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Mathematics 19 Online
OpenStudy (anonymous):

sin(-pi/12); use 1/2 * pi/6

OpenStudy (anonymous):

cookie do i get a cookie if i help

OpenStudy (anonymous):

@Jerry)123gnzcraft sure

OpenStudy (anonymous):

So exactly how do you want this to be done? Using a multiple-angle formula to do this requires you to know the value of a sine/cosine of yet another value. For example, let's say you used the double-angle formula for sine: sin(2x) = 2sin(x)cos(x). With x = π/6, this becomes: sin(π/3) = 2sin(π/6)cos(π/6). At this point, the values of sin(π/3) and cos(π/6) need to be known. The formula for sin(6x) will involve both sin(x) and cos(x) in it; there is no way to get sin(6x) in terms of sin(x) only unless we do: sin(6x) = 2sin(3x)cos(3x) = 2sin(3x)√[1 - sin^2(3x)]. Using sin(3x) = 3sin(x) - 4sin^3(x), this becomes: sin(6x) = 2[3sin(x) - 4sin^3(x)]√{1 - [3sin(x) - 4sin^3(x)]^2}. Let x = π/6 here and you get: sin(π) = 2[3sin(π/6) - 4sin^3(π/6)]√{1 - [3sin(π/6) - 4sin^3(π/6)]^2} ==> 2[3sin(π/6) - 4sin^3(π/6)]√{1 - [3sin(π/6) - 4sin^3(π/6)]^2} = 0. (Note: in order to get this far, the assumption that sin(π) = 0 is required.) Let y = sin(π/6) and this equation becomes: 2(3y - 4y^3)√[1 - (3y - 4y^3)^2] = 0. By the zero-product property, this becomes: 3y - 4y^3 = 0 or √[1 - (3y - 4y^3)^2] = 0 ==> y = -1, -√3/2, -1/2, 0, 1/2, and 1. We know that y = sin(π/6) > 0 as π/6 is in Quadrant I, so this leaves us between: y = 0, 1/2, and 1. If you can use sin(0) = 0 and sin(π/2) = 1 (again, assuming that you know these values), then y = 1/2 is the only possible answer. Thus: y = sin(π/6) = 1/2. Q.E.D. I hope this helps

OpenStudy (anonymous):

cookie 1 only 1

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