Question

Cos36×cos72×cos108×cos144=1÷16

Answer 1

Value of cos36 = [(√3+1)/2√2]
Value of sin18 = [(√3-1)/2√2]
Now,
=cos36×cos72×cos108×cos144

=cos36×sin(90-18)×cos(90+18)×cos(180-36)

=cos36×sin18×(-sin18)×(-cos36)

=cos36sin18sin18cos36

=[(√3+1)/2√2][(√3-1)/2√2}[(√3-1)/2√2][(√3+1)/2√2]

=2/8×2/8

=1/16