How do you prove trigonometric identities?
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You will have to memorize a few basic identities and then practice proving using those basics and through algebraic methods.
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Since tanx = sinx/cosx,\n\ntan(A-B) = sin(A-B)/cos(A-B)\n\nNow, write out the expression for each term on the right side of this equation.\n\nWhen you finish that, divide each term by cosAcosB and you will get the result you are seeking.
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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!
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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
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Job one, convert everything to trig functions of a single angle, namely x.\n\nsin2x=2sinxcosx\ncos2x=cos^2(x)-sin^2(x)=2cos^2(x)-1=1-2sin^2(x)\n\nThen substitute in and simplify:\n2sinxcosx+2(2cos^2(x)-1)=1+sinx-4cosx\n2sinxcosx+4cos^2(x)-2-1-sinx+4cosx=0\n4cos^2(x)+2sinxcosx-sinx+4cosx-3=0\n\nFactor this into:\n(2cosx-1)(2cosx+sinx+3)=0\n\nset each parantheses equal to zero\n2cosx-1=0   \ncosx=1/2\nx=arccos(1/2)\nx={pi/3, 5pi/3} + 2pi*n  where n=0,1,2....\n\nAnd\n\n2cosx+sinx+3=0\n\nSolved numerically gives imaginary solutions of\n{-i*ln((-2/5)-(i/5),-i*ln(-2-i)}\n\nSolved....
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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
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cos^2 (2x) is equal to 1-sin^2(2x), remember, sin^2(x)+cos^2(x)=1
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Here are a couple i learned for double angle formulas.\n\nFor sin(a+b)= sin(a)cos(b)+cos(a)sin(b)\nSay out loud: "sine cosine cosine sine"\n\nFor cos(a+b)= cos(a)cos(b)-sin(a)sin(b)\nSay out loud: "cosine cosine SIGN sine sine"\nNotice the word "SIGN". The word SIGN is there to remind you that the SIGN of the rightmost term has the opposite SIGN of the input to the cosine on the left hand side of the equation (if its a+b, then subtract. if its a-b, then add).\n\nJust say it out loud like 3 times. It has a really nice rythm:\n\n"Sine Cosine Cosine Sine"\n"Cosine Cosine SIGN Sine Sine"\n"Sine Cosine Cosine Sine"\n"Cosine Cosine SIGN Sine Sine"\n"Sine Cosine Cosine Sine"\n"Cosine Cosine SIGN Sine Sine"\n\nSee? Easy.
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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)
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Open a trigonometry book!
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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)
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Opening brackets \nLHS=SinA+ SinA. TanA+ CosA+ CosA.CotA\n=SinA+Sin^2A/CosA+CosA+Cos^2A/SinA\nGrouping\n=(SinA+Cos^2A/SinA)+(CosA+Sin^2A?CosA)\n=(Sin^2A+Cos^2)/sinA+(Cos^2A+Sin^2A)/CosA\n=1/sinA+1/CosA\n=SecA+CosecA\n\nEasy Pesy
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Here are some real tough ones.(Looked tough to me)\n\nProve that cot(90 - theta)sin(90 - theta) = sin theta\n\nFind Cos(3x)\n\nFind Sin(3x)
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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
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All you need to do is enter these on a scientific or graphing calculator.  Make sure it is in degree mode.  For the second part use the 2nd or shift button to get sin^-1 etc
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I assume you want a proof of this fact?\n\nWe know that sin(x)^2+cos(x)^2 = 1 for any x,\nand also any function of the form x^2-y^2 can be factored as (x+y)(x-y) these two facts allow us to prove the above.\n\n\nHere if you let x=sin(A)^2 and y=cos(A)^2, then\n\nsin(A)^4-cos(A)^4 = (cos(A)^2+sin(A)^2)(sin(A)^2-cos(A)^2)\n                             = (sin(A)^2-cos(A)^2)\n                             = (1-cos(A)^2-cos(A)^2)\n                             = 1-2cos(A)^2\nlet me know if you need more assistance
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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