How do you express cos( (15 pi)/ 8 ) * cos (( 5 pi) /8 ) cos(15π8)cos(5π8) without using products of trigonometric functions?

1 Answer
Apr 4, 2016

cos((15pi)/8)cos ((5pi)/8)=1/2 cos((5pi)/2)+1/2 cos((5pi)/4)=-sqrt2/2cos(15π8)cos(5π8)=12cos(5π2)+12cos(5π4)=22

Explanation:

2cos A cos B=cos(A+B)+cos(A-B)2cosAcosB=cos(A+B)+cos(AB)

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

A=(15pi)/8, B=(5pi)/8A=15π8,B=5π8

=>cos((15pi)/8)cos ((5pi)/8)=1/2 (cos((15pi)/8+(5pi)/8)+cos((15pi)/8-(5pi)/8))cos(15π8)cos(5π8)=12(cos(15π8+5π8)+cos(15π85π8))

=1/2 (cos((20pi)/8)+cos((10pi)/8)) =12(cos(20π8)+cos(10π8))

=1/2 cos((5pi)/2)+1/2 cos((5pi)/4) =0+ -sqrt2/2=-sqrt2/2 =12cos(5π2)+12cos(5π4)=0+22=22

cos((15pi)/8)cos ((5pi)/8)=1/2 cos((5pi)/2)+1/2 cos((5pi)/4)=-sqrt2/2cos(15π8)cos(5π8)=12cos(5π2)+12cos(5π4)=22