cos(5π//6) = -(√3)/2 ≈ -0.866
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
Cos(47) = Cos(313) = 0.681998 ( 6 d.p.).
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
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
To find the exact value of (\cos\left(\frac{5\pi}{6}\right)), first note that (\frac{5\pi}{6}) is located in the second quadrant, where cosine values are negative. The reference angle is (\pi - \frac{5\pi}{6} = \frac{\pi}{6}). The cosine of (\frac{\pi}{6}) is (\frac{\sqrt{3}}{2}), so (\cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2}). Thus, the exact value is (-\frac{\sqrt{3}}{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
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.
The sine of (\frac{\pi}{12}) radians (which is equivalent to 15 degrees) can be calculated using the sine subtraction formula: (\sin(a - b) = \sin a \cos b - \cos a \sin b). By letting (a = \frac{\pi}{4}) (45 degrees) and (b = \frac{\pi}{3}) (60 degrees), we find that (\sin\left(\frac{\pi}{12}\right) = \sin\left(\frac{\pi}{4} - \frac{\pi}{3}\right) = \sin\frac{\pi}{4} \cos\frac{\pi}{3} - \cos\frac{\pi}{4} \sin\frac{\pi}{3}). This evaluates to (\frac{\sqrt{2}}{2} \cdot \frac{1}{2} - \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{2}}{4} - \frac{\sqrt{6}}{4} = \frac{\sqrt{2} - \sqrt{6}}{4}).
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.
CosX = Sin2x Using Trog. Identityt CosX = 2SinXCosX CosX cancels down to '1' Hence CosX/CosX = 2 SinX 1 = 2SinX SinX = 1/2 = 0.5 SinX = 0.5 X = Sin^(-1) 0,5 X = 30 degrees = pi/6 radians
pi/6 = 30 degrees pi/6 = 0.523598775.... radians
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)