Trig Identity:  Stumped for two hours!  Please help!?
912647
Left side: \n\n{[cos^2(x)/sin^2(x) - 1/sin^2(x)]}/{cos^(x)/sin^2(x) - cos(x)/sin^2(x)}\n\nsin^2(x) is denominator of all 4 terms, so combine on each side\n\n= (cos^2(x)-1)/(cos^2(x) - cos(x) \n\n= (cos^2(x)-1)/cos(x)(cos(x)-1)  \nFactor top as diff. of two squares;\n\n= [(cos(x) + 1) (cos(x)-1)]/cos(x)(cos(x)-1)\n\nCancel factor of (cos(x)-1)\n\n= (cos(x)+1)/cos(x)\n\n= cos(x)/cos(x) + 1/cos(x)  = 1 + sec(x)
732786
Using Trig identity 2sin(x)cos(x) = sin2x\n\n(2sin3x)(cos3x) = sin6x
241372
Csc (t) = 1/sin(t)\ntan (t) = sin(t)/cos(t)\n\nTherefore, Csc^2(t) = 1/sin^2(t)\n\ncsc^2(t)tan(t)sin(t) = [1/sin^2(t)][sin(t)/cos(t)]sin(t)\n                     = [sin^2(t)]/[sin^2(t)*cos(t)]\n                     = 1/cos(t)= sec(t)\n\nI do not know why they always make it complicated, why not just right sec(t)!
331507
sorry hunny you need at least 2 side lengths in order for us to help you figure it out... plus once you have that the formula for figuring it out is using the pathagorean theorem! a2+b2=c2 (s = squared) say for instance... the 2 sides of your triangle are 7 and 3inches. 3>2+7>2=c>2\nwhich would turn into 9+49=c>2~~~~~~~ then 58=c>2 then to get c by itself(which is needed) you would take the square root of 58! your hypotenuse would equal 7.6 or rounded, 8!\nnow if you use my steps you should be able to do your problem! I'm a 10th grader in geometry and it's the most lovely subject!
72362
I forgot how to do those, but it's some sort of trig identity isn't it?  Like\n\nsqrt (1-tan [x])\n\nProbably not though.
857524
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
1102373
i wont answer.... log onto www.mathguru.com and discover solutions with explanations.
291468
Oh yes i have cracked it at last....\n\nsolution...\n\nLet x=tan-1 (y) (i mean tan inverse)\nthen tan x = y,and cot x = 1/y (coz tan-1 (x) = cot-1 (1/x))\n\nthen ur given equation's Left hand side part (LHS) will be\n\n(1/y)/(1-y) + (y)/(1-(1/y))\n\n= 1/(y-y2)+ y/(1-(1/y))\n\n= 1/(y-y2) + y2/(y-1)\n\n= y2(y-1) - 1/y(y-1)\n\n= y3/y(y-1)\n\nI hope u understood till this step.\n\nNow we have (a3-b3) = (a-b) (a2+ab+b2)\n\nthen the LHS becomes...\n\n(y-1)(y2+y+1)/y(y-1)\n\n= (y2+y+1)/y\n\n= y+1+(1/y)\n\nsubstitute back y=tanx, 1/y = cot x and leave 1 as it is...\n\nU have LHS = 1 + tan x + cot x...Thats ur RHS !!!
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!!
882948
Just make the following substitution:\nr=sqrt(x^2+y^2)\ntheta=arctan(y/x)\nand then simplify.  Bring everything over to one side and you get your answer.\n\nIf tan(theta)=y/x, then sec(theta)=sqrt(x^2+y^2)/x.\n\nThe problem is a little ugly, but it is doable and then at the end everything ends up canceling nicely.
814944
we can  Use the trig identity\nsin(a+b)=sin(a)cos(b) + sin(b)cos(a) \nand we have\n\n10[sin(w.t)cos(π/4) + sin(π/4)cos(w.t)] \nwhich can be simplified as\n 10[√(2)/2sin(w.t) + √(2)/2cos(w.t)]\n= 5(√2)[sin(w.t) + cos(w.t)] \n\n a=5√2\n b=5√2
706522
It's often handy to draw yourself out a little sketch.  From it you can see the location of the given point.\n\nTo identify the angle the given point makes with the x-axis and the y-axis use trigonometry.  Switch to radians.\n tan Θ = opp./ adj.\n Θ = tan^-1 opp./ adj.\n Θ = tan^-1 (-0∙4425./ -0∙8968)\n Θ = tan^-1 (-0∙493 210 52)\n Θ =  0∙458 370 613 rad.\n    This is the value of the angle below the x-axis.  You need to include the angle above the x-axis, which is 180º or π radians.\n Θ =  0∙458 370 613 rad. + π rad.\n Θ =  (0∙458 370 613 + 3∙14) rad.\n Θ =  3∙598 370 613 rad.\n Θ ≈  3∙6 rad.
356153
using trig identities\nsin^2x+cos^2x=1, so 1-sin^2x=cos^2x and sin^2x=1-cos^2x\nsubstitue\ncos^2x(cosx)-2sin^2cosx\ncos^3x-2sin^2xcosx\ncosx(cos^2x-2*(1-cos^2x))\ncosx(cos^2x-2+2cos^2x)\ncosx(3cos^2x-2)
562730
cos2t=cos^2t-sin^2t or 2cos^2t-1 or 1-2sin^2t\n\nso you can arrive at something like 2cost-sect (remember 1/cost=sect). that's the simplest answer that you can get.
949562
Use csx =1/sin and  1-cos^2=sin^2  and cot=cos/sin\n Ready...Go\ncscx - (cosx)(cotx) =  1/sin - cos^2/sin = (1-cos^2)/sin=sin^2/sin =sin\n       ole, ole , infree
971568
What's the point of asking the same question twice?\n\nIf the issue is that you're only allowed to work on one side of the equation or the other, then you can pick one side, follow down to the bottom, and do everything in reverse to match the other side.\n\nRHS:\n\n1 + tan²x / tanx = [1 + (sec²x - 1)] / tanx = sec²x / tanx\n= 1/cos²x / (sinx / cosx) = 1/cos²x * cosx/sinx\n= 1/(sinx cosx) = cscx secx\n\nLHS:\n\n[cscx + secx]/[sinx + cosx] = [1/sinx + 1/cosx]/[sinx + cosx]\n= [(cosx + sinx)/(sinx cosx)] * 1/[sinx + cosx]\n= 1/(sinx cosx) = cscx secx
77120
It doesn't appear to simply significantly from the original expression.  I was able to get it to the following:\n\n(tan^3-tan)*sec\n\nDo the multipling by csc-1 top-n-bottom and then use the fact that cot^2=csc^2-1.  Bring the cot^2=cos^2/sin^2 to the top to get the above.\nGood luck.