How do you express #sin(pi/12) * cos(( 5 pi)/8 ) # without products of trigonometric functions?

1 Answer
Feb 27, 2016

#sin(pi/12)*cos((5pi)/8)=-(sqrt((1-sqrt3/2)(1-sqrt2/2)))/2#

Explanation:

#[1]" "sin(pi/12)*cos((5pi)/8)#

Think of #pi/12# as #(pi/6)/2#.

#[2]" "=sin((pi/6)/2)*cos((5pi)/8)#

Half angle identity: #sin(x/2)=+-sqrt((1-cos(x))/2)#
**#pi/12# is in the first quadrant, and sine is positive in the first quadrant. Therefore, we know that #sin((pi/6)/2)=+sqrt((1-cos(pi/6))/2)#

#[3]" "=sqrt((1-cos(pi/6))/2)*cos((5pi)/8)#

Evaluate #cos(pi/6)#

#[4]" "=sqrt((1-sqrt3/2)/2)*cos((5pi)/8)#

Think of #(5pi)/8# as #((5pi)/4)/2#.

#[5]" "=sqrt((1-sqrt3/2)/2)*cos(((5pi)/4)/2)#

Half angle identity: #cos(x/2)=+-sqrt((1+cos(x))/2)#
**#(5pi)/8# is in the second quadrant, and cosine is negative in the second quadrant. Therefore, we know that #cos(((5pi)/4)/2)=-sqrt((1+cos((5pi)/4))/2)#

#[5]" "=sqrt((1-sqrt3/2)/2)*(-sqrt((1+cos((5pi)/4))/2))#

Evaluate #cos((5pi)/4)#

#[6]" "=sqrt((1-sqrt3/2)/2)*(-sqrt((1+(-sqrt2/2))/2))#

Simplify.

#[7]" "=color(blue)(-(sqrt((1-sqrt3/2)(1-sqrt2/2)))/2)#

I think you can leave it at that.