Question

find
the value
of
Cos 6° Cos 42° Cos 66° Cos 78°​

Answer 1

Answer:

(cos 6°)(cos 42°)(cos 66°)(cos 78°)

=(cos6°cos66°)(cos42°cos78°)

{multiplying and dividing by 2*2}

= (1/4)[2cos6°cos66°] [2cos42°cos78°]

{2cosAcosB= cos(A+B)+cos(A-B)}

= (1/4)(cos72° + cos60°)(cos120° + cos36°)

= (1/4)(cos72° + 1/2)(- 1/2 + cos36°)

= (1/4) [- 1/4 + (1/2) (cos36° - cos72°) + cos36° cos72°]

= (1/4) [- 1/4 + sin54°sin18° + sin54° sin18°]

{multiplying and dividing by cos18°}

= (1/4) [- 1/4 + 2sin54°sin18°cos18° / cos18°]

{2sinAcosA=sin2A}

{multiplying and dividing by 2}

= (1/4) [- 1/4 + 2sin54°sin36° / 2cos18°]

{2sinAsinB=cos(A-B)-cos(A+B)}

= (1/4) [- 1/4 + (cos18° - cos90°) / 2cos18°]

= (1/4) [- 1/4 + 1/2]

= 1/16