How do you express sin(pi/12) * cos(( 5 pi)/8 ) sin(π12)cos(5π8) without products of trigonometric functions?

1 Answer
Feb 27, 2016

sin(pi/12)*cos((5pi)/8)=-(sqrt((1-sqrt3/2)(1-sqrt2/2)))/2sin(π12)cos(5π8)=(132)(122)2

Explanation:

[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)=±1cos(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)=+ 1cos(π6)2

[3]" "=sqrt((1-cos(pi/6))/2)*cos((5pi)/8)[3] = 1cos(π6)2cos(5π8)

Evaluate cos(pi/6)cos(π6)

[4]" "=sqrt((1-sqrt3/2)/2)*cos((5pi)/8)[4] =1322cos(5π8)

Think of (5pi)/85π8 as ((5pi)/4)/25π42.

[5]" "=sqrt((1-sqrt3/2)/2)*cos(((5pi)/4)/2)[5] =1322cos(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] =1322⎜ ⎜ ⎜ 1+cos(5π4)2⎟ ⎟ ⎟

Evaluate cos((5pi)/4)cos(5π4)

[6]" "=sqrt((1-sqrt3/2)/2)*(-sqrt((1+(-sqrt2/2))/2))[6] =1322⎜ ⎜ ⎜ 1+(22)2⎟ ⎟ ⎟

Simplify.

[7]" "=color(blue)(-(sqrt((1-sqrt3/2)(1-sqrt2/2)))/2)[7] =(132)(122)2

I think you can leave it at that.