Trigonometric Identities?
542056
Start by factoring:\n\n(cos^2(x))(sin^2(x))-(cos^2(x)) = (cos^2(x))(sin^2(x) - 1)\n= -(cos^2(x))(1 - sin^2(x))\n= -(cos^2(x))(cos^2(x))\n= -cos^4(x)
702823
want the proof? Here it is ( i am assuming you know the dervatives of sinx and cosx)\n\ny = tanx\ny = sinx/cosx\ny' = [cosx(cosx) - sinx(-sinx)]/(cosx)^2 --> quotient rule\ny'= [(cosx)^2 + (sinx)^2]/(cosx)^2\ny' = 1/(cosx)^2\ny' = (secx)^2\n\nvoila
876696
1)\nRHS = (1-cot^2 x)/(1+ cot^2 x)\n= (1- cos^2 x/sin^2 x)/(1+ cos^2x/sin^2 x)\n= (sin^2x - cos^2 x)/(sin^2x + cos^2 x) muliypy numerator and denominator by sin^2 x\n= sin^2x -cos^2 x as denominator = 1\n\n2)\nRHS = sin x + cot x (cos x)\n= sin x + cos x/sinx .cos x\n= 1/sinx(sin^2 x+ cos^2x)\n= 1/sin x \n= csc x
733318
I think your identity should be :\ncsc^2(x) / [1 + tan^2(x)] = cot^2(x)\nas you forgot to square the tan(x).\n\nWorking on the 1 + tan^2(x) gives:\n1 + sin^2(x) / cos^2(x)\n= [sin^2(x) + cos^2(x)] / cos^2(x)\n= 1 / cos^2(x)\n\nNow we have : csc^2(x) / [1 / cos^2(x)]\n= [1 / sin^2(x)] / [1 / cos^2(x)]\n= cos^2(x) / sin^2(x)\n= cot^2(x), which was to be shown.
876211
2 cos^2 x + sin^2 x = 1\n\ncos^2 x + cos^2 x + sin^2 x = 1\n\ncos^2 x + (cos^2 x + sin^2 x) = 1\n\ncos^2 x + 1 = 1\n\ncos^2 x = 0\n\ncos x = 0\n\nx = pi/2 or 3 pi/2
336698
2. (2sin^2 x - 1)^2 / sin^4 x - cos^4 x = 1 - 2cos^2 x\n\nAssuming you meant\n(2sin^2 x - 1)^2 / (sin^4 x - cos^4 x)\nthen\n\n(2sin^2 x - 1)^2 / (sin^4 x - cos^4 x)\n = (2sin^2 x - 1)^2 / ((sin^2 x + cos^2 x)(sin^2 x - cos^2 x))\n = (2sin^2 x - 1)^2 / (sin^2 x - cos^2 x)\n = (2(1 - cos^2 x) - 1)^2 / ((1 - cos^2 x) - cos^2)\n = (1 - 2cos^2x)^2 / (1 - 2cos^2 x)\n = 1 - 2cos^2 x\n\n\n\nUpdate (added after wiildmunkii's answer)\n= = = = = = = = = = = = = = = = = = = = =\n\n1. sin^2 x + sin^4 x / cos^2 x + cos^4 x = tan 3x\n\nIf you meant\n(sin(2x) + sin(4x)) / (cos(2x) + cos(4x)) = tan3x\nthen yes.\n\nCombining\n\nsin(a+b) = sin(a) cos(b) + cos(a) sin(b)\ncos(a+b) = cos(a) cos(b) - sin(a) sin(b)\ntan(a + b) = sin(a + b) / cos(a + b)\n = (sin(a) cos(b) + cos(a) sin(b)) / (cos(a) cos(b) - sin(a) sin(b))\n = (tan(a) + tan(b)) / (1 - tan(a) tan(b))\n\nyou get\n\nsin(2x) = sin(x) cos(x) + cos(x) sin(x)\n = 2 sin(x) cos(x)\ncos(2x) = cos(x) cos(x) - sin(x) sin(x)\n = cos^2(x) - sin^2(x)\n = cos^2(x) - (1 - cos^2(x))\n = 2 cos^2(x) - 1\n [similarly = 1 - 2 sin^2(x)]\ntan(2x) = (tan(x) + tan(x)) / (1 - tan(x) tan(x))\n = 2 tan(x) / (1 - tan^2(x))\n\n\ne^it = (cos(t) + i sin(t))\n\nSo (e^it)^n = (cos(t) + i sin(t))^n\n\nBut (e^it)^n = (e^int) = cos(nt) + i sin(nt)\n\nSo cos(nt) + i sin(nt) = (cos(t) + i sin(t))^n\n\nFor any given n, expand the right hand side using the binomial theorem and equate real and imaginary parts to get:\n\ncos(2t) + i sin(2t) = cos^2(t) - sin^2(t) + 