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What is the exact value of sin(-π/12)? Please show all steps.

 May 2, 2020

Best Answer 

 #1
avatar+20906 
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The angle  -pi/12  is in Quadrant IV, so the answer will be negative.

 

Using the half-angle formula:  sin( a/2 )  =  +/- sqrt[ (1 - cos(a) / 2]   with  a  =  -pi/6

[we will need to choose the - sign in the formula]

 

sin( a/2 )  =  - sqrt[ (1 - cos(a) / 2]     --->     sin[ ( -pi/6 / 2 ]  =  - sqrt[ (1 - cos( -pi/6 ) / 2] 

    =  - sqrt[ (1 - sqrt(3)/2 ) / 2 ]  =  - sqrt[ (2 - sqrt(3) ) / 4 ]  =  - ½·sqrt( 2 - sqrt(3) )

 May 2, 2020
 #1
avatar+20906 
+1
Best Answer

The angle  -pi/12  is in Quadrant IV, so the answer will be negative.

 

Using the half-angle formula:  sin( a/2 )  =  +/- sqrt[ (1 - cos(a) / 2]   with  a  =  -pi/6

[we will need to choose the - sign in the formula]

 

sin( a/2 )  =  - sqrt[ (1 - cos(a) / 2]     --->     sin[ ( -pi/6 / 2 ]  =  - sqrt[ (1 - cos( -pi/6 ) / 2] 

    =  - sqrt[ (1 - sqrt(3)/2 ) / 2 ]  =  - sqrt[ (2 - sqrt(3) ) / 4 ]  =  - ½·sqrt( 2 - sqrt(3) )

geno3141 May 2, 2020

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