When trying to figure out a phase shift trig problem how do you figure out the right values for x.?
407215
For an equation of the form y = A sin (bx + c), the phase shift is -c/b where negative is left shift and positive is right shift.\n\nExample: y = 3 sin (2x - pi/6)\n\nhas a phase shift of -(-pi/6)/2 or pi/12 to the right.\n\nThis is because originally 3 sin 2x = 0 when x = 0.  But\n3 sin (2x - pi/6) = 0 when x = pi/12.  So the x intercept has been moved to the right by pi/12.
736379
for this just substitute for x and y in eqn and see if LHS=RHS\nLHS=4\nRHS=(-1/2)(-3)-5/2\n=3/2-5/2\n=-2/2\n=-1\nnot = RHS\nso pt. not on line
956961
For an equation of the form y = a cos (bx + c) + d:\n\na is the amplitude\n2 pi / b is the period\nc is the phase shift\nd is the vertical shift.\n\nTherefore, the amplitude is 2, the period is 2 pi / 3, and the vertical shift is +4.
730722
put sinx=t\ncosxdx=dt\nI=fcos(pi t)dt with limits -1,+1\nI=(sin(pi t))/pi with limits +1,-1\nI=(sin(pi)-sin(-pi))/pi=0\nthanks & good luck
638682
Domain ~ x values\nRange ~ y values\n\nYou have to work the function and figure them out.
848127
When you added x to the right side, you should have gotten:\n\n4 = x^2\n\nThen take the square root of both sides to get\n\nx = 2, -2
1171615
I think this is also the 'trial and error method'\nYou will have an equation where you do not know the value of x.\nKeep on substituting different values of x into the equation until you come up with the right answer.\nIf you are lucky, you may have been told that x lies somewhere between 2 values.\nBest way to do it is to choose a number, get the answer, then choose another number a fair bit larger than the first. If the answer you want is between these 2 values, choose a number about half way between the 2 and so on.
931616
COSx = -4/5\nx = ARCCOS(-4/5)\n\nThen just plug in the x value to the rest of the equations.
844799
Wallstream is correct -- start with a right triangle.\n\nLabel the interior angles a, b and c, with c being the 90-degree angle.\n\nNow notice that a and b are complementary angles that add up to 90-degrees.\n\nLet x = sina.  Since the angles are complementary, we also have that x = cosb.\n\nSo we can rewrite as follows:\n\nIf x = sina, then a = arcsinx\nIf x = cosb, then b = arccosx\n\nBut a+b = 90-degrees (aka pi/2).\n\nSo arcsinx + arccosx = pi/2\n\nHope that helped!
1186712
arctan x should be interpreted as "the angle whose tangent is x". So draw a right triangle and include an angle whose tangent is x. Then figure out the sine of that angle in terms of x and you're done.\n\nEasiest way to do this is to let the side opposite the angle be x and the side adjacent to the angle be 1. Then the hypotenuse will have length sqrt(1+x^2).
873572
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
830100
'x= rcost\ny=rsint\n\nand r= sqrt(x^2+y^2)\nt=arctan(y/x)\nso\nif x=2, and y=2, then r= sqrqt(4+4)=sqrt(8)\nand t=arctan(2/2) = arctan(1) = pi/4
1177030
If you start at where 3 would be on a clock and go counter-clockwise 150 degrees, you get to 30 degrees above the x axis.  The radius, the x axis and the height between the x axis and where the ray touches the circle form a special 30-60-90 right triangle.  Since the reference angle is 30 degrees, the opposite side, or the y coordinate, must be 1/2 (since the length of the 30 degree side is always 1/2 that of the 90 degree side and in a unit circle the 90 degree side is always 1).  By that same rule, the x axis length (x coordinate) must be -1/2 root 3 because it is in the 2nd quadrant, making it negative, and it is the length of the 30 degree side times root three, since it is the 60 degree side.\n\nHope this helped!
754643
im taking Trig. try finding a homework helper web page. or go to the book that ur learning from web page because the majority of the times they have practice help on there.
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
263848
cot = o/a\ncsc = r/o\n\no = opposite side\na = adjacent side\nr = radius\n\ncalculate csc(2x) in terms of o and r.\nSubstitute o/a for cot(x) and substitue the value you determine for csc(2x) \nmultiply these, then convert to a trig function in terms of x.\n\nIf you know trig identities better than I do, you can reduce this expression using identities, without resorting to o, a, and r.  However, either route will work.
929777
we have x^2+y^2 = r^2 for polar coordinates'\n\nx = r cos t\ny = r sin t\n\nwe need to eliminate t ( t for theta as I cannot key in theta)\n\n1st get in terms of cos sin\n\nr = 8/(4+ cos t)\n\n4r + r cos t = 8\n\nbut r cos t = x\n\nso 4r = 8- x\nsquare both sides\n16r^2 = 64+x^2-16x\n\nusing r^2 = x^2 + y^2\n\n16(x^2+y^2) = 64 + x^2 -16x\n\n16x^2+16y^2 = 64 + x^2 - 16x\nor \ntarnsferring RHS to left \n\n15x^2+16x+16y^2- 64 = 0