2 i sin(t) cos(t)\ncos(2t) = cos^2(t) - sin^2(t)\n = 2 cos^2(t) - 1\n = 1 - 2 sin^2(t)\nsin(2t) = 2 sin(t) cos(t)\n\ncos(3t) + i sin(3t) =\n cos^3(t) + i (3 cos^2(t) sin(t)) - (3 cos(t) sin^2(t)) - i sin^3(t)\ncos(3t) = cos^3(t) - 3 cos(t) sin^2(t)\n = 4 cos^3(t) - 3 cos(t)\nsin(3t) = 3 cos^2(t) sin(t) - sin^3(t)\n = 3 sin(t) - 4 sin^3(t)\n\ncos(4t) + i sin(4t) =\n cos^4(t) + i (4 cos^3(t) sin(t)) - (6 cos^2(t) sin^2(t)) - i (4 cos(t) sin^3(t)) + sin^4(t)\ncos(4t) = cos^4(t) - 6 cos^2(t) sin^2(t) + sin^4(t)\n = cos^4(t) - 6 cos^2(t) (1 - cos^2(t) + (1 - cos^2(t))^2\n = 8 cos^4(t) - 8 cos^2(t) + 1\nsin(4t) = 4 cos^3(t) sin(t) - 4 cos(t) sin^3(t)\n = 4 cos(t) sin(t) (cos^2(t) - sin^2(t))\n = 4 cos(t) sin(t) (2cos^2(t) - 1)\n = 4 cos(t) sin(t) (1 - 2 sin^2(t))\n\nFinally...\n\n(sin(2x) + sin(4x)) / (cos(2x) + cos(4x))\n = (2 sin(x) cos(x) + 4 cos(x) sin(x) (1 - 2 sin^2(x))) / (2 cos^2(x) - 1 + 8 cos^4(x) - 8 cos^2(x) + 1)\n = (6 sin(x) cos(x) - 8 sin^3(x) cos(x)) / (8 cos^4(x) - 6 cos^2(x))\n # divide by 2 cos(x)\n = (3 sin(x) - 4 sin^3(x)) / (4 cos^3(x) - 3 cos^(x))\n = sin(3x) / cos(3x)\n = tan(3x)
874407
sec x - 1 = tan x\nsec x = tan x + 1\nsec ^2 x = tan^2 x + 2 tan x + 1\n\n[Pythagorean identity: sin^2 x + cos^2 x = 1\nDivide through by cos^2 x to get tan^2 x + 1 = sec^2 x]\n\n(tan^2 x + 1) = tan^2 x + 2 tan x + 1\n2 tan x = 0\n\ntan x = 0 where sin x = 0, which is at x = 0, 180, or 360.
962689
To solve for #1, you need to use trigonometric substitution.  When doing trig substitution, you have to note the form of it.  They can be in one of these forms:\na) 1 - [junk]^2\nb) [junk]^2 -1\nc) 1 + [junk]^2 or [junk]^2 + 1\n\nThe above cases mirror the following trig identities.\na) 1 - (cos(x))^2 = (sin(x))^2\nb) (sec(x))^2 - 1 = (tan(x))^2\nc) 1 + (tan(x))^2 = (sec(x))^2\n\nWhen doing trig substitution, you have to keep this in mind.\n\nThe appropriate trig substitution in #1 is c).  You'll want to let x=1/(sqrt(5)tany) and work from there.  The answer is much too huge to put on here, and my goal was to get you started anyway.\n\nTo solve #2, Note that sqrt(2/x) is the same as sqrt(2)/sqrt(x).\n\nIntegral (sqrt(2/x))dx = Integral(sqrt(2)/sqrt(x))dx\n= sqrt(2) Integral (1/sqrt(x))dx (I just pulled out the constant).\n\nAnd then from here, all you have to do is note that 1/sqrt(x) is the same as 1/(x^(1/2)) which is the same as x^(-1/2).  Now it's the matter of using the reverse power rule.\n\nsqrt(2) * Integral (x^(-1/2)) = sqrt(2) * 2 * x^(1/2) + C
950350
LHS=sin^2x-cos^2x=(sinx+cosx)(sinx-cosx)\nRHS=1-2sinxcosx\n=(sin^2x+cos^2x)-2sinxcosx as sin^2x+cos^2x=1\n=(sinx-cosx)^2\nLHS and RHS can be equal only ifsinx+cosx=sinx-cosx\nso it is not an identity\nan identity must hole for all values
1159612
