I know how to solve for sin2x using the double angle formulas, but how would i solve for sin4x?
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
847590
sin (2x ) = 2.sinx.cosx so;\n6- 6(2.sinx.cosx) = 6-12sinxcosx;\n6(1 - 2sin2x) is the simplified form.
129827
3.44(sinx)^2 -23.06=25sin2x\nsquare the 2 members u'll have\n11.8336(sinx)^4 -158.6528(sinx)^2+531.7636=\n25*4[(sinx)^2 * (1-(sinx)^2)]\ncoz sin2x=2sinx*cosx &(sin2x)^2= 4(sinx)^2(cosx)^2\nredice this equation :\n111.8336(sinx)^4 -258.6528(sinx)^2 +531.7636=0\nput t=(sinx)^2\n111.8336t^2 -258.6528t+531.7636=0\ndelta=7432.23321088\nt=0.77097759 we cannot take the other solution coz t<1\n=> sin x =+0.87805329 or sinx=-0.87805329\nx=61.408418 or x=-61.408418 or x= 118.59158196 or\nx= -118.59158196\nthere r other ways to solve it but hope this helps
276083
Yes, it can be rewritten as 2sin2xcos2x.  Or, you can use the addition formula for sineto prove this.  sin(x + y) = sinxcosy + cosxsiny.  In your case, sin(2x + 2x) = sin2xcos2x + cos2xsin2x = 2sin2xcos2x.
1054686
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....
411302
cos^2 (2x) is equal to 1-sin^2(2x), remember, sin^2(x)+cos^2(x)=1
825804
First, let's simplify sin(2x) / sin(x)\nRemember that sin(2x) = 2sin(x)cos(x), so\nsin(2x) / sin(x) \n= 2sin(x)cos(x) / sin(x)\n= 2cos(x)\nThe integral of 2cos(x) dx\n= 2sin(x) + C
225231
sinx = a     with    a Є [-1;1]\n\n2sin^2 x - 5sinx + 2 = 0 \n<=> 2a^2 - 5a + 2 = 0\nWe have 2 roots: a = 2 (reject) & a= 1/2 because a Є [-1;1]\n => sinx = 1/2\n<=> x = ∏/3 + k2∏ and x = 2∏/3 + t2∏ with k,t Є Z
785756
well sin x = x - x^3/3! + x*5/5! - x^7/7! + .... + (-1)^n x^(2n-1)/(2n-1)! + ....\n= ∑(-1)^n x^(2n-1)/(2n-1)!\n\nSo x sinx = ∑(-1)^n x^(2n)/(2n-1)! (= x^2 - x^4/3! + x*6/5! - x^8/7! + ... )
824139
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
381123
Yes. I assume to want to find sin 1/2x, cos 1/2x, and so on.\nIt's quite simple:\n1. Write the expanded form of cos 2x\ncos^2x - sin^2x\n2. Write that twice, in once in a column (remember it equals cos 2x)\ncos^2x - sin^2x  = cos 2x =  cos^2x - sin^2x\n3. Rewrite cos^2x in the first column as 1-sin^2x, and in the second column rewrite sin^2x as 1-cos^2x\n(1-sin^2x) - sin^2x = cos 2x = cos^2x - (1 - cos^2x)\n4. Combine like terms\n1-2sin^2x   and -1+2cos^2x\n5. In order to find cosx and sinx, we must isolate each:\n-2sin^2x = cos 2x -1  and 2cos^2x = cos 2x + 1\n6. Divide by the coefficent of each\nsin^2x = 1/2 - 1/2(cos 2x) and cos^2x = 1/2 + 1/2(cos 2x)\n7. Find the square root of both sides\nsin x = sqrt(1/2 - 1/2(cos 2x)) and \ncos x = sqrt(1/2 + 1/2(cos 2x))\n8. We found what sin x and cos x equal, we want what 1/2x equals. Let x = 1/2x\nsin 1/2x = sqrt(1/2 - 1/2(cos 2(1/2x)) and\ncos 1/2x = sqrt(1/2 + 1/2(cos 2(1/2x))\n9. Simplify\nsin 1/2x = sqrt(1/2 - 1/2(cos x)) and\ncos 1/2x = sqrt(1/2 + 1/2(cos x))\n10. You're done. You can find tan 1/2x, csc 1/2x, sec 1/2x, and cot 1/2x by using the trigometric identities with sin 1/2x and cos 1/2x:\n\ntan 1/2x = (sin 1/2x)/(cos 1/2x)\ncsc 1/2x = 1/(sin 1/2x)\nsec 1/2x = 1/(cos 1/2x)\ncot 1/2x = (cos 1/2x)/(sin 1/2x)\n\nObviously there are other expansions for the half angle formulae. These are just the easiest.
