+0

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

0
50
1
+59

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

May 2, 2020

#1
+20906
+1

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
+20906
+1

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