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Unfortunately, limitations of the browser used by Answers.com means that we cannot see most symbols. It is therefore impossible to give a proper answer to your question. However, I am assuming the question is about sin (5pi/12). If not, please resubmit your question spelling out the symbols as "plus", "minus", "times" sin(5pi/12) = sin(pi/4 + pi/6) = sin(pi/4)*cos(pi/6) + cos(pi/4)*sin(pi/6) = √2/2*√3/2 + √2/2*1/2 = √2(√3 + 1)/4
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
You cannot because you do not know what R is.
sin(60) or sin(PI/3) = sqrt(3)/2 cos(60) or cos(PI/3)=1/2 tan(60) or tan(PI/3) = sin(60)/cos(60)=sqrt(3) But we want tan for -sqrt(3). Tangent is negative in quadrant II and IV. In Quadrant IV, we compute 360-60=300 or 2PI-PI/3 =5PI/3 tan(5PI/3) = -sqrt(3) Tangent is also negative in the second quadrant, so we compute PI-PI/3=2PI/3 or 120 degrees. tan(t)=-sqrt(3) t=5PI/3 or 2PI/3 The period of tan is PI The general solution is t = 5PI/3+ n PI, where n is any integer t = 2PI/3+ n PI, where n is any integer
5cm