what is COS^4x - SIN ^4 x =sin x?
412953
An equation that is incorrect.
749606
L.H.S. (tanx - cotx)/(tanx + cotx) + 2cos^2x = (sinx/cosx - cosx/sinx)/(sinx/cosx + cosx/sinx) + 2cos^2x = (sin^2x - cos^2x)/(sin^2x + cos^2x) + 2cos^2x = (sin^2x - cos^2x)/1 + 2cos^2x = sin^2x - cos^2x+ 2cos^2x = sin^2x + cos^2x = 1 = R.H.S.
542056
Start by factoring:\n\n(cos^2(x))(sin^2(x))-(cos^2(x)) = (cos^2(x))(sin^2(x) - 1)\n= -(cos^2(x))(1 - sin^2(x))\n= -(cos^2(x))(cos^2(x))\n= -cos^4(x)
967873
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
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
817619
you should use formula for\n\nsin 4x = sin 2(2x)\n\nassumin 2x = y only for understanding purpose\n\nsin 4x =sin 2y\n           = 2 sin y cos y\n           = 2 sin 2x cos 2x\nnow apply the formula for 2x
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
836167
is that sin^3, or sin(3x)? The equality works for sin(3x) and cos(3x), so I'll assume that:\n\nusing angle addition formulas:\n\nsin(2x+x) = sin(2x)cos(x)+sin(x)cos(2x)\n\nNoting sin(2x) = 2sin(x)cos(x)\n\nwe have sin(3x)/sin(x) = 2cos^2(x)+cos(2x)\n\ncos(2x+x) = cos(2x)cos(x)-sin(x)sin(2x)\nagain, sin(2x) = 2sin(x)cos(x),\nso cos(3x)/cos(x) = cos(2x)-2sin^2(x)\n\nThen sin(3x)/sin(x)-cos(3x)/cos(x) = cos(2x)+2cos^2(x)-cos(2x)+2sin^2(x) = 2(cos^2(x)+sin^2(x)) = 2\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.
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
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
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
901670
(sin x +cos x)^2 =1    [0=<x=<360]\nsin^2x + 2sinx * cosx +cos^2x =1 or 2sinxcosx=0      Because sin^2x+cos2^x =1 \n2sinxcosx is zero when x =0, 90, 180, or 270 degrees\n\n1-sinx = cosx     [0=<x=<360]\nsinx + cosx = 1\nThis occurs when x= 0, 360
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.
932663
Let a = A cos C\nb= A sin C\n\nsquare and add a^2+b^2 = A^2\n\nthen given expression\n= A cos C sin Bx + A sin C cos Bx = A sin(Bx+C)
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