How do I verify sine^5X*cos^2 = (cos^2X = 2cos^4X + cos^6)sinx?
Q. I can't seem to figure it out
Asked by Ky - Wed Mar 19 21:43:39 2008 - - 1 Answers - 0 Comments
A. sin (x) + cos (x) = 1 LHS = sin^5(x)cos (x) = sin(x)[sin^4(x)cos (x)] = sin(x)[(sin (x)) cos (x)] = sin(x)[(1-cos (x)) cos (x)] = sin(x)((1 - 2cos (x) + cos^4(x))cos (x)) = sin(x)(cos (x) - 2cos^4(x) + cos^6(x)) = (cos (x) - 2cos^4(x) + cos^6(x))sin(x) = RHS
Answered by gudspeling - Wed Mar 19 21:51:43 2008
Q. I can't seem to figure it out
Asked by Ky - Wed Mar 19 21:43:39 2008 - - 1 Answers - 0 Comments
A. sin (x) + cos (x) = 1 LHS = sin^5(x)cos (x) = sin(x)[sin^4(x)cos (x)] = sin(x)[(sin (x)) cos (x)] = sin(x)[(1-cos (x)) cos (x)] = sin(x)((1 - 2cos (x) + cos^4(x))cos (x)) = sin(x)(cos (x) - 2cos^4(x) + cos^6(x)) = (cos (x) - 2cos^4(x) + cos^6(x))sin(x) = RHS
Answered by gudspeling - Wed Mar 19 21:51:43 2008
What is the area of the smallest rectangle that encloses 2+2cos(2t)?
Q. Oh, btw, It's in polar coordinates. Sorry I left that out. Hah SO the equation is r=2+2cos(2(theta))
Asked by Fanclub - Sun Apr 26 12:32:21 2009 - - 2 Answers - 0 Comments
Q. Oh, btw, It's in polar coordinates. Sorry I left that out. Hah SO the equation is r=2+2cos(2(theta))
Asked by Fanclub - Sun Apr 26 12:32:21 2009 - - 2 Answers - 0 Comments
How can you prove cot^2cos^2 = cot^2-cos^2 ?
Q. I know the fundamentals of solving these trig problems, but I'm stuck on this one. Just as a note, each trig function is followed with a 'theta' symbol, I just don't know how to insert it onto here.
Asked by Kyle - Fri Aug 7 14:57:59 2009 - - 2 Answers - 0 Comments
A. cot^2 cos^2 = cos^2 * cos^2/sin^2 = (1-sin^2) * cos^2/sin^2 [ as sin^2 + cos^2 = 1] = (cos^2 - sin^2*cos^2)/sin^2 = cos^2/sin^2 - cos^2 = cot^2 - cos^2
Answered by Robin T - Fri Aug 7 15:03:15 2009
Q. I know the fundamentals of solving these trig problems, but I'm stuck on this one. Just as a note, each trig function is followed with a 'theta' symbol, I just don't know how to insert it onto here.
Asked by Kyle - Fri Aug 7 14:57:59 2009 - - 2 Answers - 0 Comments
A. cot^2 cos^2 = cos^2 * cos^2/sin^2 = (1-sin^2) * cos^2/sin^2 [ as sin^2 + cos^2 = 1] = (cos^2 - sin^2*cos^2)/sin^2 = cos^2/sin^2 - cos^2 = cot^2 - cos^2
Answered by Robin T - Fri Aug 7 15:03:15 2009
I'm stuck. How do you use Newton's method, to find the positive root of 2cos(x)=x^4?
Q. I need to answer it correct to 6 decimal places. thanks for your help.
Asked by Jessica B - Wed Apr 18 16:33:26 2007 - - 4 Answers - 0 Comments
A. let f(x) = 2 cosx - x^4 You could tell that there is a root just after x = 1, but we don't know exactly where. So we could use x =1 as a starting point. Newton's method states that if we have an initial guess x_old then an improved estimate x_new as follows: x_new = x_old - f(x_old) / f'(x_old) f'(x) = -2sinx - 4*x^3 x_new = x_old - [2 cos(x_old) - (x_old)^4] / [-2sin(x_old) - 4*(x_old)^3] Start with x_old = 1. On the first iteration, you get x_new = 1.014 Now set x_old = 1.014 and go again until there is no longer any significant change. In about 3 iterations you should get x = 1.013958 That's your answer. NOTE: Make sure you are working in radians and not degrees.
