Question

Prove that : sin pi/10 + sin 3 pi/10=1/4

Answer 1

LHS of the given equation is:
sinπ10.sin13π10=sinπ10.sin(π+3π10)=sinπ10.[−sin3π10]=−12.[2sinπ10.sin3π10]=−12.[cos2π10−cos4π10]   [since 2sinAsinB=cos(A−B)−cos(A+B)=−12.[cosπ5−cos2π5]=−12.[cosπ5−2cos2π5+1]....................(1) [since cos2θ=2cos2θ−1
now we will try to find the value of cosπ5.
let θ=π5⇒5θ=π
sin(5θ)=sinπsin(3θ+2θ)=0⇒sin3θ.cos2θ+cos3θ.sin2θ=0 [since sin(A+B)=sinAcosB+cosAsinB⇒(3sinθ−4sin3θ)(2cos2θ−1)+(4cos3θ−3cosθ).(2sinθcosθ)=0sinθ.[(3−4sin2θ)(2cos2θ−1)+2cos2θ.(4cos2θ−3)]=0since sinθ≠0⇒(3−4sin2θ)(2cos2θ−1)+2cos2θ.(4cos2θ−3)=0(3−4{1−cos2θ))(2cos2θ−1)+2cos2θ.(4cos2θ−3)=0(−1+4cos2θ)(2cos2θ−1)+8cos4θ−6cos2θ=0⇒−2cos2θ+1+8cos4θ−4cos2θ+8cos4θ−6cos2θ=0
⇒16cos4θ−12cos2θ+1=0⇒cos2θ=12±144−64√32=12±80√32cos2θ=12±45√32=6±25√16cos2θ=(1±5√)242cosθ=±1±5√4 since θ=π5⇒cosθ>0⇒cosθ=1+5√4
⇒cosπ5=1+5√4..............(2)
from eq(1) and eq(2):
LHS  of the given equation is:
−12.[cosπ5−2cos2π5+1]=−12.[1+5√/4−2.(1+5/√4)2+1]=−12.[14+5√4−2.(1+5+25/√16)+1]=−12.[1/4+5/√4−6+2√5/8+1]=−1/2.[14+5√4−34−5√4+1]=−1/2.[−1/2+1]=−1/2.1/2=−1/4
= RHS

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