# Sin^2(π/12)+sin^2(3π/12)+sin^2(5π/12)+sin^2(7π/12)+sin^2(9π/12)+sin^2(11π/12)=? #

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Answer 1 · 2018-05-15T11:31:02

#3#

We know that,

#sin(pi/2+theta)=costheta=>sin^2(pi/2+theta)=cos^2theta#

So we take,

#color(red)(sin^2((11pi)/12)=sin^2(pi/2+(5pi)/12)=cos^2((5pi)/12)...to(1)#

#color(blue)(sin^2((9pi)/12)=sin^2(pi/2+(3pi)/12)=cos^2((3pi)/12)...to(2)#

#color(violet)(sin^2((7pi)/12)=sin^2(pi/2+(pi)/12)=cos^2((pi)/12)...to(3)#

Given that,

#X=sin^2((pi)/12)+sin^2((3pi)/12)+sin^2((5pi)/12)#

#color(white)(..........)+sin^2((7pi)/12)+sin^2((9pi)/12)+sin^2((11pi)/12)#

Using #(1),(2), and (3)# ,we get

#X=sin^2((pi)/12)+sin^2((3pi)/12)+sin^2((5pi)/12)#

#color(white)(..........)+color(violet)(cos^2((pi)/12))+color(blue)(cos^2((3pi)/12))+color(red)(cos^2((5pi)/12)#

#X={sin^2((pi)/12)+cos^2((pi)/12)}+{sin^2((3pi)/12)+cos^2((3pi)/12)}#

#color(white)(..........)+{sin^2((5pi)/12)+cos^2((5pi)/12)}#

#=>X={1}+{1}+{1}...to [as, sin^2theta+cos^2theta=1 ]#

#=>X=3#