Want this question answered?
Be notified when an answer is posted
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
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
Problem: find three solutions to z^3=-1. DeMoivre's theorem is that (cos b + i sin b)^n = cos bn + i sin bn So we can set z= (cos b + i sin b), n = 3 cos bn + i sin bn = -1. From the last equation, we know that cos bn = -1, and sin bn = 0. Three possible solutions are bn=pi, bn=3pi, bn=5pi. This gives three possible values of b: b=pi/3 b=pi b = 5pi/3. Now using z= (cos b + i sin b), we can get three possible cube roots of -1: z= (cos pi/3 + i sin pi/3), z= (cos pi + i sin pi), z= (cos 5pi/3 + i sin 5pi/3). Working these out gives -1/2+i*sqrt(3)/2 -1 -1/2-i*sqrt(3)/2
x = sin-1 (4/15) ( sin -1 is [SHIFT] [sin] on a calculator ) = 15.5
sin(pi/4) and cos(pi/4) are both the same. They both equal (√2)/2≈0.7071■
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
-5pi/2
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
tan(x) is the same as sin(x) / cos(x). Domain is all the real numbers, except those numbers where the cos(x) = 0. That is, the domain does not include pi/2, 3pi/2, 5pi/2, etc. The range includes all real numbers.tan(x) is the same as sin(x) / cos(x). Domain is all the real numbers, except those numbers where the cos(x) = 0. That is, the domain does not include pi/2, 3pi/2, 5pi/2, etc. The range includes all real numbers.tan(x) is the same as sin(x) / cos(x). Domain is all the real numbers, except those numbers where the cos(x) = 0. That is, the domain does not include pi/2, 3pi/2, 5pi/2, etc. The range includes all real numbers.tan(x) is the same as sin(x) / cos(x). Domain is all the real numbers, except those numbers where the cos(x) = 0. That is, the domain does not include pi/2, 3pi/2, 5pi/2, etc. The range includes all real numbers.
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.