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.