Trigonometic Transformation?
397638
Use these identities:\nsin(2x) = 2*sin(x)*cos(x)\nsin(4x) = 4*cos^3(x)*sin(x) - 4*cos(x)*sin^3(x)\ncos(2x) = -1 + 2*cos^2(x)\ncos(4x) = 1 - 8*cos^2(x) + 8*cos^4(x)\ntan(3x) = (3*tan(x)-tan^3(x))/(1-3*tan^2(x))\n\nAfter you expand the left hand side of the equation with the above identities, you can factor out a 2*sin(x)*cos(x) from the numerator and a 2*cos^2(x) from the denominator, which will turn into a tan(x). Now divide the numerator and denominator of the fraction by cos^2(x). You should get the right hand side of the tan(3x) identity above.\n\nTry this and if you can't figure it out, I'll write it out. (I don't really want to since equations look like garbage on this site).\n\nCheck out this site for some usefull identities.
381123
Yes. I assume to want to find sin 1/2x, cos 1/2x, and so on.\nIt's quite simple:\n1. Write the expanded form of cos 2x\ncos^2x - sin^2x\n2. Write that twice, in once in a column (remember it equals cos 2x)\ncos^2x - sin^2x  = cos 2x =  cos^2x - sin^2x\n3. Rewrite cos^2x in the first column as 1-sin^2x, and in the second column rewrite sin^2x as 1-cos^2x\n(1-sin^2x) - sin^2x = cos 2x = cos^2x - (1 - cos^2x)\n4. Combine like terms\n1-2sin^2x   and -1+2cos^2x\n5. In order to find cosx and sinx, we must isolate each:\n-2sin^2x = cos 2x -1  and 2cos^2x = cos 2x + 1\n6. Divide by the coefficent of each\nsin^2x = 1/2 - 1/2(cos 2x) and cos^2x = 1/2 + 1/2(cos 2x)\n7. Find the square root of both sides\nsin x = sqrt(1/2 - 1/2(cos 2x)) and \ncos x = sqrt(1/2 + 1/2(cos 2x))\n8. We found what sin x and cos x equal, we want what 1/2x equals. Let x = 1/2x\nsin 1/2x = sqrt(1/2 - 1/2(cos 2(1/2x)) and\ncos 1/2x = sqrt(1/2 + 1/2(cos 2(1/2x))\n9. Simplify\nsin 1/2x = sqrt(1/2 - 1/2(cos x)) and\ncos 1/2x = sqrt(1/2 + 1/2(cos x))\n10. You're done. You can find tan 1/2x, csc 1/2x, sec 1/2x, and cot 1/2x by using the trigometric identities with sin 1/2x and cos 1/2x:\n\ntan 1/2x = (sin 1/2x)/(cos 1/2x)\ncsc 1/2x = 1/(sin 1/2x)\nsec 1/2x = 1/(cos 1/2x)\ncot 1/2x = (cos 1/2x)/(sin 1/2x)\n\nObviously there are other expansions for the half angle formulae. These are just the easiest.
748572
sin(x)tan(x)=(cos(x)csc^2(x))/(cot^2(x)+cot^4(x))\nStep 1. Starting with the right side (the obviously more complicated one) factor out a cotangent squared out of the denominator.\n\n(cos x csc^2 x)/ cot^2 x (1+cot^2 x)\n\nStep 2. Use the identity csc^2 x= cot^2 x+1 in the numerator  and separate the cosine x in the top and the cotangent squared x in the bottom\n\n(cosx/cot^2 x) * (cot^2 x+1/ cot^2 x +1)\n\nStep 3. This leaves you with a cosine x divided by cotangent squared x times 1.  Use the identity that says cot x = cos x/ sin x\n\n(cos x) / (cos^2 x/ sin^2 x)\n\nStep 4. Divide the double fraction and get\n\n(sin^2 x cosx)/ (cos^2 x)\n\nStep 5. Separate your multiple factors from the squares and separate the fraction into three parts, one will be the sine x, the other will be the tangent x (because tan x = sin x/ cos x), and the cosine x divided by cosine x will go to 1.\n\nsin x * (sin x/ cos x) *(cos x/ cos x)\n\nStep 6.  Write final answer\n\nsin x tan x = sin x tan x
982428
LHS\n= (sin^2B-tan^2B)/(1-sec^2B)\n= (sin^2B-tan^2B)/(-tan^2B)\n= (sin^2B/-tan^2B) + (-tan^2B/-tan^2B)\n[splitting the numerator up]\n= (-cos^2B) + 1\n[cosx= sinx/tanx]\n= 1 - cos^2B\n= sin^2B\n= RHS
263848
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.
983394
Depends\n\nlim θ→π/2- (ie from below) tanθ = +∞\n\nlim θ→π/2+ (ie from above) tanθ = -∞ \n\nThis is because in the 1st quadrant tanθ >0 and in the second quadrant tanθ <0
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.
