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Q: What is sin inverse of sin of 3 pi over 8?
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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


How do you solve sin parenthesis arcsin parenthesis 2 over 3 closed parenthesis and another closed parenthesis?

sin(arcsin(2/3)) = 2/3, since sin is the inverse function of arcsin.


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.


What is the additive inverse of a over 3?

It is -a/3