What's the method for solving for a variable inside of a sin()?\nFor example:\nsin(300/A)=B/C\nHow do you solve for A?  What steps are taken?  Thanks.
204361
sin(300/A) = B/C\n300/A = arcsin(B/C)\nA = 300/arcsin(B/C)\n\nwhere y=arcsin(x) is function inverse to x=sin(y).\n\nWhat's a problem?
506147
It looks like you are trying to solve a literal equation, which is an equation with more than one variable, and you have to solve for a specified variable.  You can use the SAME rules to solve this kind of equation as you do with regular equations (those with numbers).  For example, if you had 8=2(b)(9), you would divide by 2 and then again by 9 (which is like dividing by 18).  \n\nIf your problem is a=(1/2)bc^2, then you would divide by 1/2 and then again by c^2.  Remember that to divide by a fraction, you multiply by the reciprocal.
506182
-------------- edit ---------------\n\nThe function\n\n     sin(πx/180) - x/60\n\nis 0 when x = 0 and x = 30, and can be shown, using derivatives, to have a maximum between 0 and 30\n\n∴     sin(πx/180) > x/60 if 0 < x < 30\n\n∴     sin(xº) > x/60 if 0 < x < 30\n\nPutting x = 6 and 10 get\n\nsin 6º > 1/10\nsin 10º > 1/6\n\nand similarly, also get\n\nsin 1º > 1/60\nsin 2º > 1/30\nsin 3º > 1/20\nsin 4º > 1/15\nsin 5º > 1/12\nsin 12º > 1/5\nsin 15º > 1/4\nsin 20º > 1/3\n\n---------------- end of edit ------------------------------\n\nAlternatively, prove the special cases as follows.\n\nProof that sin 6º > 1/10\n-----------------------------\n\nSee Source 1 to calculate:\n\n     sin 36º = ¼√( 2 ( 5 - √5 ) ) \n     cos 36º = ¼√5\n\n     sin 30º = ½  and cos 30º = ½√3\n\nUse difference formula to get\n\n     sin 6º = sin( 36º - 30º ) = sin 36º cos 30º - sin 30º cos 36º\n\nSimplify to get (and check with Source 2):\n\n     sin 6º  =  [√6 √( 5 - √5) - (√5 + 1 ) ] / 8\n\nTo prove this > 1/10 rearrange, square, etc to get\n\n     30 - 6√5 > 206 / 25 + 18 / 5 √5\n\n     √5 < 34 / 15\n\n     1125 < 1156 which is true.\n\nA simpler method is similar to the following for 10º. Use the multiple angle formula for sin (5 × 6º) and show that when s = 1/10\n\n     32 s^5 - 40 s³ + 10 s - 1 < 0\n\nProof that sin 10º > 1/6\n-----------------------------\n\nUse multiple angle formula (Source 3) to get\n\n     sin 30º = 3 cos² 10º sin 10º - sin³ 10º\n\nLet s = sin 10º and simplify to\n\n     f(s) = 8s³ - 6s + 1 = 0\n\nUse calculus to show that this function decreases from 0 to ½ then increases. So sin 10º < 1/6 if f(1/6) > 0 which is true.
543602
sin^2(x)=1/2sin(2x)\nsin^2(x)*2sin(2x)=0\nsin^2x=0\nsinx=0\nx=npi\nsin(2x)=0\n2x=npi\nx=npi/2
342755
sin[x-(pi/2)]=kcosx\n-sin[(pi/2)-x]=kcosx\n-cosx=kcosx\nk=-1\ncos[x+{pi/2)]=ksinx\n=cos[(pi/2)+x]=ksinx\n=-sinx=ksinx\nk=-1
464676
14 sin² y = 13\nsin² y = (13/14)\n|sin y| = √(13/14)\nsin y = ±√(13/14)\ny = arcsin [±√(13/14)]\n\nThere are four solutions for this equation such that 0° ≤ y < 360°... one in each quadrant.  They are (to the nearest tenth of a degree) 68.2°, 111.8°, 248.2°, and 291.8°.
353522
One important rule is that cos^2+sin^2=1\n\nFirst of all, do this:\nsin^2c(1+(cos^2c/sin^2c))\n+cos^2c(1+(sin^2c/cos^2c))=\nsin^2c+cos^2c+cos^2c+sin^2c=\ncos^2c+sin^2c+cos^2c+sin^2c=\n1+1=2\n2=2\n\nI hope this helps!
483991
Did they tell you that\nIf x is very close to 0, then sin(x)=tan(x)=x;\n(sin(x/2))^2 / x^2 = (x/2)^2/x^2=1/4\n(tan(2x))^3 / x^3 = (2x)^3/x^3=8
451172
It's a product:\n\nif f(x) = a*b\nthen f'(x) = a*(b') + b*(a')\n\nAlso, you must multiply by the derivative of "what's inside."  For this problem, you must take the derivative of "what's inside" the cosine parentheses (2 + 3x - 5x^7).\n\nSo for this one, a = x^5 - 3x^3 and b = cos(2 + 3x - 5x^7).  "What's inside" is (2 + 3x - 5x^7).  Therefore:\n\n--- --- --- --- --- --- --- --- --- ---- ---- --- "what's inside"\nf'(x) = ( (x^5 - 3x^3)(-sin(2 + 3x - 5x^7)*(3 - 35x^6)) + (cos(2 + 3x - 5x^7)*(5x^4 - 9x^2))
347644
Wait a minute, did you want to know when the function is zero or when its derivative is zero?  \n\nFor derivative:\ny' = 2x + sin x, right?\n\nIt looks like there's one place where it's 0.  Requires and numerical solution.\n\nWell, it's 0 at x=0, right?\nI assume that x is in radians.  1 radian is about 57 deg.  As x decreases from 0 both parts become more negative.\n\nIt looks like x = 0 is the only zero of the dirivative.