1) You meant to have a minus sign not "over". I've used x in place of "theta"\n    secx = 1/cosx and tanx = sinx/cosx \nSo 1 becomes (1/cosx)*sinx - tanx = 0 \nor tanx - tanx = 0, which is true.\n\n2) tanx = 1/cotx, so tanx*cotx = 1 \nSo 2 becomes 1- tanx*cosx = 1-sinx\nBut tanx = sinx/cosx so we have:\n1- (sinx/cosx)*cosx = 1-sinx, so\n1-sinx =1-sinx, which is true.\n\nI'll do one more:\n\n3) tan^2 x - 3sinx*tanx*secx = - 2tan^2 x\nAgain secx = 1/cosx and tanx = sinx/cosx\nSo 3. becomes:\ntan^2 x - 3(sinx/cosx)*tanx = - 2tan^2 x, so\ntan^2 x - 3tan^2 x = -2tan^2 x, so\ntan^2 x = tan^2 x, which is true.\n\nYou need to learn to do these yourself: Here is a link which gives you all the trig identities to solve the rest. They are not difficult.\n \nhttp://www.sosmath.com/trig/Trig5/trig5/trig5.html
160524
This is not a calculus problem, this is merely a trigonometric identity.  No derivatives, integrals, or numerical approximation were used.\n\nDefinition:\n\ntanh(2x) = sinh(2x)/cosh(2x)\n\nYou know the double angle formulas for sinh and cosh:\n\nsinh(x + x) = sinh(x)cosh(x)+cosh(x)sinh(x) = 2cosh(x)sin(h)\n\ncosh(x + x) = cosh(x)cosh(x)+sinh(x)sinh(x) = cosh(x)^2 + sinh(x)^2\n\nDivide:\n\n(2cosh(x)sinh(x))/(cosh(x)^2 + sinh(x)^2)\n\nDivide numerator and denominator by cosh(x)^2:\n\n((2cosh(x)sinh(x))/cosh(x)^2) / ((cosh(x)^2 + sinh(x)^2)/cosh(x)^2)\n\nSimplify:\n\n(2 sinh(x)/cosh(x))/(1 + sinh(x)/cosh(x)) \n\nSimplify more to solution:\n\n2tanh(x)/(1 + tanh(x)
870598
(sin x+ cos x)^2 = sin^2x + 2sin x cos x + cos^2 x using (a+b)^2\n= sin^2 x + cos ^2 x + 2 sinx cos x\n= 1 + sin 2x\nbecuase sin^2x+ cos^2x =1 \nand sin 2x = sin(x+x) = sin x cos x + cosx sin x = 2sin xcos x\nusing the formula sin(A+B) = sin A cos B + cos A sin B
676590
Sorry, I can't see the photo...but:\n\nSOH - sin = opp/hyp\nCAH - cos = adj/hyp\nTOA - tan = opp/adj
935340
= cosx[cosxcosy+sinxsiny]+sinx[sinxcosy-cosxsiny]\n\n= (Cosx)^2cosy+cosxsinxsiny+(sinx)^2cosy-cosxsinxsiny\n\n= (Cosx)^2cosy + (Sinx)^2cosy\n\n= Cosy [ (cos)x^2 + (sinx)^2]\n\n= Cos(y)
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 <--
324333
Look at this link.\n\nUse the first two "sum to product" identities.\n\nOn the right you will have:\n\n= (2 / 2) (sin a / cos a) (cos b / sin b)\n\n= 1 (tan a) (cot b)\n\na = (x + y) / 2,  b = (x - y) / 2\n\nso\n\n= (tan (x+y)/2) (cot (x-y)/2)\n\nOh, darn!  You're right, the signs are reversed from this.  I wonder if the problem was given to you incorrectly.  I don't see how you could go from our result to the problem the way it was given.  Ugh!\n\nI'm sorry, but I can't go any further with this.  Good luck, and let us know if the problem was stated wrong, please.
908517
On the calc. you'll see SIN, COS, and TAN\n\nSin- Sine\nCos- Cosine\nTan- Tangent\n\n\nAll used for trigonometry.