Answered by Dr D - Wed Apr 18 16:49:27 2007
Q. I need to answer it correct to 6 decimal places. thanks for your help.
Asked by Jessica B - Wed Apr 18 16:33:26 2007 - - 4 Answers - 0 Comments
A. let f(x) = 2 cosx - x^4 You could tell that there is a root just after x = 1, but we don't know exactly where. So we could use x =1 as a starting point. Newton's method states that if we have an initial guess x_old then an improved estimate x_new as follows: x_new = x_old - f(x_old) / f'(x_old) f'(x) = -2sinx - 4*x^3 x_new = x_old - [2 cos(x_old) - (x_old)^4] / [-2sin(x_old) - 4*(x_old)^3] Start with x_old = 1. On the first iteration, you get x_new = 1.014 Now set x_old = 1.014 and go again until there is no longer any significant change. In about 3 iterations you should get x = 1.013958 That's your answer. NOTE: Make sure you are working in radians and not degrees.
Answered by Dr D - Wed Apr 18 16:49:27 2007
Find all solutions to the equation 2cos(x) + 3 = sin^2(x) using the quadratic formula?
Q. I'm working on some trig problems, and I came across one that's confusing me. I would be extremely grateful if someone could explain how to do this problem ^_^ Find all solutions to the equation 2cos(x) + 3 = sin^2(x) using the quadratic formula. Thank you in advance!! =]
Asked by madxmoogle - Wed May 6 19:29:14 2009 - - 1 Answers - 0 Comments
A. 2cos(x) + 3 = sin (x) 2cos(x) + 3 = 1 cos (x) cos (x) + 2cos(x) + 2 = 0 cos x = { 2 [2 (4)(1)(2)]}/2(1) cos x = { 2 [ 4]}/2 cos x is not a real number. There are no solution for x in the real world.
Answered by Jerome J - Wed May 6 19:36:58 2009
Q. I'm working on some trig problems, and I came across one that's confusing me. I would be extremely grateful if someone could explain how to do this problem ^_^ Find all solutions to the equation 2cos(x) + 3 = sin^2(x) using the quadratic formula. Thank you in advance!! =]
Asked by madxmoogle - Wed May 6 19:29:14 2009 - - 1 Answers - 0 Comments
A. 2cos(x) + 3 = sin (x) 2cos(x) + 3 = 1 cos (x) cos (x) + 2cos(x) + 2 = 0 cos x = { 2 [2 (4)(1)(2)]}/2(1) cos x = { 2 [ 4]}/2 cos x is not a real number. There are no solution for x in the real world.
Answered by Jerome J - Wed May 6 19:36:58 2009
How do u solve:If f(x)=2cos^2X + sinx - 1, find the value of f(pi/2)?
Q. How do u solve:If f(x)=2cos^2X + sinx - 1, find the value of f(pi/2)?
Asked by sun8979 - Thu Mar 27 16:17:51 2008 - - 2 Answers - 0 Comments
A. REplace cosx by 1-sin^2x getting f(x) = 2(1-sin^2x) +sinx -1 f(x) = -2sin^2x +sinx +1 Now you have a quadratic equation in sinx sinx = [-1 +/- sqrt(1+8)]/-4 sinx = [-1 +/- 3]/-4 sinx = 1 or sinx = -.5 x = arcsin(1) and arcsin(-.5) f(pi/2) = 2*0 + 1-1 =0
Answered by ironduke8159 - Thu Mar 27 16:28:26 2008
Q. How do u solve:If f(x)=2cos^2X + sinx - 1, find the value of f(pi/2)?
Asked by sun8979 - Thu Mar 27 16:17:51 2008 - - 2 Answers - 0 Comments
A. REplace cosx by 1-sin^2x getting f(x) = 2(1-sin^2x) +sinx -1 f(x) = -2sin^2x +sinx +1 Now you have a quadratic equation in sinx sinx = [-1 +/- sqrt(1+8)]/-4 sinx = [-1 +/- 3]/-4 sinx = 1 or sinx = -.5 x = arcsin(1) and arcsin(-.5) f(pi/2) = 2*0 + 1-1 =0
Answered by ironduke8159 - Thu Mar 27 16:28:26 2008
2cos(t+1)=(-2) How do I solve this without a graph? How many solutions are there, and how do I determine this?