812461
As I mentioned in another post here the other day, there is one technique that *always* works, even if it isn't the fastest.\n\nChange everything to sin and cos. Use sin^2 x + cos^2 x = 1. Simplify. Thats it.\n\nSo, the LHS is \n1/tan x - tan x\n= cos x / sin x  - sin x / cos x\n= (cos^2 x - sin^2 x) / (sin x cos x).\nNow, we have the right denominator, but the numerator isn't right yet. What else can we simplify? We know how to change cos2x into something just involving cos x or sin x:\nYou may already know that cos 2x = cos^2 x - sin^2 x. If not, maybe you know this one:\ncos 2x = 2 cos^2 x - 1\n= 2 cos^2 x - (sin^2 x + cos^2 x)\n= cos^2 x - sin^2 x.\nAnd thus the two sides are equivalent.
247689
Use analytical trig (hint: sin function).  That should give you the answer you are looking for.
459338
Go to http://mathworld.wolfram.com/Cosine.html
826654
we start on RHS as it is more complex\n\n(a-b)^2 (cos C/2)^2 + (a+b)^2 (sin C/2)^2 ... 1\n\ncos t = 2 cos^2 t/2 -1 so cos^2 t/2 = (cos t+1)/2\ncos t = 1- 2 sin^2 t/2 so sin ^2 t/2 = (1-cos t)/2\nputting in1 RHS\n= (a-b)^2 (cos C +1 )/2 + (a+b)^2 (1- cos C)/2\n= cos C((a-b)^2-(a+b)^2)/2 + ((a+b)^2 +(a-b)^2)/2 (rearranging terms)\n= -2ab cos C + a^2 + b^2\n= c^2 \n\nproved
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.
96766
1) tan A = x/y\n2) sin A = x/r\n3) cos A = y/r\n4) sin A/cos A = (x/r)/(y/r)\n5) sin A/cos A = xr/yr\n6) sin A/cos A = x/y\n7) sin A/cos A = tan A\n8) tan A = sin A/cos A\nReasons\n1-3)DEFINITION OF CIRCULAR FUNCTIONS\n4)MPE (STATEMENTS 2 AND 3)\n5)SIMPLIFYING COMPLEX FRACTIONS\n6)CANCELLATION\n7)SUBSTITUTION (STATEMENT 1)\n8)SYMMETRIC PROPERTY\n\nQED
676590
Sorry, I can't see the photo...but:\n\nSOH - sin = opp/hyp\nCAH - cos = adj/hyp\nTOA - tan = opp/adj
107615
t=theta\n^=to the power of...\n\ncos (2t) + 2 sin ^2 (t) = cos (2t) + 2 - 2 cos ^2 (t)
1081956
Multiply both sides by r thus:\nr^2=2*r[sin(theta)]\nso    X^2+y^2=2y\nNow rearrange.
1093073
A = 45 degrees or (pi / 4) radians\n\n1/sin^2A-1/cos^2A-1/tan^2A-1/cot^2A-1/sec^2A-1/csc^2A = -3\nIf i understand the way you wrote the problem above, it's 1/ each term, subtract each other right?\n1/ ((sinA)^2) - 1/ ((cosA)^2) - 1/ ((tanA)^2) - 1/ ((cotA)^2) - 1/ ((secA)^2) - 1/ ((cscA)^2) = -3\nI'm going to drop the A part, i assume you know that it will still be officially added in the final version\n\n[1/ (sin^2)] + [(-1)/ (cos^2)] + [-(cot^2)] + [-(tan^2)] + [-(cos^2)] +  [-(sin^2)] = -3\n[1/ (sin^2)] + [(-1)/ (cos^2)] + [-(cos^2) / (sin^2)] + [-(sin^2) / (cos^2)] + [-(cos^2) - (sin^2)] = -3\ncombine denominators\n[( 1 - (cos^2) ) / (sin^2)] + [( (-1) - (sin^2) ) / (cos^2)] + (-1) [(cos^2) + (sin^2)] = -3\n[(sin^2) / (sin^2)] + (-1)[( 1 + (sin^2) ) / (cos^2)] + (-1) [ 1 ] = -3\nTrig identity [(cos^2) + (sin^2)] = 1\n[ 1 ] + (-1)[( 1 + (sin^2) ) / (cos^2)] + [ -1 ] = -3\n(-1)[( 1 + (sin^2) ) / (cos^2)] = -3\n[( 1 + (sin^2) ) / (cos^2)] = +3\n1 + (sin^2) = 3 * (cos^2)\nTrig identity 1 = [(sin^2) + (cos^2)] =\n[(sin^2) + (cos^2)] + (sin^2) = 3 * (cos^2)\n2 * (sin^2) = 2 * (cos^2)\nsin^2 = cos^2\nsinA = cosA\nA= 45 degrees of (pi/4) radians