How do you evaluate #\cos \frac { 8\pi } { 15} \cos \frac { 3\pi } { 10} - \sin \frac { 8\pi } { 15} \sin \frac { 3\pi } { 10}#?

Question

724 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-21T14:45:13

#cos((8pi)/15)cos((3pi)/10)-sin((8pi)/15)sin((3pi)/10)=-sqrt3/2#

We may use the identity #cosAcosB-sinAsinB=cos(A+B)#

Hence #cos((8pi)/15)cos((3pi)/10)-sin((8pi)/15)sin((3pi)/10)#

= #cos((8pi)/15+(3pi)/10)#

= #cos((16pi+9pi)/30)#

= #cos((25pi)/30)#

= #cos((5pi)/6)#

= #cos(pi-pi/6)#

= #-cos(pi/6)#

= #-sqrt3/2#