Como você encontra o valor exato de #sin (pi / 12) #?

We want to find replacement angles for # pi/12" that will produce exact values " #

These must come from : # pi/6 , pi/3 , pi/4 #

# rArr sin(pi/12) = sin(pi/3 - pi/4 ) #

Using the appropriate #color(blue)" Addition formula " #

#color(red)(|bar(ul(color(white)(a/a)color(black)( sin(A ± B) = sinAcosB ± cosAsinB )color(white)(a/a)|)))#

#rArr sin(pi/3 - pi/4) = sin(pi/3)cos(pi/4) - cos(pi/3)sin(pi/4) #

Extract #color(blue)" exact values from triangles " #
#sin(pi/3) = (sqrt3)/2 , sin(pi/4) = 1/(sqrt2)#
and # cos(pi/3) = 1/2 , cos(pi/4) = 1/(sqrt2)#
now substitute into the right side of the expansion.

# = (sqrt3)/2xx1/(sqrt2) - 1/2xx1/(sqrt2) = (sqrt3)/(2sqrt2)-1/(2sqrt2)#

# = (sqrt3 - 1)/(2sqrt2) " and rationalising the denominator "#

gives #((sqrt3 - 1)xxsqrt2)/(2sqrt2xxsqrt2)= (sqrt6 - sqrt2)/4#