proof

cos(3π/14)sin(π/7)cos(π/14) + sin(π/14)cos(π/7)cos(3π/14) + sin(3π/14)sin(π/14)sin(π/7)

(knowing that sin (3π/14) is on the right so combine the terms not having sin(3π/14)

= cos(3π/14)[sin(π/7)cos(π/14) + sin(π/14)cos(π/7)] + sin(3π/14)sin(π/14)sin(π/7)

(using sin A cos B + cos A sin B = sn (A+B) in next line

= cos(3π/14)[sin(π/7+π/14] + sin(3π/14)sin(π/14)sin(π/7)

= cos(3π/14)sin(3π/14) + sin(3π/14)sin(π/14)sin(π/7)

= sin(3π/14)[ cos(3π/14) + sin(π/14)sin(π/7)]

= sin(3π/14)[ cos(π/7+π/14)+ sin(π/14)sin(π/7)] (as cos (A+B) shall cancel sin A sin B by contributing –ve of it)

= sin(3π/14)[ cos(π/7) cos (π/14) - sin(π/14)sin(π/7+ sin(π/14)sin(π/7)]

= cos(π/7)sin(3π/14)cos(π/14)

Fun with maths