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How do you evaluate #cos^2(pi/3)-cot(pi/4)+sin(pi/6 )#?

1 Answer

May 7, 2016

#=1/4#

Explanation:

To evaluate #cos^2(pi/3)-cot(pi/4)+sin(pi/6 )#

As we know from table #cos(pi/3)=sqrt3/2,cot(pi/4)=1 &sin(pi/6 )=1/2#

#cos^2(pi/3)-cot(pi/4)+sin(pi/6 )#
#=>(sqrt3/2)^2-1+1/2#
#=>3/4-1+1/2=(3-4+2)/4=1/4#

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