253584
sin^2(a) = 3/4 gives sin(a) = -sqrt(3)/2 or +sqrt(3)/2\n\nIf a it's in the third quadrant, sin(a) must be negative, so\nsin(a) = -sqrt(3)/2\n\nThis yields a = 210 degrees (7*pi/6 radians)\n\ntan(7pi/6) = sqrt(3)/3
290097
I hope that this will help: \nsin(70) = 0.7739,sin(110) = -0.0442,sin(47) = 0.1236,sin(133) = 0.8689, sin(69) = -0.1148,\nsin(111) = -0.8645,cos(70) = 0.6333, cos(110) = -0.9990,cos(47) = -0.9923,cos(133) = 0.4948,\ncos(69) = 0.9933, cos(111) = -0.5025
418786
sin A + sin B = sin½(A + B)cos½(A - B)\n\nSo sin 50 + sin 10 = sin½(50 + 10)cos½(50 - 10)\n= sin30cos20\n= ½cos20 \n≠cos20
399370
-2 sin^2 x-3 sin x-1=0\n-(2 sin^2 x+3 sin x+1)=0\n2 sin^2 x+3 sin x+1=0\n(2 sin x+1)(sin x+1)=0\nsin x=-1/2 or -1\nx=7pi/6, 11pi/6, and 3pi/2\nI hope that helps!
30640
It says use the sum and difference formulas to find sin(165). So you have two questions to answer.\n1) What do the sum and difference formulas look like.\n2) How can I rewrite 165 degrees as something that I can easily use in the sum and difference formulas.\n\nFor part 1, if you look in your book, you should find that sin(a+b) = sin(a)cos(b) + cos(a)sin(b)\n\nFor part 1, you should know what the sin, cos, tan, etc. are for 0, 30, 45, 60, 90, 120, 135, 150, 180, etc.\n\nSo 165 = a + b = ? + ?, and you should be able to take it from here.
593903
EDIT: I was looking at the answer above mine, and the cos(0) and sin(0) values are reversed, so it is not x=0.\n\nOK!! No problem. First of all, let me just say I teach math, but there is no way I would want to memorize all of the trig identities, so open up your precalc book, and if it is a good book, the major identities will be located on the front or back cover. If not, then go to the applicable page.\n\nStart with what is given:\n(sq root:3)sinx + cosx = 1\n\nAll I did was go through the identity lists and see what seemed to match closest to what you have there, and I noticed that:\n\ntan(x/2) = (1 - cosx) / sinx\n\nThen I visualized doing a little bit of algebra with your problem, and saw that if you subtracted cosx from both sides you would have:\n\nsqrt(3) sinx = 1 - cosx\n\nWhich is kind of similar to tan(x/2), except we need a sinx in the denominator. And if you divide both sides by sinx, that is what you get! So now we have:\n\nsqrt(3) = (1 - cosx) / sinx\n\nBut we already know that this is the same as tan(x/2)\n\nSo, we have:\n\nsqrt(3) = tan(x/2)\n\nFrom here, since you want to find the actual value for x which will be an angle measure, you go to inverse tangent:\n\ntan(-1) sqrt(3) = (x/2)\n\nI also have a handy table of all the major trig values open (it is in the back cover of my precalc book) and notice that tan(60) = sqrt(3)\n\nWe are almost done:\n\n60 degrees = x / 2\n\nMultiply by 2,\n\nx = 120 degrees.\n\nOf course, you should check that answer and make sure it is correct!!
470996
Hello...\n\nWell the first part is a bit tricky to explain and I'm not sure what range of x you want your answer in, because with inequalities and trig functions the solution lies in many intervals. So I'll explain the second one first\n\nIt's actually not that bad, you just need to be familiar with some trig identities\n\nIf you recall, I'm sure you learned that 2sinxcosx = sin(2x)\n\ntherefore, you can say that sin(2x)=sqrt(5)/3, and solve for x by taking the inverse sin of both sides.\n\nNow for the first problem, you can work the inequality a bit further and get cosx greater than or equal to -1/2\n\nIf you're looking for a simple interval, then x is between -2pi/3 and 2pi/3. But the actual answer also includes (4pi/3,8pi/3), (10pi/3,14pi/3)....\n\nThe best way to do the inequality is to draw the cosine function on your paper, draw y=-1/2 (horizontal line), and you can find the first intersection by taking the inverse cosine of (-1/2), which is 2pi/3. You'd have to study the symmetry of the cosine function, so since you know that it's symmetric about x=pi, you'll figure that the next point is 4pi/3 and so on and so forth and you'll see a pattern really quickly. \n\nThis may be more complicated than what you're looking for but I hope it helps you out. Good luck on your test