sin(-pi) = sin(-180) = 0 So the answer is 0
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
No, for example if A = Pi/3. Then sin3A = sin pi = 0, but sinA = sin Pi/3 = 1/2. So for A = Pi/3, the sum is 1/2, not zero. It can't be proved because the statement is false. For example if A = Pi/3. Then sin3A = sin pi = 0, but sinA = sin Pi/3 = 1/2. So for A = Pi/3, the sum is 1/2, not zero.
sin x - cos x = 0sin x = cos x(sin x)^2 = (cos x)^2(sin x)^2 = 1 - (sin x)^22(sin x)^2 = 1(sin x)^2 = 1/2sin x = ± √(1/2)sin x = ± (1/√2)sin x = ± (1/√2)(√2/√2)sin x = ± √2/2x = ± pi/4 (± 45 degrees)Any multiple of 2pi can be added to these values and sine (also cosine) is still ± √2/2. Thus all solutions of sin x - cos x = 0 or sin x = cos x are given byx = ± pi/4 ± 2npi, where n is any integer.By choosing any two integers , such as n = 0, n = 1, n = 2 we can find some solutions of sin x - cos x = 0.n = 0, x = ± pi/4 ± (2)(n)(pi) = ± pi/4 ± (2)(0)(pi) = ± pi/4 ± 0 = ± pi/4n = 1, x = ± pi/4 ± (2)(n)(pi) = ± pi/4 ± (2)(1)(pi) = ± pi/4 ± 2pi = ± 9pi/4n = 2, x = ± pi/4 ± (2)(n)(pi) = ± pi/4 ± (2)(2)(pi) = ± pi/4 ± 4pi = ± 17pi/4
The derivative with respect to 'x' of sin(pi x) ispi cos(pi x)
y = 3 sin x The period of this function is 2 pi.
Do you mean Sin(pi/2) = 1 or [Sin(pi)] /2 = 0.0274....
sin(pi) = 0
(cos(pi x) + sin(pi y) )^8 = 44 differentiate both sides with respect to x 8 ( cos(pi x) + sin (pi y ) )^7 d/dx ( cos(pi x) + sin (pi y) = 0 8 ( cos(pi x) + sin (pi y ) )^7 (-sin (pi x) pi + cos (pi y) pi dy/dx ) = 0 8 ( cos(pi x) + sin (pi y ) )^7 (pi cos(pi y) dy/dx - pi sin (pi x) ) = 0 cos(pi y) dy/dx - pi sin(pi x) = 0 cos(pi y) dy/dx = sin(pi x) dy/dx = sin (pi x) / cos(pi y)
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
Sin(3pi/2) = Sin(2pi - pi/2) Double angle Trig. Identity. Hence Sin(2pi)Cos(pi/2) - Cos(2pi) Sin(pi/2) Sin(2pi) = 0 Cos(pi/2) = 0 Cos(2pi) = 1 Sin(pi/2) = 1 Substituting 0 x 0 - 1 x 1 = 0 - 1 = -1 The answer!!!!!
sin(pi) = 0 so 4*sin(pi) = 0 so Y = 0
2*pi is one complete revolution, i.e. 360 degrees. Sin of 2*pi = sin 360º = 0
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
Assuming you mean that the pi is not within the sin(2pi), its a vertical shift of +pi
sin(pi/4) and cos(pi/4) are both the same. They both equal (√2)/2≈0.7071■
No, for example if A = Pi/3. Then sin3A = sin pi = 0, but sinA = sin Pi/3 = 1/2. So for A = Pi/3, the sum is 1/2, not zero. It can't be proved because the statement is false. For example if A = Pi/3. Then sin3A = sin pi = 0, but sinA = sin Pi/3 = 1/2. So for A = Pi/3, the sum is 1/2, not zero.
Do you mean Sin(pi/7) or {Sin(pi)} / 7 Note the position of the '7' be it inside or outside the Sine Function. Clarity is needed, in the position of the '7'. However, Maker sure you calculator is in 'Radian' mode. Sin(pi/7) = Sin(0.44879895...) =0.0078329.... or {Sin(pi)}/7 = 0.054803665.../7 = 0.007829095021.... Note the two different answers.