sin^3xcos^2x=cos^2xsinx-cos^4xsinx (help with trig identities)?
814691
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.
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~ ♥ ~
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,...
873429
cos5x = sinx\ncos5x = cos(pi/2 - x)\n5x = 2npi +- (pi/2 - x)\n6x = (4n + 1)pi/2   ,   4x= (4n-1)pi/2\nx= (4n+1)pi/12    ,  x = (4n-1)pi/8
226513
csc(x) is the same as 1/sin(x) and sec(x) is the same as 1/cos(x) so the left side of the equation can be simplified to \n\nsin(x)sin(x) + cos(x)cos(x) \nor sin^2(x) + cos^2(x)    {"sin squared x plus cos squared x"}\n\nUsing the same steps, the right side of the equation becomes sin(x)/sin(x) or 1.\n\nSo the identity has been reduced to  sin^2(x) + cos^2(x) = 1 which is another common trigonometry identity!
846581
Expand the exponent: (cos (x/2) - sin(x/2))^2 = (cos x/2)^2 - 2*cos(x/2)*sin(x/2) + (sin x/2)^2 = 1 - 2*sin(x/2)*cos(x/2) = 1-sin(x)\n\nusing the fact (cos a)^2 + (sin a)^2 = 1 and 2*sin(a)*cos(a) = sin(2a)\n\nSteve
355137
first, this equation is wrong, it should look likle this:\nsin2x=2(sinx (multiplied by) cosx). If yu want, i'll try to solve the true equation for yu.
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)
433978
Simply equate the functions as it asks you to;\nso you get 1-sinx = 2sin2x\nThen use the trig identity sin2x=2sinxcosx\nso you get 1-sinx=4sinxcosx\n\nCan you solve it from there?\nI'd next use cosx = sqrt (1-(sinx)^2)\nThen square the equation and solve it for x as you've probably done a million times before.\n\nYou'll probably get a quadratic equation in sinx. Use the quadratic formula to get some values out, and remember that you have to give answers between -360 and 360 degrees.
160524
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)
953609
2cos(x) - sin(2x) = 0 \nremembering that sin(2x) = 2sin(x)cos(x), \n2cos(x) -2sin(x)cos(x) = 0\n2cos(x) * [1 - sin(x)] = 0\nor cos(x) = 0 or 1-sin(x) = 0\ncos(x) = 0 or sin(x) = 1\n\nπ/2, 3π/2
733318
I think your identity should be :\ncsc^2(x) / [1 + tan^2(x)] = cot^2(x)\nas you forgot to square the tan(x).\n\nWorking on the 1 + tan^2(x) gives:\n1 + sin^2(x) / cos^2(x)\n= [sin^2(x) + cos^2(x)] / cos^2(x)\n= 1 / cos^2(x)\n\nNow we have : csc^2(x) / [1 / cos^2(x)]\n= [1 / sin^2(x)] / [1 / cos^2(x)]\n= cos^2(x) / sin^2(x)\n= cot^2(x), which was to be shown.
847590
sin (2x ) = 2.sinx.cosx so;\n6- 6(2.sinx.cosx) = 6-12sinxcosx;\n6(1 - 2sin2x) is the simplified form.
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
859883
cosx(1-2sin^2x)=0\ncosx(cos2x)=0\ncosx=0\nx=pi/2,3pi/2\ncos2x=0\n2x=pi/2 or 3pi/2\nso x=pi/4 or 3pi/4
68963
If it's an identity, then you aren't allowed to move terms from one side to the other. Instead you must manipulate one side (leaving the other untouched) until it is identical to the other.\n\nSince the right side only depends on sinx (it has no cosx's), a good strategy would be to replace the cosx's on the left side.\n\nWe know (sinx)^2 + (cosx)^2 = 1, so\n(cosx)^2 = 1 - (sinx)^2\n\nSubstitute this for (cosx)^2 on the left side, then simplify by combining like terms. Voila - you'll have the right side.
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)