How do you express sin(pi/ 4 ) * sin( ( 15 pi) / 8 ) sin(π4)sin(15π8) without using products of trigonometric functions?

1 Answer
Mar 10, 2016

sin((5pi)/4)sin((15pi)/8)=1/2(cos((5pi)/8)+cos(pi/8))sin(5π4)sin(15π8)=12(cos(5π8)+cos(π8))

Explanation:

For this one uses the identity

cos(A+B)=cosAcosB-sinAsinBcos(A+B)=cosAcosBsinAsinB and
cos(A-B)=cosAcosB+sinAsinBcos(AB)=cosAcosB+sinAsinB.

Subtracting upper identity from lower, one gets

cos(A-B)-cos(A+B)=2sinAsinBcos(AB)cos(A+B)=2sinAsinB or

sinAsinB=1/2(cos(A-B)-cos(A+B))sinAsinB=12(cos(AB)cos(A+B))

Hence sin((5pi)/4)sin((15pi)/8)sin(5π4)sin(15π8)

= 1/2(cos((15pi)/8-(5pi)/4)-cos((15pi)/8+(5pi)/4))12(cos(15π85π4)cos(15π8+5π4))

= 1/2(cos((5pi)/8)-cos((25pi)/8))12(cos(5π8)cos(25π8))

= 1/2(cos((5pi)/8)-cos(3pi+pi/8))12(cos(5π8)cos(3π+π8))

= 1/2(cos((5pi)/8)+cos(pi/8))12(cos(5π8)+cos(π8))