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What is the exact value of the expression cos 7pi over 12 cos pi over 6 -sin 7pi over 12 sin pi over 6?
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
Is sin3A plus sinA equal to 0?
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 0 0?
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
What is the derivative of sinΟx?
The derivative with respect to 'x' of sin(pi x) ispi cos(pi x)
What is the period for y-3 sin x?
y = 3 sin x The period of this function is 2 pi.
What is the sin of pi?
sin(pi) = 0
What is the sin of pi divided by 2?
Do you mean Sin(pi/2) = 1 or [Sin(pi)] /2 = 0.0274....
What is the derivative of cos pi x plus sin pi y all to the 8th power equals 44?
(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)
What is the sin pi over 2?
sin(pi/2)=1
What is sin of 3 pi over 2?
sin pi/2 =1 sin 3 pi/2 is negative 1 ( it is in 3rd quadrant where sin is negative
What is the exact value of the expression cos 7pi over 12 cos pi over 6 -sin 7pi over 12 sin pi over 6?
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
Y equals 4 sin x for x equals pi?
sin(pi) = 0 so 4*sin(pi) = 0 so Y = 0
What is the exact values of sin 2 pi?
2*pi is one complete revolution, i.e. 360 degrees. Sin of 2*pi = sin 360º = 0
What is the exact value using a sum or difference formula of the expression cos 11pi over 12?
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
What is the phase shift for sin 2pi plus pi?
Assuming you mean that the pi is not within the sin(2pi), its a vertical shift of +pi
What is the derivative of sin pi x?
pi cos(pi x)
What is the derivative of cos pi x?
-(pi)*sin(pi*x)