689360
sinx/(1+cosx)\n= sinx(1-cosx)/[(1+cosx)(1-cosx_]\n= sinx(1-cosx)/(1-cos^2x)\n= sinx(1-cosx)/sin^2x\n=(1-cosx)/sinx\n= 1/sinx - cosx/sinx\n\n9+8cosx/sinx + 8sinx/1+cosx\n= 9+8[cosx/sinx + sinx/1+cosx]\n= 9 + 8[cosx/sinx + 1/sinx - cosx/sinx]\n= 9 + 8/sinx
756591
Given\n\nsin 3x = 1/2\n\n\n\nGet arcsin of both sides\n\narcsin (sin 3x) = arcsin (1/2)\n\n\n\nThe left side simplifies as 3x.  The right side varies.  \nLet k be any integer.  Thus,\n\n3x = π/4 + 2kπ and\n3x = 3π/4 + 2kπ\n\n\n\nThus,\n\n3x = (π + 8kπ)/4 and\n3x = (3π + 8kπ)/4\n\n\n\nTherefore,\n\nx = π/12 (8k + 1) and\nx = π/12 (8k + 3)\nwhere k is ANY integer. Enumerating, we have,\n\nx = ... -5π/4, -7π/12, π/12, 3π/4, 17π/12, 25π/12, 11π/4, ...\nx = ... -13π/12, -5π/12, π/4, 11π/12, 19π/12, 9π/4, 35π/12, ...\n\n^_^
765649
switch the variables.\n\nStart with\n\ny = sin(0.5x-3)\n\nand solve for x.\n\nto get rid of  the sin, we apply arcsin or sin^-1 (on calculators)\n\narcsin(y) = arcsin(sin(.5x - 3))\n\nthis gives\n\narcsin(y) = .5x - 3\n\nx = 2*[arcsin(y) + 3]
838146
x  =  3pi/2  is one possible solution to the equation.  3pi/2 (or 270° if you're using degrees) is on the boundary between the third and fourth quadrants, so its sine has the same value as sin pi/2 (90°) but the opposite sign; i.e. sin 3pi/2  =  -1 .\n\nNow, (-1)³  -  4(-1)²  -2(-1)   =   -1  -  4(+1)  +2   =   -3\nSo  3pi/2 is a solution.\n\nHowever, it's not the only solution.  You could do a complete revolution into the 7th/8th quadrant (x  =  7pi/2) and you would get the same value for sin x.\nIn general, x  =  (4n + 3)pi/2  will satisfy the equation, where  n  is zero or any whole number, positive or negative.
881174
use formulas for cos(2x) and sin(2x) depending on cos(x) and sin (x) and then use the fact that\nsin(2x) = sqrt(1-cos(2x)^2).\nyou will get a quadratic equation that you solve with quadratic formula
640359
let sinx=u\nthen cosxdx=du\n\nand ∫[(cosx)/(9+sin^2x)]dx=\n\n∫du/(9+u^2)=        \n\n\n ( now use this formula\n∫du/(a^2+u^2)= 1/a arctan(u/a)+C)\n\n\n=1/3 arctan(u/3) +C=\n\n\n=(1/3)arctan(sinx/3)+C