Need Help With a Maths Question

Given that x is sufficiently small for x³ and higher powers of x to be neglected, show that

[(8+x)^(1/3)]/cos2x = 2 + x/12 + kx²

Where k is a fraction to be determined.

I obtained a weird answer for k in the end. Can you please help me? Thanks!

Student X

[(8+x)^(1/3)]/cos2x =[(8+x)^(1/3)]/[1-2sin²(x)]

When x is small, the above approximates to

[(8+x)^(1/3)]/[1-2x²]=[(8+x)^(1/3)]*[(1-2x²)^(-1)] -----------------(1)

Considering the expansions of (8+x)^(1/3) and (1-2x²)^(-1) up to and including x² separately,

(8+x)^(1/3)= 2(1+x/8 )^(1/3)

= 2[1+x/24+(1/3)*(-2/3)/2!*(x/8)²]

= 2[1+x/24 - x^2/576]

= 2+ x/12 - x²/ 288

(1-2x^2)^(-1) = 1+2x²

Substituting both these expansion back into (1):

[(8+x)^(1/3)]*[(1-x^2)^(-1)]=( 2+ x/12 - x^2/ 288) (1+2x²)

= 2 + x/12 - x²/ 288 + 4x² (ignoring all x³ and higher order terms)

= 2 + x/12 -1151/288 x²

By comparison k= -1151/288 (shown)

Indeed k isn’t a neat value. Hope this helps. Peace.

Best Regards,

Mr Koh