What is #(11pi)/6# in degrees and hence find #cos((11pi)/6)#?

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Answer 1 · 2017-03-28T14:40:32

#cos((11pi)/6)=sqrt3/2#

As #pi# radians are equal to #180^@#

#1# radians are equal to #180^@/pi#

#(11pi)/6# radians are #(cancel(180)30^@)/cancelpixx(11cancelpi)/(cancel6^1)#

= #30^@xx11=330^@#

and #cos((11pi)/6)=cos330^@#

= #cos(360^@-30^@)#

= #cos(-30^@)#

= #cos30^@#

= #sqrt3/2#