[1]" "sin(pi/12)*cos((5pi)/8)[1] sin(π12)⋅cos(5π8)
Think of pi/12π12 as (pi/6)/2π62.
[2]" "=sin((pi/6)/2)*cos((5pi)/8)[2] =sin(π62)⋅cos(5π8)
Half angle identity: sin(x/2)=+-sqrt((1-cos(x))/2)sin(x2)=±√1−cos(x)2
**pi/12π12 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)sin(π62)=+
⎷1−cos(π6)2
[3]" "=sqrt((1-cos(pi/6))/2)*cos((5pi)/8)[3] =
⎷1−cos(π6)2⋅cos(5π8)
Evaluate cos(pi/6)cos(π6)
[4]" "=sqrt((1-sqrt3/2)/2)*cos((5pi)/8)[4] =√1−√322⋅cos(5π8)
Think of (5pi)/85π8 as ((5pi)/4)/25π42.
[5]" "=sqrt((1-sqrt3/2)/2)*cos(((5pi)/4)/2)[5] =√1−√322⋅cos(5π42)
Half angle identity: cos(x/2)=+-sqrt((1+cos(x))/2)cos(x2)=±√1+cos(x)2
**(5pi)/85π8 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)cos(5π42)=−
⎷1+cos(5π4)2
[5]" "=sqrt((1-sqrt3/2)/2)*(-sqrt((1+cos((5pi)/4))/2))[5] =√1−√322⋅⎛⎜
⎜
⎜⎝−
⎷1+cos(5π4)2⎞⎟
⎟
⎟⎠
Evaluate cos((5pi)/4)cos(5π4)
[6]" "=sqrt((1-sqrt3/2)/2)*(-sqrt((1+(-sqrt2/2))/2))[6] =√1−√322⋅⎛⎜
⎜
⎜⎝−
⎷1+(−√22)2⎞⎟
⎟
⎟⎠
Simplify.
[7]" "=color(blue)(-(sqrt((1-sqrt3/2)(1-sqrt2/2)))/2)[7] =−√(1−√32)(1−√22)2
I think you can leave it at that.