Help proving trig identity:  cscx(cscx + sin-x)=cot^2x?
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Change cscx to 1/sinx and sin(-x) = -sinx\ncscx(cscx + sin-x)\n= (1/sinx)(1/sinx - sinx)\nGet a common denominator\n= (1/sinx)([1-(sinx)^2]/sinx)\n1-(sinx)^2 = (cosx)^2\nSo\n(1/sinx)([1-(sinx)^2]/sinx)\n= (1/sinx)((cosx)^2/sinx)\n= (cosx)^2/(sinx)^2\n= (cosx/sinx)^2\n= (cotx)^2
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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.
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(cot(20))^2 + (cot(40))^2 + (cot(80))^2 = 9\n\nor you can think of it like this\n\n(1/tan(20))^2 + (1/tan(40))^2 + (1/tan(80))^2\n\nMultiply everything by (tan(20))^2 * (tan(40))^2 + (tan(80))^2\n\n(((tan(40)^2)(tan(80)^2)) + ((tan(20)^2)(tan(80)^2)) + ((tan(20)^2)(tan(40)^2))/((tan(20)^2)(tan(40)^2)(tan(80)^2)\n\nJust in case you don't know how to do cot\n\nor replace cot(x)^2 with csc(x)^2 - 1\n\ncsc(x)^2 = (1/sinx)^2\n\n(((1/sin(20))^2) - 1) + (((1/sin(40))^2) - 1) + (((1/sin(80))^2) - 1)\n\n((1/sin(20))^2) - 1 + ((1/sin(40))^2) - 1 + ((1/sin(80))^2) - 1\n\n((1/sin(20))^2) + ((1/sin(40))^2) + ((1/sin(80))^2) - 3\n\n((1/sin(20))^2) + ((1/sin(40))^2) + ((1/sin(80))^2) = 12\n\n12 - 3 = 9
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sinx+cosecx=2\nsin+(1/sinx)=2\nfrom here try to cross multiply!
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cos(x) +sin(x)tan(x)=sec(x)\nmultiply by cos(x) on both sides\ncos^2(x)+sin^2(x)=1\n1=1\nI believe they are probably asking you to prove the identity
463304
(sinx)^(-1) - sqrt(1-x^2) + C\nStart by multiplying (1+x)/(1-x) by (1+x) on the top and bottom\n[(1+x)(1+x)]/[(1-x)(1+x)]\nMultiply the denominator out but not the numerator.\n[(1+x)(1+x)]/[(1-x)(1+x)] = [(1+x)(1+x)]/(1-x^2)\nNow\nsqrt[(1+x)/(1-x)] \n= sqrt[[(1+x)(1+x)]/(1-x^2)]\n= (1+x)/sqrt(1-x^2)\nNow use trig substitution again letting x = sin(u)\nthen dx = cos(u) du\n(1+x)/sqrt(1-x^2) dx \n= [(1+sin(u))/sqrt(1-(sinu)^2)]cos(u) du\n= [(1+sin(u))/sqrt((cosu)^2)]cos(u) du\n= [(1+sin(u))/cos(u)]cos(u) du \nthe cosines cancel\n= [1 + sin(u)]du\n= u - cos(u) + C\nNow x= sin(u) so u = (sin x)^(-1)\nand using a triangle you can see that cos(u) = sqrt(1-x^2)
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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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Did you get the question correct?\n\nStart from right-hand side:\n   cos^2x sinx - cos^4x\n= cos^2x [sinx - cos^2x]\n= cos^2x [sinx - (1 - sin^2x)]\n= cos^2x [sin^2x + sinx -1]\n\nYou can't transform (sin^2x + sinx - 1) into sin^3x.
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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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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
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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
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(1+tanx)/(1+cotx)\n=(1+sinx/cosx)/(1+cosx/sinx)\n=[(cosx+sinx)/cosx]/(sinx+cosx)/sinx]\n=(cosx+sinx)(sinx)/(cosx+sinx)cosx)\n=sinx/cosx\n=tanx\nhence proved
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Using Trig identity 2sin(x)cos(x) = sin2x\n\n(2sin3x)(cos3x) = sin6x
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Unfortunately the above answer isn't right. The correct answer is:\n\n(-5/128)cos(4x) + (1/64)cos(8x) + (-1/384)cos(12x) +C\n\nThe trick with these problems is you need to use the power reducing trig identity to get the power of either \nsin(2x) or cos(2x) to be even. For example\n\nInt sin^4(2x)cos(2x) dx = (1/10) sin^5(2x) + C and\nInt cos^4(2x)sin(2x) dx = (-1/10) cos^5(2x) +C\n\nIf you have an odd power it screws up:\n\nInt sin^3(2x)cos(2x) = (1/64)(cos(8x) - 4cos(4x)) + C\n\nThis is because of the relationship sin^2(x)+cos^2(x)=1\n\nUse the trig identities:\n\nsin^2(2x) = (1-cos(4x))/2 or cos^2(2x) = (1+cos(4x))/2  \n\nThis will reduce the power by 1 and convert sine to cos or cos to sine. Since one can always be the derivative or the negative derivative of the other, for integration purposes you can do either. In your case use the sine squared reduction formula. Multiply the terms out and integrate termwise. \n\nHere's a link to useful trig identities:\n\nhttp://www.sosmath.com/trig/Trig5/trig5/trig5.html
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let that be x insted of *, it's easier to write\nsinx/(1+cosx) +1 + cosx/sinx = \n[(sinx)^2 + sinx + sinxcosx + cosx + (cosx)^2]/[(1+cosx)*sinx]=\n(sinx+cosx+sinxcosx+1)/(1+ cosx)sinx=\n[sinx+ 1 + cosx(1+sinx)]/[(1+cosx)sinx)]=\n[(1+sinx)(1+cosx)]/[(1+cosx)\nsinx]=\n(1+sinx)/sinx=\n1/sinx +1 = cscx+1\nI don't know maybe I have done something wrong but it doesn't look like 2cscx.
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Consider the left-hand side(LHS) [cot(x) + 1] / [cot(x) – 1]. We want to prove that this is equal to the right-hand side (RHS), that is, [1 + tan(x)] / [1 – tan(x)]. Thus, we have the following:\n\nLHS = [cot(x) + 1] / [cot(x) + 1]\n        =  { [cot(x) + 1] / [cot(x) – 1] }* { tan(x) / tan(x) }.\n\nNOTE: We multiply tan(x) to cot(x) + 1 and the other tan(x) is multiplied to cot(x) – 1.\n\n        =  { tan(x) [cot(x) + 1] } / { tan(x) [cot(x) – 1] }\n\n        =  { tan(x) cot(x) + tan(x)] } / { tan(x) cot(x) – tan(x)] }, by distributive property of multiplication over addition (subtraction)\n\n        =  { 1 + tan(x)] } / { 1 – tan(x)] }, since tan(x) cot(x) = 1 by reciprocal identity.\n\n        =  RHS.\n\nIndeed, [cot(x) + 1] / [cot(x) – 1] = [1 + tan(x)] / [1 – tan(x)]
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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