what is this trig graphing question ... y= -sin 4x?
100190
plot these coordinates:\n(0,0),(22.5,-1),(45,0),(67.5,1),(360,0), etc., etc.\n\nconnect them together in a curve and you'll have it.\n\nsine graphs are a series of curved lines. they start at zero, go up to one, go down to zero, go down to -1, and go up to 0 again.\n\nthe negative in front of the sine means that the graph is reflected along the x-axis. so that the 0, 1, 0, -1, 0 sequence is switched to 0, -1, 0, 1, 0.\n\nthe 4 in front of the x means that the normal period for the graph to make a complete curve is divided by for. so instead of the graph following the sequence every 90 degrees, it is done every 22.5 degrees.
1090403
x = r .cos ө\ny = r. sin ө\nsince x=y\nr.cos ө = r.sin ө\nsin ө/cos ө =1\nө = tan-1 ( 1)\nө = ∏/4\nr^2 = x^2 + y^2\nr^2 = 2.x^2\nr = x√2\n\npolar form would be ( r.CiS ө)  { i.e. r.cos ө + i.sin ө }\n\nso the polar form is x√2.CiS (∏/4)
1127640
I'm not sure if you have learned the unit circle yet, but the sin and cos functions follow it perfectly...For any position coordinate on the unit circle, sin=y and cos=x... Simply look at your unit circle and draw draw out the sin and cos functions according to the values given for each one by the coordinates found on the unit circle. You should have clear intersections.
932480
sin(A+B)sin(A-B)/(cos²Acos²B)\n= (sinAcosB + cosAsinB)(sinAcosB - cosAsinB)/(cos²Acos²B)\n= (sin²Acos²B - cos²Asin²B)/(cos²Acos²B) (since (x + y)(x - y) = x² - y²)\n= sin²A/cos²A - sin²B/cos²B\n= tan²A - tan²B ← Option c)\n\nPlease note: sin(A + B) = sinAcosB + cosAsinB\nand sin(A - B) = sinAcosB - cosAsinB
294550
Graph sin(2x) and sin(x). Once graphed you will see that y=sin(2x)the top function. You will integrate (sin2x-sinx)dx from 0 to pi. you can split the intergrals\nint(sin2x)dx = -1/2cos2x \nminus\nint(sinx)dx = -cosx\n\n-1/2cos2x+cosx evaluated at pi and 0.\n\n(-1/2 cos(2(pi))+ cos(Pi))-(-1/2 cos(0)+cos(0))
1155939
It's something like:\n\n2x + 4 = 2(x +2)\n\nWhen you distribute the right hand side and then put the x's together and the constants together, you end up with 0=0.\n\nThat means any value of x would make this a true equation. There is no specific solution to this equation.\n\nLater, when you get into trigonometry, identities become very important and aren't as simply and obvious as this example. For example you learn that tan(x)=sin(x)/cos(x) and that is always true. (Of course, you don't know what that means yet.) These identities mean we can always substitute what's on one side of the equation for what's on the other side of the equation without changing anything.
814753
The x-intercepts are where y is zero, hence, crossing the x-axis.\ny = sin(pi/2x) + 1\nSet y = 0\nthen, sin(pi/2x) = -1\nSo what value of x will make sin()=-1\nWell, when the argument pi/2x is some multiple of pi/2, like 3pi/2, 7pi/2, into + then - infinity.  Let's see, we want 3, 7, 11,... on the positive side and -3, -7, -11,... on the negative side.  Thinking about it yields an odd number plus 4; an odd number is 2n+1, then adding 4, 2n + 5.  Let's try it:  for n=-1, 2n+5=3; for n=0, 2n+5=5; for n=1, 2n+5=7; and the same for the negative side.\nBack to the original argument of sine:\npi/2x = 2n+5 for n = -infinity to + infinity\n(2n+5)2x = pi\nx=pi/[2(2n+5)] for 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]
1135417
u can factor it out to be (2sinx+1)(cosx)=0.. then find x for which either sinx=-1/2 or cosx=0\n\n(sinx+1)cosx=0 same method applies
735129
they would look like cross (X).. the sin(-x) graph loos like  / and     -sinx would look like &#92;. \n\nIf you dont get it please visit this figure uploaded on my webpage\n\nhttp://www2.lut.fi/~vasava/untitled.jpg.
766977
4cos2x +3cosx +3 = 0  [x=theta]\nLet y=2x\nThen 4cos y +3 cos(y/2) +3 = 0\nNow use the cos 1/2 angle formula to get:\n4cos y +3 [sqrt (1+cosy)/2] +3 =0\n3 [sqrt (1+cos y)/2] = -3 -4cos y\n9(1+cos y)/2= 9 +24 cos y +16cos^2y\n9+9cosy =18 +48cos y +32cos^2y\n32cos^2y +39cos y +9 =0\ncos y = [-39 + or - sqrt(39^2- 4*32*9)]/2\ncos y = -39/2 + 3/2sqrt (41), and \ncos y = -39/2 -3/2sqrt(41)\ny=arccos[-39/2 + 3/2sqrt (41)] , and \ny=arccos[-39/2 - 3/2sqrt (41)]\nx= y/2 =1/2*arccos[-39/2 + 3/2sqrt (41)], and\nx=1/2*arccos[-39/2 - 3/2sqrt (41)] \nUse your calculator to find final answer
593616
It's not hard if you draw the graph of the functions y=x and y=5*sin(Pi*x), and see how many times they cross.\n\nNow, you know that sin(pi*x) is always less than 1, so there are only solutions when -5<x<5.  And you know that at x=5, sin(pi*x) =0. (Same at x=-5)\n\nIf you plot the graph of sin(pi*x), you see that it's positive if 0<x<1, 2<x<3, and 4<x<5.  Since the sine curve goes up and down, you can see that there are two solutions in each interval.  Counting these up gives 6 solutions in each interval, or 6 solutions.  Now, look at x<0-- there are 6 solutions there also, but the x=0 is included in both sets, so you have to ignore one of them-- there are 11 solutions to the equation x=5sin(pi*x)
147089
y intercept: (0,4)\nx intercept: (3,0)
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.
885632
Yes you are correct.\n\nx=tan t = (sin t)/(cos t)\nso cos t = (sin t)/x\nSo y^2 = (sin^2 t)/x^2 = (1 - cos^2 t)/x^2 = (1 - y)/x^2\n\nRearranging this you get y = 1/(x^2 +1)
1050391
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!!
838763
if y=0\n\n2x+4(0)=-8\n2x+0=-8\nx=-4\n\n\nif x=0\n\n2(0)+4y=-8\n0+4y=-8\ny=-2\n\nIf your doing co-ordinate geometery, two points on the line are (0,-2) and (-4,0)