(1-sin^2x)cosx-2sin^2xcosx?
356153
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)
952344
(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
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
696270
For the first one, let's first rewrite "cscx secx" as "1 / (sinx cosx)"\n\nSo we have:\n\ntanx - (1-2cos²x)/(sinx cosx) =\n\ntanx + (2cos²x - 1)/(sinx cosx) =\n\n(sinx/cosx) + (2cos²x - 1)/(sinx cosx) =\n\n(sin²x/sinx cosx) + (2cos²x - 1)/(sinx cosx) =\n\n(sin²x + cos²x + cos²x - 1)/(sinx cosx) =\n\n(1 + cos²x - 1)/(sinx cosx) =\n\ncos²x / (sinx cosx) =\n\ncosx/sinx = cotx\n\n~ ~ ~ ~ \n\nFor the second one, you started off in the right direction!  Let's do it here:\n\n( sinx + tanx ) / (cotx + cscx) = \n\n( sinx + sinx/cosx) / (cosx/sinx + 1/sinx) = \n\n( (sinxcosx + sinx) / cosx ) * ( sinx / (1 + cosx)) =\n\n-- now I'm just going to rearrange the terms a bit --\n\n( (sinxcosx + sinx) / (1 + cosx ) * ( sinx / cosx ) =\n\n( sinx(1+cosx) / (1+cosx) ) * sinx/cosx = \n\nsinx * tanx\n\n~ ~ ~ ~\n\nTips for solving -- experience, trial and error (sorry that's not too helpful!).   What I did for the second one was I started out like you did, then I rewrote where I wanted to end up (sinx tanx) as sinx * sinx/cosx to see if I could manipulate where I had gotten so far with my problem to where I wanted to go, and I was able to see how to connect the two.  Hope that helps somewhat.  I had a hard time with trig identities when I first learned them too.  They really just take lots and lots of practice.\n\n~ ♥ ~
912647
Left side: \n\n{[cos^2(x)/sin^2(x) - 1/sin^2(x)]}/{cos^(x)/sin^2(x) - cos(x)/sin^2(x)}\n\nsin^2(x) is denominator of all 4 terms, so combine on each side\n\n= (cos^2(x)-1)/(cos^2(x) - cos(x) \n\n= (cos^2(x)-1)/cos(x)(cos(x)-1)  \nFactor top as diff. of two squares;\n\n= [(cos(x) + 1) (cos(x)-1)]/cos(x)(cos(x)-1)\n\nCancel factor of (cos(x)-1)\n\n= (cos(x)+1)/cos(x)\n\n= cos(x)/cos(x) + 1/cos(x)  = 1 + sec(x)
408737
Need to use the product rule here:\n\nd/dx(1/2*tanxsinx)\n\n= [d/dx(1/2tanx)]sinx + 1/2tanx*d/dx(sinx)\n= (1/2sec^2x)sinx + 1/2tanx*cosx\n= sinx/2cos^2x + sinx/2cos^2x\n= tanx/cosx
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
546258
cos2x=cos(x+x)=cos^2x-sin^2x=1-sin^2x-sin^2x=1-2sin^2x\ncos2x=1-2sin^2x  \ncos2x+2sin^2x-sinx=1-2sin^2x+2sin^2x-sinx=1-sinx=0\nsinx=1\nx=2kπ+π/2  ,   k=...,-1,0,1,...
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
1132938
I can help you with the part of  "Identify the expression that completes the equasion so that it is an identity.\nsin^2x/(1-cosx)"\n\nYou got sin^2x/(1-cosx). Then to complete the equation you multiply[(1+cosx )/(1+cosx)] to the equation.  \n\nThen you get {[(sin^2x)(1+cosx)]/[(1-cosx)(1+cosx)]}...Then by factorization you get that {[(sin^2x)(1+cosx)]/(1-cos^2)]}....The (1-cos^2) is the same as to sin^2x.  \n\nThen you cancel the terms sin^2x/(1-cos^2)...It leaves the equation as following ==> (1+cosx)(1) = (1+cosx)\n\nHope someone helps you with the rest SOrry
885716
cot(-x)cos(-x) + sin(-x) = -csc x\n\nLHS = cot(-x)cos(-x) + sin(-x)\n= cos(-x)/sin(-x) * cos(-x) + sin (-x)\n= cos²(-x)/sin(-x) + sin²(-x)/sin(-x)\n= (cos²(-x) + sin²(-x))/sin(-x)\n= 1/sin(-x)\n= 1/-sin(x)\n= -csc(x)\n= RHS
1182723
sinx+cosecx=2\nsin+(1/sinx)=2\nfrom here try to cross multiply!
868863
tan^2x+sec^2x-2tanxsecx\n=(sin^2x/cos^2x)+(1/cos^2x)-2(sinx/cosx)(1/cosx)\n=(sin^2x+1+2sinx/cos^2x\n=(1+sinx)^2/(1-sin^2x)\n=(1+sinx)/(1-sinx)
591905
1+2(tanx)^2+(tanx)^4\n= (1+(tanx)^2)^2\n= ((secx)^2)^2\n= (secx)^4\n= (cosx)^-4
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)
901088
4sinxcosx =1 (assumed for 0 ≤ x ≤ 2π)\nSo 2 sin2x = 1 (since sin2x = 2sinxcosx)\nSo sin2x = ½ (0 ≤ 2x ≤ 4π)\n2x = π/6, 5π/6, 13π/6, 17π/6\nThus x = π/12, 5π/12, 13π/12, 17π/12\n\nsin4x= -2sin2x (assumed for 0 ≤ x ≤ 2π)\nSo 2sin2xcos2x + 2sin2x = 0\nie 2sin2x(cos2x + 1) = 0\nso sin 2x = 0 or cos 2x = -1/2 (0 ≤ 2x ≤ 4π)\n2x = 0, π, 2π, 3π, 4π or 2x = 2π/3, 4π/3, 8π/3, 10π/3\nwhence x = 0, π/3, π/2, 2π/3, π, 4π/3, 3π/2, 5π/3, 2π
909136
Let I = ∫e^s sin(t - s) ds\n= e^s sin(t - s) - ∫e^s *( -1) * cos(t - s)ds\n= e^s sin(t - s) + ∫e^s cos(t - s)ds\n= e^s sin(t - s) + e^s cos(t - s) - ∫e^s * (-1) * -sin(t - s)ds\n= e^s sin(t - s) + e^s cos(t - s) - ∫e^s sin(t - s) ds\n= e^s sin(t - s) + e^s cos(t - s) - I\nSo 2I = e^s (sin(t - s) + cos(t - s)\nThus I = ½(e^s (sin(t - s) + cos(t - s)) + K\nSo ∫[s = 0 to s = t] e^s sin(t - s) ds\n= ½[e^s (sin(t - s) + cos(t - s)] from s= 0 to s = t\n= ½[(e^t (sin(t - t) + cos(t - t)) - (e^0 (sin(t - 0) + cos(t - 0))]\n= ½[(e^t(1) - 1(sint + cost)]\n= ½(e^t - sint - cost)