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`cos ((24Pi-Pi)/12)*sin (Pi/12) = ?`

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gillfourre | Student, College Freshman | eNoter

Posted May 17, 2013 at 5:43 PM via web

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`cos ((24Pi-Pi)/12)*sin (Pi/12) = ?`

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted May 17, 2013 at 6:01 PM (Answer #1)

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You need to replace `(24pi - pi)/12` for `(23pi)/12` such that:

`cos ((23pi)/12)* sin (pi/12) = cos ((24pi - pi)/12)* sin (pi/12)`

`cos ((24pi - pi)/12) = cos ((24pi)/12 - pi/12) = cos (2pi - pi/12)`

You need to use the following trigonometric identity, such that:

`cos(2pi - alpha) = cos alpha`

Reasoning by analogy, yields:

`cos (2pi - pi/12) = cos(pi/12)`

Replacing `cos(pi/12)` for `cos ((23pi)/12)` yields:

`cos ((23pi)/12)* sin (pi/12) = cos(pi/12)*sin(pi/12)`

You should use the double angle identity, such that:

`cos 2alpha = 2 cos alpha*sin alpha `

Reasoning by analogy, yields:

`cos ((23pi)/12)* sin (pi/12) = (sin (2*(pi/12)))/2`

`cos ((23pi)/12)* sin (pi/12) = (sin(pi/6))/2 = 1/4`

Hence, evaluating the product `cos ((23pi)/12)* sin (pi/12)` , using the specified trigonometric identities, yields `cos ((23pi)/12)* sin (pi/12) = 1/4.`

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oldnick | Valedictorian

Posted May 18, 2013 at 1:44 AM (Answer #2)

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`cos((24pi-pi)/12) sin(pi/12)= cos(pi 23/12)sin(pi/12)=`

`=1/2[ sin(pi 23/12 + pi/12) -sin(pi 23/12-pi/12)]=` `1/2[sin2pi -sin pi 11/6 ]` `=-1/2 sin pi 11/6 =` `-1/2 sin(2pi- pi/6)=`

`=-1/2(-sin pi/6)=` `-1/2(-1/2)=1/4`

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oldnick | Valedictorian

Posted May 18, 2013 at 2:07 AM (Answer #3)

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oldnick | Valedictorian

Posted May 18, 2013 at 2:14 AM (Answer #4)

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`cos((24pi-pi)/12)sin(pi/12)=cos (23pi/12) sin (pi/12)=`

`=1/2[sin((23pi+pi)/12)-sin((23pi-pi )/12)]=` `1/2[sin(2pi) -sin (pi11/6)]=` `-1/2 sin (pi 11/6)=` `-1/2 sin(2pi-pi/6)=-1/2sin(pi/6)=-1/2(-1/2)= 1/4`

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