How do you express #cos( (15 pi)/ 8 ) * cos (( 5 pi) /12 ) # without using products of trigonometric functions?

Question

1318 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-23T17:24:24

The answer is #=1/2(cos(55/24pi)+cos(35/24pi))#

#cos(A+B)=cosAcosB-sinAsinB#

#cos(A-B)=cosAcosB+sinAsinB#

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

Therefore,

#cosAcosB=1/2(cos(A+B)+cos(A-B))#

Here,

#A=15/8pi# and #B=5/12pi#

So,

#cos(15/8pi)cos(5/12pi)=1/2(cos(15/8pi+5/12pi))+cos(15/8pi-5/12pi))#

#=1/2(cos(55/24pi)+cos(35/24pi))#