Q. 2cos(t+1)=(-2) How do I solve this without a graph? How many solutions are there, and how do I determine this?
Asked by Amedeo Avogadro - Sat Apr 5 22:12:39 2008 - - 5 Answers - 0 Comments
A. divide both sides by 2: cos(t+1)=-1 cos180 or cos(pi)=-1 so t+1= 180 or pi thus t= 179 or (pi-1) depending on what units you're using.
Answered by bitchinho - Sat Apr 5 22:18:13 2008
Q. 2cos(t+1)=(-2) How do I solve this without a graph? How many solutions are there, and how do I determine this?
Asked by Amedeo Avogadro - Sat Apr 5 22:12:39 2008 - - 5 Answers - 0 Comments
A. divide both sides by 2: cos(t+1)=-1 cos180 or cos(pi)=-1 so t+1= 180 or pi thus t= 179 or (pi-1) depending on what units you're using.
Answered by bitchinho - Sat Apr 5 22:18:13 2008
How would you prove the identity of 2cos^2 x = (1+sec 2x) cos 2x?
Q. How would you prove the identity of 2cos^2 x = (1+sec 2x) cos 2x?
Asked by Mstack429. - Sun Jan 18 15:49:40 2009 - - 1 Answers - 0 Comments
A. The right hand side is (1 + sec2x)cos2x = (1 + 1/cos2x)cos2x = cos2x + 1 But, cos2x = cos^2x - sin^2x So, the right hand side becomes = cos^2x - sin^2x + 1 But 1 - sin^2x = cos^2x So, the right hand side becomes = cos^2x + cos^2x = 2cos^2x = left hand side
Answered by Math H - Sun Jan 18 15:54:37 2009
Q. How would you prove the identity of 2cos^2 x = (1+sec 2x) cos 2x?
Asked by Mstack429. - Sun Jan 18 15:49:40 2009 - - 1 Answers - 0 Comments
A. The right hand side is (1 + sec2x)cos2x = (1 + 1/cos2x)cos2x = cos2x + 1 But, cos2x = cos^2x - sin^2x So, the right hand side becomes = cos^2x - sin^2x + 1 But 1 - sin^2x = cos^2x So, the right hand side becomes = cos^2x + cos^2x = 2cos^2x = left hand side
Answered by Math H - Sun Jan 18 15:54:37 2009
How do I simplify and solve this equation? (2cos^2)(x) - cos(x) = 1 ?
Q. (2cos^2)(x) - cos(x) = 1 My book says to solve the equation for x but I'm confused as to how??? Can someone please help me?
Asked by anonymity - Sun Oct 4 10:16:03 2009 - - 2 Answers - 0 Comments
A. There are two ways to solve this equation. Because I come from a different country I will write both of them, just in case there is a difference in the education system. 1. You can rewrite the equation in the following way: cos^2(x)+cos^2(x) - cos(x) -1 =0 Now you will group the first term with the last and the second with the third: cos^2(x) -1+cos^2(x) - cos(x) =0 (cos(x)-1)(cos(x)+1) + cos(x)(cos(x)-1)=0 (cos(x)-1)(cos(x)+1+cos(x ))=0 (cos(x)-1)(2cos(x)+1)=0 and from this you have: cos(x)=1 from where x= 2npi, where n is an integer or cos(x)= - 1/2 from where x= 2/3pi+ npi 2. You can note cosx=a for instance You will have then the equation: 2a^2-a-1=0 delta=1+8=9 a= (1+ - sqrt(delta))/4= (1+ - 3)/4 from where, a1=1 that… [cont.]
Answered by Stephany C - Sun Oct 4 10:41:20 2009
Q. (2cos^2)(x) - cos(x) = 1 My book says to solve the equation for x but I'm confused as to how??? Can someone please help me?
