Showing Trigonometric Identity. How do I demonstrate the next identity? (sin3θ/cosθ)+(cos3θ/sinθ)=2cot2θ?
830578
Adding these fraction expressions (common denominator  etc) gives\n(sin3θsinθ + cos3θcosθ)/(cosθsinθ)\n\nThe top is the expansion of\ncos(3θ - θ) which is cos2θ\n\nThe bottom is half of \nsin2θ = 2sinθcosθ,\nso multiply top and bottom by 2 and you have the result, since\ncos2θ/sin2θ = cot2θ
760807
Please be careful with your symbols. I believe you mean to prove\n\n    tan²x\n------------ = sin²x\n1 + tan²x\n\nwritten as tan²x / ( 1 + tan²x ) = sin²x\n\nReplace tan x with sin x / cos x to get\n\n( sin²x / cos²x ) / ( 1 +  ( sin²x / cos²x ) )\n\nMultiply numerator and denominator by cos²x to get\n\nsin²x / ( cos²x + sin²x ) = sin²x / 1 = sin²x
880711
sin(2A)=sin(A+A)=sinAcosA + sinAcosA=2sinAcosA
200427
tangent = opposite/adjacent (toa)\nsine = opposite/hypotenuse (soh)\ncosine = adjacent/hypotenuse (cah)\n\nTom's old aunt sat on her coffin and howled.\n\nI learned that 10 years ago and I still remember it for some reason. Then cotangent is just the flip of tangent, secant is the flip of cosine, and cosecant is the flip of sine.\n\nAll, sine, tangent, cosine (all students take calculus) tells you which of the trig functions are positive in the various quadrants I-IV, in that order.\n\nThese are the only things in trig that you need to sit down and memorize (more or less). Almost else can be justified using this information, specifically the trig identities. The best way to remember these guys is to derive them yourself. If you try to memorize them all, you're bound to forget them since they don't have convenient mnemonics, but if you derive them yourself and understand why they hold, you tend to remember them.\n\nSpecifically you can prove easily that tan = sin/cos, and that sin^2 + cos^2 = 1. Once you've proven these, you can prove a ton of other identities.
797886
Have you learned trig identities yet?  I'd check to see if there was a trig identity you could substitute here.\n\nBTW, I am assuming you meant cos²x rather than cosx².\n\nTry this:  cos²x + sin²x = 1, so substitute 1-sin²x for cos²x:\n\n6cos²x + sinx - 4 = 0\n\n6(1-sin²x) + sinx - 4 = 0\n\n-sin²x + sinx + 2 = 0\n\n(multiply by -1 to make it a little easier)\n\nsin²x - sinx - 2 = 0\n\n(sinx-2)(sinx+1) = 0\n\nSo, sinx = 2 or sinx = -1\n\nSolving for x, this means x = sin^-1(2) or sin^-1(-1)
484958
sin 2x=2 sinx . cos x\ncos 2x=(cos x)^2-(sin x)^2\ncos 2x=2 (cos x)^2 -1\ncos 2x=1- 2 (sin x)^2\ntan 2x=(sin 2x)/(cos 2x)\n\nor you can \nsin 2x=1\nlet 2x=y\nsin y=1\ny=90 degree + 360 degree\n2x=90 degree + 360 degree\nx=45 degree + 180 degree\nso x={45,225,....}degree\n\ntan x+cot x=2/sin 2x:\ntan x+cot x\n=(sin x/cos x )+(cos x/sin x)\n=((sin x)^2+(cos x)^2)/(sin x . cos x)\n=1/(sin x . cos x)\n=2/2(sin x . cos x)\n=2/sin 2x\nso it is an identity
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
592160
i think the answer wil be (tanX)/(tanY)\n\nSIN(X+Y)=sin x cos y+cos x sin y\n\nsin(x-y)=sin x cos y-cos x sin y\n\n use cos x cos y to divide the numerator and denominator... there is the answer...
906876
2cos^2x-1+1/2sinxcosx\n=2cos^2x/2sinxcosx\n=cosx/sinx\n=cotx
741518
1+(cotx)^2=(cscx)^2\n\nLeft hand side = 1 + cot²x\n= 1 + cos²x/sin²x\n= (sin²x + cos²x)/sin²x\n= 1/sin²x\n=csc²x\n= right hand side ......... QED
278456
So you know the ones for sin(2a) and cos(2a). Cool.\n\nNow, tan(2a) is the same as sin(2a)/cos(2a), since tan = sin/cos (definition of tangent).\n\nSub in the formulae for sin(2a) and cos(2a) :\n\n2sin(a)cos(a)\n----------------\ncos(a)^2-sin(a)^2\n\nDivide both numerator and denominator by (cos(a))^2\n\n2tan(a)\n----------\n1-tan(a)^2
842198
I think you mean\n\n(sin^4 t - cos^4 t )/ sin 2t where t is for O\n= (sin ^2 t+ cos^2t)(sin^2 - cos ^ 2 t) as (a+b)(a-b) = a^2 - b^2\n= 1. (- cos 2t)/sin 2t  using sin^2t + cos^2 t = 1 and cos^2 t- sin^2 t = cos 2t\n= - cot 2t
145134
I will use Q for theta here:\ncosQ = 2 - 2cosQ\nNow we add 2 cosQ to both sides to get rid of it on the right.\nSo we get:\n3cosQ = 2\nDividing both sides by 3 gives:\ncosQ = 2/3\nAnd solving for Q, we take the inverse cosine of both sides:\nQ = cos(-1)(2/3)\nHope this is helpful :-)
714157
One important rule is that cos^2+sin^2=1\n\nFirst of all, do this:\nsin^2c(1+(cos^2c/sin^2c))\n+cos^2c(1+(sin^2c/cos^2c))=\nsin^2c+cos^2c+cos^2c+sin^2c=\ncos^2c+sin^2c+cos^2c+sin^2c=\n1+1=2\n2=2\n\nI hope this helps!
880627
Do you mean verifying trig identities?\n\nTry: ask dr math\n\nIf you have a specific trig question, write back.\n\nGuido
913234
it's "pi" not "pie" dear ... :-)\n\nAnd, yes I do know how to prove those identities.\n\nAnd, no I won't do them all, as it will deprive you of the wonderful opportunity to learn and practice for your test. \n\nBut, I will do one for you!  I'll do #8:\n\n(sinx + cosx) ² =\nsin²x + 2sinxcosx + cos²x = \nsin²x + cos²x + 2sinxcosx = \n1 + 2sinxcosx  {{ using trig ID #1 }} = \n1 + sin2x {{ using trig ID #2 }}\n. . . . .\nTrig ID #1 :  sin²x + cos²x = 1\nTrig ID #2 :  sin2x = 2sinxcosx\n\nNRT
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