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

2 Answers
Ratnaker Mehta
Aug 24, 2016

#1/2(2-sqrt3)#

Explanation:

Let us note that, because, #cos(-x)=cosx......(star)#, the given expression

#=sin^2(pi/6)-2sin(pi/6)cos(pi/6)+cos^2(pi/6)#

#=(sin(pi/6)-cos(pi/6))^2#

#=(1/2-sqrt3/2)^2={(1-sqrt3)/2}^2#

#=1/4(1-2sqrt3+3)#

#=1/4(4-2sqrt3)#

#=1/2(2-sqrt3)#

Alternatively, we can directly plug in the respective values, keeping

#(star)# in mind, to get,

The Reqd. Value#=(1/2)^2-2(1/2)(sqrt3/2)+(sqrt3/2)^2#

#=1/4-sqrt3/2+3/4#

#=1-sqrt3/2=1/2(2-sqrt3)#, as before!

Enjoy Maths.!

Nghi N.
Aug 24, 2016

#1 - sqrt3/2#

Explanation:

Since
#sin^2 (pi/6) + cos^2 (- pi/6) = 1#
#2sin (pi/6).cos (pi/6) = sin (pi/3)# (identity: sin 2a = 2sin a.cos a)
There for, the given expression is equal to:
#S = 1 - sin (pi/3) = 1 - sqrt3/2#

Related questions
See all questions in Trigonometric Functions of Any Angle
Impact of this question
4938 views around the world
You can reuse this answer
Creative Commons License