Asked by anonymity - Sun Oct 4 10:16:03 2009 - - 2 Answers - 0 Comments
A. There are two ways to solve this equation. Because I come from a different country I will write both of them, just in case there is a difference in the education system. 1. You can rewrite the equation in the following way: cos^2(x)+cos^2(x) - cos(x) -1 =0 Now you will group the first term with the last and the second with the third: cos^2(x) -1+cos^2(x) - cos(x) =0 (cos(x)-1)(cos(x)+1) + cos(x)(cos(x)-1)=0 (cos(x)-1)(cos(x)+1+cos(x ))=0 (cos(x)-1)(2cos(x)+1)=0 and from this you have: cos(x)=1 from where x= 2npi, where n is an integer or cos(x)= - 1/2 from where x= 2/3pi+ npi 2. You can note cosx=a for instance You will have then the equation: 2a^2-a-1=0 delta=1+8=9 a= (1+ - sqrt(delta))/4= (1+ - 3)/4 from where, a1=1 that… [cont.]
Answered by Stephany C - Sun Oct 4 10:41:20 2009
What is the simplest form of the following? sin^4 x + 2cos^2 x cos4x?
Q. What is the simplest form of the following? sin^4 x + 2cos^2 x cos4x Thanks in advance for any help! :)
Asked by Yoshi K - Wed Jan 28 19:54:01 2009 - - 1 Answers - 0 Comments
A. y = sin^4(x) + 2cos^2(x) - cos(4x) Noting that sin^4(x) = [sin^2(x)]^2 = [1 - cos^2(x)]^2 = 1 - 2cos^2(x) + cos^4(x), and that cos(4x) = 8cos^4(x) - 8cos^2(x) + 1, we have : y = [1 - 2cos^2(x) + cos^4(x)] + [2cos^2(x)] - [8cos^4(x) - 8cos^2(x) + 1] = 8cos^2(x) - 7cos^4(x) Can't really make it any simpler than this, although some manipulations give : cos^2(x)[1 + 7sin^2(x)], but this should be multiplied out. or, (7/4)sin^2(2x) + cos^2(x)
Answered by falzoon - Wed Jan 28 21:31:19 2009
Q. What is the simplest form of the following? sin^4 x + 2cos^2 x cos4x Thanks in advance for any help! :)
Asked by Yoshi K - Wed Jan 28 19:54:01 2009 - - 1 Answers - 0 Comments
A. y = sin^4(x) + 2cos^2(x) - cos(4x) Noting that sin^4(x) = [sin^2(x)]^2 = [1 - cos^2(x)]^2 = 1 - 2cos^2(x) + cos^4(x), and that cos(4x) = 8cos^4(x) - 8cos^2(x) + 1, we have : y = [1 - 2cos^2(x) + cos^4(x)] + [2cos^2(x)] - [8cos^4(x) - 8cos^2(x) + 1] = 8cos^2(x) - 7cos^4(x) Can't really make it any simpler than this, although some manipulations give : cos^2(x)[1 + 7sin^2(x)], but this should be multiplied out. or, (7/4)sin^2(2x) + cos^2(x)
Answered by falzoon - Wed Jan 28 21:31:19 2009
How do I solve the equation in the interval [0,2) that is... 2cos^2(x)+sin(x)=1?
Q. How do I solve the equation in the interval [0,2) that is... 2cos^2(x)+sin(x)=1?
Asked by Dylon M - Mon Apr 21 17:52:30 2008 - - 1 Answers - 0 Comments
A. Given: 2cos^2(x) + sin(x) = 1 First, rewrite the following identity: cos^2(x) + sin^2(x) = 1 cos^2(x) = 1 - sin^2(x) Then substitute into original and simplify: 2cos^2(x) + sin(x) = 1 2(1 - sin^2(x)) + sin(x) = 1 2 - 2sin^2(x) + sin(x) = 1 -2sin^2(x) + sin(x) + 2 = 1 -2sin^2(x) + sin(x) + 1 = 0 Note that this is a quadratic in "sin(x)". Either use the quadratic equation or manually factor. Since it is pretty easy, I'll factor: [2sin(x) + 1] [-sin(x) + 1] = 0 Thus: 2sin(x) + 1 = 0 2sin(x) = -1 sin(x) = -1/2 x = arcsin(-1/2) x = 7*pi/6 or 11*pi/6 x = 3.6651914 or 5.7595865 or: -sin(x) + 1 = 0 -sin(x) = -1 sin(x) = 1 x = arcsin(1) x = pi/2 x = 1.570796327 Note, however, that the question restricts the answer to the range [0,2), and… [cont.]
