Question #d6ecb

I have edited the Problem so as to get the desired result.

However, I request the Questioners to be very careful while posing the Problem and avoid typographical errors.

`cosx=sqrt{(m^2-1)/3} :. {(1-tan^2(x/2))/(1+tan^2(x/2))}=sqrt{(m^2-1)/3}.`

But `tan^3(x/2)=tana=sina/cosa rArr tan(x/2)=sin^(1/3)a/cos^(1/3)a.`

Therefore, `{(1-sin^(2/3)a/cos^(2/3)a)/(1+sin^(2/3)a/cos^(2/3)a)}=sqrt{(m^2-1)/3}.`

`:. (cos^(2/3)a+sin^(2/3)a)/(cos^(2/3)a-sin^(2/3)a)=sqrt3/sqrt(m^2-1).`

By Componendo-Dividendo, `cos^(2/3)a/sin^(2/3)a=(sqrt3+sqrt(m^2-1))/(sqrt3-sqrt(m^2-1)), or,`

`cos^(2/3)a/(sqrt3+sqrt(m^2-1))=sin^(2/3)a/(sqrt3-sqrt(m^2-1))=k, say.`

`:. cos^(2/3)a=k(sqrt3+sqrt(m^2-1)), sin^(2/3)a=k(sqrt3-sqrt(m^2-1)).`

`rArr 1=cos^2a+sin^2a=(cos^(2/3)a)^3+(sin^(2/3)a)^3`

`=k^3{(sqrt3+sqrt(m^2-1))^3+(sqrt3-sqrt(m^2-1))^3}`

Here, knowing that, `(l+n)^3+(l-n)^3=2l(l^2+3n^2),` we find, that,

`1=k^3[(2sqrt3){(sqrt3)^2+3(sqrt(m^2-1))^2}]=k^3(6sqrt3m^2),`

giving, `k^3=1/(6sqrt3m^2)...................(star).`

Finally, then, `cos^(2/3)a+sin^(2/3)a`

`=k{(sqrt3+sqrt(m^2-1))+(sqrt3-sqrt(m^2-1))}=(2sqrt3)k.`

`rArr {cos^(2/3)a+sin^(2/3)a}^3=(2sqrt3)^3k^3=(8*3sqrt3)/(6sqrt3m^2).......[because, (star)]`

`, i.e, {cos^(2/3)a+sin^(2/3)a}^3=4/m^2=(2/m)^2.`

Accordingly, `cos^(2/3)a+sin^(2/3)a=(2/m)^(2/3).`

Enjoy Maths.!

Making `cosx=1-2sin(x/2)` and

considering the system of equations

`{(b=1-2sin^2(x/2)),(((sin ^2a)/(cos^2a))^(1/3) = sin^2(x/2)/(cos^2(x/2))),(sin^2(x/2)+cos^2(x/2)=1),(sin^2a+cos^2a=1):}`

and making

`u=sin^2(x/2), v = cos^2(x/2), p = sin^2a, q = cos^2a`

we have the system of four equations to four unknowns.

`{(b = 1 - 2 u), (p/q=(u/v)^3), (u+v=1), (p+q=1):}`

Solving for `u,v,p,q` we have

`{(u=(1-b)/2),(v=(1+b)/2),(p=(1-b)^3/(2+6b^2)),(q=(1+b)^3/(2+6b^2)):}`

so we have

`p^(1/3)+q^(1/3) = 2/root(3)(2+6b^2)`

making now

`2/root(3)(2+6b^2) = (2/m)^(2/3)` and solving for `b` we get

`b = sqrt((m^2-1)/3)`

and then

`p^(1/3)+q^(1/3) = (2/m)^(2/3)`