Integral from 0 to pi 6sin2xdx: integral of 6sin2xdx (-3)cos2x+c. (-3)cos(2 x pi) - (-3)cos(2 x 0) -3 - -3 0
Since circumference equals pi times the diameter, if the diameter of a circle is 6, then its circumference is going to be pi times 6 or 6 pi.
5/6
It is: cos(15) = (sq rt of 6+sq rt of 2)/4
You can calculate that on any scientific calculator. Presumably, for any expression that involves "pi" the angle should be in radians, so be sure to set the calculator to radians first.
cos(a)cos(b)-sin(a)sin(b)=cos(a+b) a=7pi/12 and b=pi/6 a+b = 7pi/12 + pi/6 = 7pi/12 + 2pi/12 = 9pi/12 We want to find cos(9pi/12) cos(9pi/12) = cos(3pi/4) cos(3pi/4)= cos(pi-pi/4) cos(pi)cos(pi/4)-sin(pi)sin(pi/4) cos(pi)=-1 sin(pi)=0 cos(pi/4) = √2/2 sin(pi/4) =√2/2 cos(pi)cos(pi/4)-sin(pi)sin(pi/4) = - cos(pi/4) = -√2/2
11pi/12 = pi - pi/12 cos(11pi/12) = cos(pi - pi/12) cos(a-b) = cos(a)cos(b)+sin(a)sin(b) cos(pi -pi/12) = cos(pi)cos(pi/12) + sin(pi)sin(pi/12) sin(pi)=0 cos(pi)=-1 Therefore, cos(pi -pi/12) = -cos(pi/12) pi/12=pi/3 -pi/4 cos(pi/12) = cos(pi/3 - pi/4) = cos(pi/3)cos(pi/4)+sin(pi/3) sin(pi/4) cos(pi/3)=1/2 sin(pi/3)=sqrt(3)/2 cos(pi/4)= sqrt(2)/2 sin(pi/4) = sqrt(2)/2 cos(pi/3)cos(pi/4)+sin(pi/3) sin(pi/4) = (1/2)(sqrt(2)/2 ) + (sqrt(3)/2)( sqrt(2)/2) = sqrt(2)/4 + sqrt(6) /4 = [sqrt(2)+sqrt(6)] /4 Therefore, cos(pi/12) = (sqrt(2)+sqrt(6))/4 -cos(pi/12) = -(sqrt(2)+sqrt(6))/4 cos(11pi/12) = -(sqrt(2)+sqrt(6))/4
Integral from 0 to pi 6sin2xdx: integral of 6sin2xdx (-3)cos2x+c. (-3)cos(2 x pi) - (-3)cos(2 x 0) -3 - -3 0
y = 2(x) - (pi/3) + (sqrt(3)/2)
1.25
Cos(x) = Sin(2x) Using angle-addition, we have Sin(a+b) = Sin(a)Cos(b) + Sin(b)Cos(a). From that, we see Sin(2x) = Sin(x)Cos(x)+Sin(x)Cos(x) = 2Sin(x)Cos(x) Cos(x) = 2Sin(x)Cos(x) If Cos(x) = 0, then the two sides are equal. This occurs at x= Pi/2 + nPi, where n is an integer and Pi is approximately 3.14. If Cos(x) doesn't equal 0, then we can divide it out. Then, 1 = 2 Sin(x) , or 1/2 = Sin(x) This occurs when x = Pi/6 or 5Pi/6, plus or minus any multiples of 2 Pi.
Using the identity, sin(X)+sin(Y) = 2*sin[(x+y)/2]*cos[(x-y)/2] the expression becomes {2*sin[(23A-7A)/2]*cos[(23A+7A)/2]}/{2*sin[(2A+14A)/2]*cos[(2A-14A)/2]} = {2*sin(8A)*cos(15A)}/{2*sin(8A)*cos(-6A)} = cos(15A)/cos(-6A)} = cos(15A)/cos(6A)} since cos(-x) = cos(x) When A = pi/21, 15A = 15*pi/21 and 6A = 6*pi/21 = pi - 15pi/21 Therefore, cos(6A) = - cos(15A) and hence the expression = -1.
Sin(2*pi/6) = sin(pi/3) which, by definition, is 0.5 If you wish, you can calculate y/1! - y^3/3! + y^5/5! - y^7/7! + ... where y = pi/3.
There is an infinite number, ex. 2/4, 3/6, 4/8, 5/10 6/12, ext. (-4) / (-8) 14/28 (pi)/(2 pi) 792/1584 sin(x)cos(x)/sin(2x)
It is (5*pi/6)*8 = 20*pi/3 m.
Unfortunately, limitations of the browser used by Answers.com means that we cannot see most symbols. It is therefore impossible to give a proper answer to your question. However, I am assuming the question is about sin (5pi/12). If not, please resubmit your question spelling out the symbols as "plus", "minus", "times" sin(5pi/12) = sin(pi/4 + pi/6) = sin(pi/4)*cos(pi/6) + cos(pi/4)*sin(pi/6) = √2/2*√3/2 + √2/2*1/2 = √2(√3 + 1)/4
mass of proton is 6 x pi raised to power 5 times mass of electron