Skip to main content

Euler's theorem for cosec⁻¹ function

Short answer

Let z = cosec u; it is homogeneous of degree 1/6. By Euler’s theorem x z_x + y z_y = (1/6)z. Substituting z=cosec u yields −cosec u cot u (x u_x + y u_y) = (1/6) cosec u, so x u_x + y u_y = −(1/6) tan u.

If u = cosec⁻¹((x^(1/2)+y^(1/2))/(x^(1/3)+y^(1/3))), show that x ∂u/∂x + y ∂u/∂y = −(1/6) tan u.

Euler's theorem

If f(x, y) is homogeneous of degree n, then x ∂f/∂x + y ∂f/∂y = n f.

Proof

Given u = cosec⁻¹((x^(1/2) + y^(1/2))/(x^(1/3) + y^(1/3))). Let

z = cosec u = (x^(1/2) + y^(1/2))/(x^(1/3) + y^(1/3)).

The numerator has degree 1/2 and the denominator has degree 1/3: replacing x, y by tx, ty gives z(tx, ty) = t^(1/2 − 1/3) z = t^(1/6) z. So z is homogeneous of degree n = 1/6.

By Euler's theorem:

x ∂z/∂x + y ∂z/∂y = (1/6) z

With z = cosec u: ∂z/∂x = −cosec u cot u ∂u/∂x (and similarly for y). Hence

−cosec u cot u ( x ∂u/∂x + y ∂u/∂y ) = (1/6) cosec u
x ∂u/∂x + y ∂u/∂y = −(1/6) / cot u
                   = −(1/6) tan u

Hence proved. ∎

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Mathematics chapter 1 questions →