How do you evaluate #sin(pi/5)#?

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Answer 1 · 2016-07-02T04:10:42

#sin(pi/5)=sqrt(10-2sqrt5)/4#

Let #theta=pi/5#, then #5theta=pi#

and #3theta=pi-2theta#. Note #theta) is an acute angle.

Hence #sin3theta=sin(pi-2theta)# but as #sin(pi-A)=sinA#

This can be written as

#sin3theta=sin2theta# expanding them

or #3sintheta-4sin^3theta=2sinthetacostheta#

as #theta=pi/5# we have #sintheta!=0# and dividing by it we get

#3-4sin^2theta=2costheta# or

#3-4(1-cos^2theta)=2costheta#

or #4cos^2theta-2costheta-1=0#

and using quadratic formula #costheta=(2+-sqrt(2^2-4*4*(-1)))/(2*4)#

= #(2+-sqrt(20))/8=(1+-sqrt5)/4#.

But as #(1-sqrt5)/4# is negative and #costheta# cannot take this value, hence

#costheta=(1+sqrt5)/4# and

#sintheta=sqrt(1-((1+sqrt5)/4)^2)=sqrt(1-((1+5+2sqrt5)/16))#

= #sqrt((16-6-2sqrt5)/16)=sqrt(10-2sqrt5)/4#

As #theta=pi/5#

#sin(pi/5)=sqrt(10-2sqrt5)/4#