Answered by morgan - Mon Apr 21 18:53:55 2008
Q. How do I solve the equation in the interval [0,2) that is... 2cos^2(x)+sin(x)=1?
Asked by Dylon M - Mon Apr 21 17:52:30 2008 - - 1 Answers - 0 Comments
A. Given: 2cos^2(x) + sin(x) = 1 First, rewrite the following identity: cos^2(x) + sin^2(x) = 1 cos^2(x) = 1 - sin^2(x) Then substitute into original and simplify: 2cos^2(x) + sin(x) = 1 2(1 - sin^2(x)) + sin(x) = 1 2 - 2sin^2(x) + sin(x) = 1 -2sin^2(x) + sin(x) + 2 = 1 -2sin^2(x) + sin(x) + 1 = 0 Note that this is a quadratic in "sin(x)". Either use the quadratic equation or manually factor. Since it is pretty easy, I'll factor: [2sin(x) + 1] [-sin(x) + 1] = 0 Thus: 2sin(x) + 1 = 0 2sin(x) = -1 sin(x) = -1/2 x = arcsin(-1/2) x = 7*pi/6 or 11*pi/6 x = 3.6651914 or 5.7595865 or: -sin(x) + 1 = 0 -sin(x) = -1 sin(x) = 1 x = arcsin(1) x = pi/2 x = 1.570796327 Note, however, that the question restricts the answer to the range [0,2), and… [cont.]
Answered by morgan - Mon Apr 21 18:53:55 2008
How to change r: 2cos(theta) into x^2 + y^2 = r^2 on the conic app for TI-84 graphing calculator? ?
Q. My calculator seems to be going crazy; it's in polar form, I think, but when I hit 2nd zoom, it's in rectangular. What is making my conics application look like r=2cos(theta) not x^2 + y^2 = r^2?
Asked by musicxisxlovex3 - Thu Dec 18 15:51:04 2008 - - 1 Answers - 0 Comments
A. Hi, Let x = r*cos(theeta) and y = r*sin(theeta) squaring and adding: => x^2 + y^2 = r^2*cos^2(theeta) + r^2*sin^2(theeta)) => x^2 + y^2 = r^2*(cos^2(theeta) + sin^2(theeta)) => x^2 + y^2 = r^2 As: cos^2(theeta) + sin^2(theeta) = 1 Hope to answer you well, keep smiling, bye.
Answered by asad meer - Mon Dec 22 07:42:30 2008
Q. My calculator seems to be going crazy; it's in polar form, I think, but when I hit 2nd zoom, it's in rectangular. What is making my conics application look like r=2cos(theta) not x^2 + y^2 = r^2?
Asked by musicxisxlovex3 - Thu Dec 18 15:51:04 2008 - - 1 Answers - 0 Comments
A. Hi, Let x = r*cos(theeta) and y = r*sin(theeta) squaring and adding: => x^2 + y^2 = r^2*cos^2(theeta) + r^2*sin^2(theeta)) => x^2 + y^2 = r^2*(cos^2(theeta) + sin^2(theeta)) => x^2 + y^2 = r^2 As: cos^2(theeta) + sin^2(theeta) = 1 Hope to answer you well, keep smiling, bye.
Answered by asad meer - Mon Dec 22 07:42:30 2008
I need help integrating the limacon r = 2cos(Q) - 1 in polarcoordinates?
Q. I'm asked to find the area between the inner and outer loop. But the part that is confusing to me is once I integrate this equation, how do I know that i'm integrating the area between the inner and outer loop and not the area between the origin and the inner loop?
Asked by Wormhart - Thu Mar 26 10:30:08 2009 - - 3 Answers - 0 Comments
A. What you need to do is calculate two areas and then subtract them to get the area wanted. Area1 = the area inside the inner loop Area2 = the area inside the outer loop Area = Area2 - Area1 If you look at a graph of the curve you will see that the inner area can be found by integrating the curve between Q = 0 and Q = pi/3 and then multiplying this by 2 (since the integration only gives the area in the first quadrant and we also need the contribution from the fourth quadrant). The area enclosed by the outer loop can be found by integrating the curve from Q = pi to Q = (5/3)pi and then multiplying this by 2 (in this case the integration gives the area in quads 1 and 2 and we also need the contribution from quads 3 and 4). Determining… [cont.]
