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Calculate sin(11pi/12).

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jolenta | Student, College Freshman | (Level 1) Honors

Posted June 27, 2010 at 2:12 AM via web

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Calculate sin(11pi/12).

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted June 27, 2010 at 2:13 AM (Answer #1)

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We'll write 11pi/12 = 12pi/12 - pi/12 = pi - pi/12

The angle (pi - pi/12) belongs to the second quadrant, so the sine function in the second qudrant, has positive values.

So, sin (pi - pi/12) = sin pi/12

We'll consider sin pi/12 = sin[(pi/6)/2]

But sin x/2 = sqrt[(1-cosx)/2]

sin[(pi/6)/2] = sqrt {[1-(cos pi/6)]/2}

cos pi/6 = sqrt3/2

sin[(pi/6)/2] = sqrt [(1 - sqrt3/2)/2]

sin[(pi/6)/2] = sqrt(2-sqrt3)/2

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neela | High School Teacher | (Level 3) Valedictorian

Posted June 27, 2010 at 2:19 AM (Answer #2)

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To calculate sin11pi/12

Solution:

sin x= sin(pi-x) .

So sin(11pi/12) = Sin (pi-11pi/12) = sinpi

=sin (pi/3-pi/4) = sinpi/3cos pi/4 -cospi/3sinpi/4

= (sqrt3/2)(1/sqrt2)-(1/2)(1/sqrt2)

={ ( sqrt3)-1 }/ (2sqrt2) = sqrt2(sqrt3-1)/4 = (sqrt6-sqrt2)/4

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted June 27, 2010 at 3:43 AM (Answer #3)

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sin(11pi/12)

Let us rewrite 11pi/12 = 12pi/12 -pi/12= pi-pi/12

==> SIN(11PI/12)=sin (pi-pi/12)

But, sin pi-x= sinx

==> sin(pi-pi/12)= sin(pi/12)=

==> sin(pi/12)= sin(1/2)*(pi/6)

But we know that :

sin(x/2)= +-sqrt[1-cosx)/2]

==> sin(pi/6)/2)= +-sqrt{[1-cos(pi/6)]/2}

                     = +-sqrt{1-[sqrt(3)/2]/2}

                     = +- sqrt[2-sqrt(3)]/2

 

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