Question

`"If" sin^4θcos^3θ = acosθ + bcos3θ + ccos5θ + dcos7θ "then find"  a,b,c,d.` 

Answer 1

We know,` sinθ=(e^iθ-e^-iθ)/(2!)` and cosθ=`( e^iθ+e^-iθ)/2`                                                                …(1) 

Consider, `sin^4θ cos^3θ=(e^(iθ)-e^-iθ)/2i xx(eiθ+e^-10)/2`                            (From 1) 

= `1(/2^4i^4 2^3)xx(e^(iθ)-e^-(iθ)) (e^(iθ)-e^-(iθ) ^3(e^(iθ)+e^-(iθ))^3` 

`1/2^7 xx (e^(iθ)-e^(-iθ)[(e^(iθ)^2 -(e^-(iθ))^2]^3 `

`1/2^7 xx (e^(iθ)-e^(-iθ))[(e^(2iθ))-(e^-( 2iθ ))]^3`

`1/2^7xx (e^(iθ)-e^(-iθ)) [(e^(2iθ))^3-3(e^(2iθ))^2(e^(-2iθ))+3(e^(2iθ))^2(e^(-2iθ))-(e^(2iθ))^3 ]`

`1/2^7xx(e^(iθ)-e^(-iθ)) [3e^(6iθ)-3e^(2iθ)+3 e^-(2iθ)-e^-(6iθ)]` 

`1/2^7xx[e^(7iθ)-3e^(3iθ)+3e^(iθ)-e^(-5iθ)-e^(5iθ)+3e(iθ)-3e^(-3iθ)+e^(-7iθ)]` 

`1/2^7[(e^(7iθ)+e^(-7iθ)-(e^(5iθ+e^(-5iθ)))-3(e^(3iθ)+e^(-3iθ))+(e^(iθ)+e^(-iθ))]`

=`1/128xx2cos7θ-1/128xx2cos5iθ-1/128xx6cos3iθ+1/128xx6cosiθ`

∴ `sin^4θcos^3θ=3/64 cosθ-3/64 cos3θ-1/64 cos5θ+1/64 cos 7θ `…(2)

But , given, `sin^4θcos3^θ = acosθ + bcos3θ + ccos5θ + dcos7θ`                            …(3) 

Comparing (2) & (3), `a=3/64; b=(-3)/64 ; c=(-1)/64; d=1/64;`