Answered by Captain Mephisto - Thu Mar 26 11:42:10 2009
Q. I'm asked to find the area between the inner and outer loop. But the part that is confusing to me is once I integrate this equation, how do I know that i'm integrating the area between the inner and outer loop and not the area between the origin and the inner loop?
Asked by Wormhart - Thu Mar 26 10:30:08 2009 - - 3 Answers - 0 Comments
A. What you need to do is calculate two areas and then subtract them to get the area wanted. Area1 = the area inside the inner loop Area2 = the area inside the outer loop Area = Area2 - Area1 If you look at a graph of the curve you will see that the inner area can be found by integrating the curve between Q = 0 and Q = pi/3 and then multiplying this by 2 (since the integration only gives the area in the first quadrant and we also need the contribution from the fourth quadrant). The area enclosed by the outer loop can be found by integrating the curve from Q = pi to Q = (5/3)pi and then multiplying this by 2 (in this case the integration gives the area in quads 1 and 2 and we also need the contribution from quads 3 and 4). Determining… [cont.]
Answered by Captain Mephisto - Thu Mar 26 11:42:10 2009
How do you find the derivative of G(t)=1+2cos[2 /5(t-3)]?
Q. How do you find the derivative of G(t)=1+2cos[2 /5(t-3)]?
Asked by MsBootyliscious - Wed Feb 25 20:43:04 2009 - - 1 Answers - 0 Comments
A. Derivative by parts first (add both later) derivative of 1 is 0, so only the last part counts Constant rule: multiply the whole thing by two later chain rule! Derivative of outside times derivative of the inside -sin(2pi/5(t-3) * derivative of the inside. Now is t-3 on the denominator or the numerator?
Answered by lv21swd - Wed Feb 25 20:48:50 2009
Q. How do you find the derivative of G(t)=1+2cos[2 /5(t-3)]?
Asked by MsBootyliscious - Wed Feb 25 20:43:04 2009 - - 1 Answers - 0 Comments
A. Derivative by parts first (add both later) derivative of 1 is 0, so only the last part counts Constant rule: multiply the whole thing by two later chain rule! Derivative of outside times derivative of the inside -sin(2pi/5(t-3) * derivative of the inside. Now is t-3 on the denominator or the numerator?
Answered by lv21swd - Wed Feb 25 20:48:50 2009
How to integrate definate integral of 39(sin(t)^2) (2cos(t)^4) dt from 0 to pi?
Q. Thanks!
Asked by fle - Sat Feb 28 20:16:59 2009 - - 1 Answers - 0 Comments
A. 39 sin t [2cos^4(t)] dt = recall the half-angle identities: sin t = (1/2)[1 - cos(2t)] cos t = (1/2)[1 + cos(2t)] thus the integral becomes: 39 sin t [2(cos t) ] dt = 39 (1/2)[1 - cos(2t)] 2{(1/2)[1 + cos(2t)]} dt = that is, expanding the square: 39 (1/2)[1 - cos(2t)] 2{(1/4)[1 + 2cos(2t) + cos (2t)]} dt = 39 (1/2)[1 - cos(2t)] (1/2)[1 + 2cos(2t) + cos (2t)] dt = pull out the constants and expand the integrand: (39/4) [1 + 2cos(2t) + cos (2t) - cos(2t) - 2cos (2t) - cos (2t)] dt = (39/4) [1 + cos(2t) - cos (2t) - cos (2t)] dt = break it up into: (39/4) dt + (39/4) cos(2t) - (39/4) cos (2t) - (39/4) cos (2t) dt = (39/4)t + (39/4) (1/2)sin(2t) - (39/4) cos (2t) - (39/4) cos (2t) dt = as for the… [cont.]
