√2+2cosθ = √4cos2θ/2 = 2cosθ/2
√2+2cosθ/2 = 2cosθ/22 and so on
for nth square root, we get the final term as 2cos(θ/2n).
correct me if i am wrong !
easy.....
if 0<θ<Ï€ then√2+√2+√2+.......√2+2cosθ is equal to??????????
if there are n number of √ signs
√2+2cosθ = √4cos2θ/2 = 2cosθ/2
√2+2cosθ/2 = 2cosθ/22 and so on
for nth square root, we get the final term as 2cos(θ/2n).
correct me if i am wrong !