Science

Math

Humanities

... and beyond

Socratic Meta
Featured Answers

Topics

How do you find the value of #sin ((2pi)/3) Cos pi Tan ((2pi)/3)Cot ((5pi)/6)#?

1 Answer

Apr 27, 2016

#=>sin ((2pi)/3) Cos pi Tan ((2pi)/3)Cot ((5pi)/6) #
#=1/2 * -1 * -1/sqrt3*-1/sqrt3 =-1/6 #

Explanation:

#cos pi = cos180 = -1#

#sin((2pi)/3)= sin 120 = sin (90 +30) = sin 30 = 1/2#

#Tan ((2pi)/3) = tan 120 = -tan 30 = -1/sqrt3#

#Cot ((5pi)/6) = cot 150 = - cot 60 = -tan(90-60)#
#= -tan30 =- 1/sqrt3#

#=>sin ((2pi)/3) Cos pi Tan ((2pi)/3)Cot ((5pi)/6) #
#=1/2 * -1 * -1/sqrt3*-1/sqrt3 =-1/6 #

Related questions

See all questions in Solving Trigonometric Equations

Impact of this question

1430 views around the world

You can reuse this answer
Creative Commons License