting softwares.. like gnuplot (which is a free software).. define function as f(x)=2x (f(x) is equivalent to y here) and say plot f(x).. it will plot you the function.
540250
I was able to draw 16 different forms (8 with their symmetrical reversed forms).  There may be more but I don't have the time to work on it.\n\nI took another look at the problem and realized that in addition to the left-right symmetry there is also a top-bottom symmetry that adds at least another 14 answers to the 16 that I came up with earlier.  So there are at least 30.\n\nCould be more, but it is my bed time.
621156
For a fixed a and n, and allowing b and c to vary, the curve would be a hyperbolar centered at the origin. You will only be taking the rational points on that curve, though.
383806
Sine = opposite/hyp\nCosine = adjacent/hyp\nTangent = opposite/adjacent