Value of cos(-3888)?

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Answer 1 · 2018-07-18T03:57:11

# (sqrt5-1)/4#.

We have, #(1):cos(-theta)=costheta,#

#(2): cos(360^@k+theta)=costheta, k in ZZ,#,

#(3):cos(270^@+theta)=sintheta#.

Therefore, #cos(-3888^@)=cos(3888^@)#,

#=cos((3600^@+(270^@+18^@))#,

#=cos(270^@+18^@)#,

#=sin18^@#.

# rArr cos(-3888^@)=(sqrt5-1)/4#.