Answered by germano - Sun Mar 1 09:13:43 2009
Q. Thanks!
Asked by fle - Sat Feb 28 20:16:59 2009 - - 1 Answers - 0 Comments
A. 39 sin t [2cos^4(t)] dt = recall the half-angle identities: sin t = (1/2)[1 - cos(2t)] cos t = (1/2)[1 + cos(2t)] thus the integral becomes: 39 sin t [2(cos t) ] dt = 39 (1/2)[1 - cos(2t)] 2{(1/2)[1 + cos(2t)]} dt = that is, expanding the square: 39 (1/2)[1 - cos(2t)] 2{(1/4)[1 + 2cos(2t) + cos (2t)]} dt = 39 (1/2)[1 - cos(2t)] (1/2)[1 + 2cos(2t) + cos (2t)] dt = pull out the constants and expand the integrand: (39/4) [1 + 2cos(2t) + cos (2t) - cos(2t) - 2cos (2t) - cos (2t)] dt = (39/4) [1 + cos(2t) - cos (2t) - cos (2t)] dt = break it up into: (39/4) dt + (39/4) cos(2t) - (39/4) cos (2t) - (39/4) cos (2t) dt = (39/4)t + (39/4) (1/2)sin(2t) - (39/4) cos (2t) - (39/4) cos (2t) dt = as for the… [cont.]
Answered by germano - Sun Mar 1 09:13:43 2009
How would you find the area inside the circle r=3 but outside the circles r=cos(theta) and r=-2cos(theta)?
Q. Calculus 2 problem.
Asked by jamartinkw2011 - Mon Aug 4 10:28:16 2008 - - 2 Answers - 0 Comments
A. there are 3 circles circle 1 .. . . r = 3, circle with radius 3, centered at the origin. circle 2 .. . . r = cos , circle with radius 1/2, right of y-axis, tangent to origin. circle 3 .. . . r = -2cos , circle with radius 1, left of y-axis, tangent to origin. note that circles 2 & 3 are completely inside circle 1. we can use geometry instead. Area of circle 1 = 9 Area of circle 2 = /4 area of circle 3 = total area needed = 9 - 5 /4 = 31 /4
Answered by Alam Ko Iyan - Mon Aug 4 23:09:35 2008
Q. Calculus 2 problem.
Asked by jamartinkw2011 - Mon Aug 4 10:28:16 2008 - - 2 Answers - 0 Comments
A. there are 3 circles circle 1 .. . . r = 3, circle with radius 3, centered at the origin. circle 2 .. . . r = cos , circle with radius 1/2, right of y-axis, tangent to origin. circle 3 .. . . r = -2cos , circle with radius 1, left of y-axis, tangent to origin. note that circles 2 & 3 are completely inside circle 1. we can use geometry instead. Area of circle 1 = 9 Area of circle 2 = /4 area of circle 3 = total area needed = 9 - 5 /4 = 31 /4
Answered by Alam Ko Iyan - Mon Aug 4 23:09:35 2008
How do I do this? Find all values of in the interval [0, 2 ] that satisfy the equation 2cos(x) + sin(2x)=0?
Q. Find all values of in the interval [0, 2 ] that satisfy the equation 2cos(x) + sin(2x)=0 ?
Asked by Shane V - Sun Jan 11 18:24:25 2009 - - 2 Answers - 0 Comments
A. I think it would help to know a trig identity for this: sin(2x)=2sinx *cos x Let's use the right-hand side expression instead of sin(2x): 2cos(x)+2*sin(x)*cos(x)=0 There might be a way to keep using trig identities and solve for x, but I think there is an easier way to do this qualitatively (without calculations). To do this we must keep in mind when cos(x) and sin(x) are equal to zero and one in our given interval: cos(x)=1 when x=0, x=2*pi cos(x)= -1 when x=pi cos(x)=0 when x= pi/2, x= (3*pi)/2 sin(x)=0 when cos(x)= plus or minus one, or whenx=0, x=pi, x=2*pi sin(x)=1 when x= pi/2 sin(x)= -1 when x=(3*pi)/2 Let me write our equation again: 2cos(x)+2*sin(x)*cos(x)=0 Since 2cos(x) plus the other quantity can be zero, we can… [cont.]
Answered by Mr. Math - Sun Jan 11 18:49:53 2009
Q. Find all values of in the interval [0, 2 ] that satisfy the equation 2cos(x) + sin(2x)=0 ?
Asked by Shane V - Sun Jan 11 18:24:25 2009 - - 2 Answers - 0 Comments
A. I think it would help to know a trig identity for this: sin(2x)=2sinx *cos x Let's use the right-hand side expression instead of sin(2x): 2cos(x)+2*sin(x)*cos(x)=0 There might be a way to keep using trig identities and solve for x, but I think there is an easier way to do this qualitatively (without calculations). To do this we must keep in mind when cos(x) and sin(x) are equal to zero and one in our given interval: cos(x)=1 when x=0, x=2*pi cos(x)= -1 when x=pi cos(x)=0 when x= pi/2, x= (3*pi)/2 sin(x)=0 when cos(x)= plus or minus one, or whenx=0, x=pi, x=2*pi sin(x)=1 when x= pi/2 sin(x)= -1 when x=(3*pi)/2 Let me write our equation again: 2cos(x)+2*sin(x)*cos(x)=0 Since 2cos(x) plus the other quantity can be zero, we can… [cont.]
