Is this when I use trig?
958095
resultant of the force?\nsounds like you need a Jedi knight
1176396
sin=opposite/hypotenuse\ncos=adjacent/hypotenuse\ntan=opposite/adjacent
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.
104581
It takes practice!  And its like a puzzle.  Just use the main trig identities as reference and a good math tutor.  Unfortunately, this place isn't the best place to ask that kind of question because we are limited by text explainations and trig identities need a lot more than that!
528735
tan by definition is equal to sine/cosine.\n\nProof:\n\ntan = opp/adj\nsin = opp/hyp\ncos = adj/hyp\n\nsin/cos = (opp/hyp)/(adj/hyp) \n= opp/hyp*hyp/adj \n= opp/adj\n= tan\n\nSorry about that, knew that was too simple :P!\n\nNEW ANSWER:   You nedd to draw out the triangle and solve from there.  Let there be a triangle with adjacent side a,  and opp side b.  \n\nThen if tanbeta = 2, opp= 2, adj = 1 and hyp = sqrt(5)   from the pythagorean theorem.\n\nThus cos = 1/sqrt(5) and sin = 2/sqrt(5)\n\nThe quadrant only tells you the sign of cos and sin or how the angle is measured.  Since you're in QI, both sin and cos are positive so no modification to the above answers are needed.\n\nNow you might wonder why I said opp was 2 and adj was 1.  It doesn't matter (can't remember the proof for that right now).  As long as you pick a solution that works, say 4 and 1 or 3 and 1.5, it will ALWAYS simplify to the answer I just gave you.
1185586
this has happened to me before on my 89. it's usually when the expression is too complicated--that is, it would take the calculator more memory than it has to solve it. so i'd recommend trying to first simplify the expression and then enter it. For instance, don't leave things in terms like sin(2) when you can just get an approximate number for that. maybe that's it?
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
623185
Kind of sounds like Chinese Trigonometry to me! Is this a question or an answer???
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.
946513
Subject: Re: Proving trig identiies...small error..thisis the right one\nFrom: \nDate: Monday June 14, 2004 at 08:35:54\n\nRHS= 2csc^2x=2(cot^2x+1) \n\nnote :-(csc^2x=cot^2x+1) \n\n=2cot^2x+2 \n\n=2cos^2x/sin^2x +2 \n\nnote :-((cot^2x=cos^2x/sin^2x) \n\n=2cos^2x/(1-cos^2x) +2 \n\nnote :-((sin^2x=1-cos^2x) \n\n=2cos^2x +2(1-cos^2x) /(1-cos^2x) \n\n=2cos^2x +2 -2cos^2x) /(1-cos^2x) \n\n=2/(1-cos^2x) \n\n=2/(1-cosx)(1+cosx) \n\nnote :-((1-cos^2x=(1-cosx)(1+cosx)) \n\n\nTHEREFORE lHS =2/(1-cosx)(1+cosx) (1-cos^2x=(1-cosx)(1+cosx)) \nnow for the RHS we get \n\nRHS= 1/1+COSX + 1/1-COSX \n\n=(1-cosx) + ( 1 + cosx)/(1-cosx)(1-cosx) \n\nnote :-(( through expansion) \n\n=2/(1-cosx)(1+cosx) \n\nnote :-(cosx on the numerator cancel out) \n\ntherefore RHS= LHS \n\nTHEREFORE 2csc^2x= 1/1+cosx + 1/1-cosx \n\nas required\n\n\n\nCurrent Thread\n   Proving trig identiies   a girl - Monday June 14, 2004 at 08:06:52 \n   Re: Proving trig identiies   - Monday June 14, 2004 at 08:31:28 \n--> Re: Proving trig identiies...small error..thisis the right one   - Monday June 14, 2004 at 08:35:54 <--
791247
You are correct about the conversion; however, the use of 'pi' is indicative of radians.
736311
It equals (2/3)*pi.   It does not equal (-1/3)*pi  -- that would be (-inverse cosine (-1/2)).
649598
Because you have two possibilities\nit can be either sinx or -sinx (depends on the quadrant in which the angle is)\nthat's because of the definition of absolute value\n\nand yes, I now see that you have sin x^2 and not sin^2x- so those are completely different, \nsin^2x=sinx*sinx\nand sinx^2 is simply sinx^2, try any angle and you'll see the difference,\nso onlt sqrtsin^2x=sinx
20199
The basic idea behind trigonometry is that once you know the angles of a triangle, you'll know the ratios of the sides of the triangle.  And so if you know one side, you can use trig to find out the two other sides.\n\nSine, cosine, and tangent deal with right triangles.  They tell you the ratios of different pairs of sides in the right triangle.\n\nHere's a brief intro:\nhttp://opera.answers.com/trigonometry
596750
The Pythagorean theorem only works with right triangles.  Try using SIN, COS or TAN.
316686
the process u follow is correct, but actually it is wrong. fundamentally wat u did is ok, but technically it is wrong. because this becomes a quadratic equation which has more than 1 root. so this equation has more than 1 answers for which it is true.
895625
This may be unexpected, but a good site with the formulas you need.\n\nhttp://en.wikipedia.org/wiki/Trig