compute exactly a sinpi12 b sin1125 degrees thanksSolutiona

compute exactly a. sin(-pi/12) b. sin(-112.5 degrees) thanks

a) sin(-pi/12) = -sin(pi/12) -----( sin(-x) = -siinx)

Now sin(pi/12) = sin(pi/3 -pi/4)

use the sin(a-b) formula : sinacosb -cosasinb

So,sin(pi/3 -pi/4) = sinpi/3cospi/4 -cospi/3sinpi/4

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

= (sqrt3 -1)/sqrt2

sin(-pi/12) = -sin(pi/12) = ( 1-sqrt3)/sqrt2

b) sin(-112.5) = -sin(112.5)

sin(112.5) = sin(90+22.5) =cos22.5 ( by trig property)

find cos22.5 :

cos 45/2 = sqrt[ (1+cos 45)/2]
= sqrt [ (1 +1/sqrt2)/2 }

= sqrt [ (1 + sqrt 2) /2 sqrt 2 ]

= sqrt [ (2 + sqrt 2) /4 ]

= (1/2)sqrt (2+sqrt 2)