0
Want this question answered?
Be notified when an answer is posted
Still curious? Ask our experts.
Chat with our AI personalities
Coach
Steve
RafaAdd your answer:
Earn +20 pts
What kind of angle is 2 pi over 3?
It's an obtuse angle.
What is the solution to the equation 4 sin x - 3 equals 0 to the nearest degree over the domain 0degrees360degrees?
4 sin(x) - 3 = 0 Therefore sin(x) = 3/4 And so the primary solution is x = sin-1(3/4) = 49 deg The second solution in the domain is 180 - 49 = 131 deg.
How do you solve 3 sin theta 1.5 when 0 isless than or equal to theta which is less than or equal to 4 radians?
I am assuming that the equation is 3*sin(t) = 1.5 even though the equality sign is not visible - due to the browser limitations. Then sin(t) = 1.5/3 = 0.5 So t = sin-1(0.5) which gives the principal value of t = 0.5236. The next value of t, in the domain, is pi - 0.5236 = 2.618 radians. There are no further values in the specified domain.
How do you prove that 2 sin 3x divided by sin x plus 2 cos 3x divided by cos x equals 8 cos 2x?
You need to know the trigonometric formulae for sin and cos of compound angles. sin(x+y) = sin(x)*cos(y)+cos(x)*sin(y) and cos(x+y) = cos(x)*cos(y) - sin(x)*sin(y) Using these, y = x implies that sin(2x) = sin(x+x) = 2*sin(x)cos(x) and cos(2x) = cos(x+x) = cos^2(x) - sin^2(x) Next, the triple angle formulae are: sin(3x) = sin(2x + x) = 3*sin(x) - 4*sin^3(x) and cos(3x) = 4*cos^3(x) - 3*cos(x) Then the left hand side = 2*[3*sin(x) - 4*sin^3(x)]/sin(x) + 2*[4*cos^3(x) - 3*cos(x)]/cos(x) = 6 - 8*sin^2(x) + 8cos^2(x) - 6 = 8*[cos^2(x) - sin^2(x)] = 8*cos(2x) = right hand side.
What is the amplitude of the function y 3 sin 3x?
3
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 answer to inverse tangent square root 3 over 3?
30 degrees or pi/6
What is the value of sin 3 pi over 2?
sin(3π/2) = -1
Why does sin 2pi over 6 radians equal .5?
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.
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 sin of pi divided by 3?
The question is ambiguous and the two possible answers are: sin(pi)/3 = 0 and sin(pi/3) = sqrt(3)/2 It is assumed, of course, that since the angles are given in terms of pi, they are measured in radians and not degrees!
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.
Who can Solve sin(-1)sin((5pi )(7)) with step by Step please )?
Unfortunately, the browser used for posting questions is hopelessly inadequate for mathematics: it strips away most symbols. All that we can see is "sin(-1)sin((5pi )(7))". From that it is not at all clear what the missing symbols (operators) between (5pi ) and (7) might be. There is, therefore no sensible answer. It makes little sense for me to try and guess - I may as well make up my own questions and answer them!All that I can tell you that the principal sin-1 is the inverse for sin over the domain (-pi/2, pi/2). Thus sin-1(sin(x) = x where -pi/2 < x
What is the additive inverse of -2 over -3?
-5
What is the period for y-3 sin x?
y = 3 sin x The period of this function is 2 pi.
How do you solve sin2x plus sinX equals 0?
sin(2x) + sin(x) = 0 2sin(x)cos(x) + sin(x) = 0 sin(x)[2cos(x) + 1] = 0 sin(x) = 0 OR 2cos(x) + 1 = 0 sin(x) = 0 OR cos(x) = -1/2 x = n*pi OR x = 2/3*pi + 2n*pi OR x = -2/3*pi + 2n*pi x = pi*[2n + (0 OR 2/3 OR 1 OR 4/3)] Note that n may be any integer. The solutions in [-2pi, 2pi] are: -2pi, -4/3pi, -pi, -2/3pi, 0, 2/3pi, pi, 4/3pi, 2pi
What is the additive inverse of a over 3?
It is -a/3
