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How do you evaluate #Sin(π/3)*tan(π/6)#?

1 Answer

Nov 27, 2016

#1/2#

Explanation:

#sin(pi/3)xxtan(pi/6)#
#" "#
#=sqrt3/2xxsqrt3/3#
#" "#
#=sqrt(3^2)/(2xx3)#
#" "#
#=3/(2xx3)#
#" "#
#=1/2#

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