Question

Obtain tan 5𝜽 in terms of tan 𝜽 & show that `1-10tan^2  x/10+5tan^4  x/10=0`

Answer 1

we have tan `50=(sin 50)/cos 50` 

`(cos θ+i sinθ)^n=cos nθ=cos n θ+i sin nθ`

put n=5, 

∴ `cos 5θ+i sin 5θ=(cos θ+i sin θ)^5`

=`cos^5 θ+5cos^4 θ.i sin θ+10 cos^3 θ. (isin θ)^2`

`+10cos^2 θ.(isin θ)^3+5 cosθ.(isinθ)^4 + isin^5 θ`

=`[cos^5θ-10 cos^3θ.(sinθ)^2+5cosθ.(sinθ)^4]+[cos^4θ.isinθ-10 icons^2θ.(sinθ)^3+isin^5θ]`

Compare real and imaginary parts 

`cos5θ=[cos^5θ-10cos^3θ.(sinθ)^2+5cosθ.(sinθ)^4]`

`sin 5θ=+[5cos^4θ.sinθ-10cos^2θ.(sinθ)^3+sin^5θ]`

tan 5θ=`[[5cos^4θ.sinθ-10cos^2θ.(sinθ)^3+sin^5θ]]/[[cos^5θ-10cos^3θ.(sinθ)^2+5cosθ.(sinθ)^4]]` 

`tan 5θ=(5tanθ-10tan^3θ+tan^5θ)/(1-10tan^2θ+5tan^4θ)`

`"put"  θ=pi/10` 

`1-10tan^2  x/10+5tan^4  x/10=0`