Answered by Mr. Math - Sun Jan 11 18:49:53 2009
What is the maximum point of the graph: y = 2cos(x) + 1?
Q. What is the maximum point of the graph: y = 2cos(x) + 1?
Asked by Sterling Silver - Wed Apr 15 22:56:56 2009 - - 2 Answers - 0 Comments
A. The answer is 3.
Answered by Sterling - Wed Apr 15 23:01:02 2009
Q. What is the maximum point of the graph: y = 2cos(x) + 1?
Asked by Sterling Silver - Wed Apr 15 22:56:56 2009 - - 2 Answers - 0 Comments
A. The answer is 3.
Answered by Sterling - Wed Apr 15 23:01:02 2009
help me find the intersection points for the functions f(x)=2cos^2x and g(x)=sinx+1?
Q. i need to find the intersection and contact points for these two functions but only by analytical means i have to prove my answers from my graphics calc are the only intersection and contact points for the two functions. please help i really dont understand the answer to my question before was really helpful thanks... but please help again!
Asked by Madonna - Tue May 19 01:47:19 2009 - - 2 Answers - 0 Comments
A. 2cos^2x = sinx + 1 2(1-sin^2x) = sinx + 1 2sin^2x + sinx -1 = 0 (2sinx-1)(sinx+1) = 0 sinx = 1/2 => x = pi/6, 5pi/6 or sinx = -1 => x = 3pi/2 --- Attention: The solutions obtained above are in [0, 2pi].
Answered by sahsjing - Wed May 20 19:43:14 2009
Q. i need to find the intersection and contact points for these two functions but only by analytical means i have to prove my answers from my graphics calc are the only intersection and contact points for the two functions. please help i really dont understand the answer to my question before was really helpful thanks... but please help again!
Asked by Madonna - Tue May 19 01:47:19 2009 - - 2 Answers - 0 Comments
A. 2cos^2x = sinx + 1 2(1-sin^2x) = sinx + 1 2sin^2x + sinx -1 = 0 (2sinx-1)(sinx+1) = 0 sinx = 1/2 => x = pi/6, 5pi/6 or sinx = -1 => x = 3pi/2 --- Attention: The solutions obtained above are in [0, 2pi].
Answered by sahsjing - Wed May 20 19:43:14 2009
2cos^2(x/2) - 3 cos(x) = 0 What is the correct solution set?
Q. 2cos^2(x/2) - 3 cos(x) = 0 What is the correct solution set?
Asked by lkscmsejflkes - Sat Aug 29 15:25:57 2009 - - 1 Answers - 0 Comments
A. 2cos^2(x/2) - 3 cos(x) = 0 ==> 2[(1 + cos(2 * x/2)) / 2] - 3 cos x = 0 ==> (1 + cos x) - 3 cos x = 0 ==> 1 - 2 cos x = 0 ==> 1 = 2 cos x ==> cos x = 1/2. Thus, x = pi/3 + 2 pi k, or -pi/3 + 2 pi k for any integer k. I hope that helps!
Answered by kb - Sat Aug 29 15:43:46 2009
Q. 2cos^2(x/2) - 3 cos(x) = 0 What is the correct solution set?
Asked by lkscmsejflkes - Sat Aug 29 15:25:57 2009 - - 1 Answers - 0 Comments
A. 2cos^2(x/2) - 3 cos(x) = 0 ==> 2[(1 + cos(2 * x/2)) / 2] - 3 cos x = 0 ==> (1 + cos x) - 3 cos x = 0 ==> 1 - 2 cos x = 0 ==> 1 = 2 cos x ==> cos x = 1/2. Thus, x = pi/3 + 2 pi k, or -pi/3 + 2 pi k for any integer k. I hope that helps!
Answered by kb - Sat Aug 29 15:43:46 2009
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