How do you write the expression #cos(pi/7)cos(pi/5)-sin(pi/7)sin(pi/5)# as the sine, cosine, or tangent of an angle?

Question

9021 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-05T20:22:06

The answer is #=cos(pi/35)#

We use

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

Let #A=pi/7#

and #B=pi/5#

#cos(pi/7)cos(pi/5)-sin(pi/7)sin(pi/5)=cos(pi/7+pi/5)#

#=cos((5pi)/35+(7pi)/35)#

#=cos(pi/35)#