Question

Prove that from the equality $\frac{sin^4\alpha}{a}+\frac{cos^4\alpha}{b}=\frac{1}{a+b}$ follows the relation; $\frac{sin^8\alpha}{a^3}+\frac{cos^8\alpha}{b^3}=\frac{1}{(a+b)^3}$.

Answer 1

Proven

Answer 2

To prove the given relation, we start with the initial equality and deduce the values of $\sin^2\alpha$ and $\cos^2\alpha$.

Given equality:

$$\frac{\sin^4\alpha}{a}+\frac{\cos^4\alpha}{b}=\frac{1}{a+b}$$

We know that $\sin^2\alpha + \cos^2\alpha = 1$. Let $x = \sin^2\alpha$ and $y = \cos^2\alpha$. Then $x+y=1$. The given equality can be rewritten as:

$$\frac{x^2}{a} + \frac{y^2}{b} = \frac{(x+y)^2}{a+b}$$

This is a specific case of the identity: If $\frac{P^2}{A} + \frac{Q^2}{B} = \frac{(P+Q)^2}{A+B}$, then it implies $\frac{P}{A} = \frac{Q}{B}$. This identity can be proven as follows: Starting from $\frac{P^2}{A} + \frac{Q^2}{B} = \frac{(P+Q)^2}{A+B}$: Multiply by $AB(A+B)$: $B(A+B)P^2 + A(A+B)Q^2 = AB(P+Q)^2$ $(AB+B^2)P^2 + (A^2+AB)Q^2 = AB(P^2+2PQ+Q^2)$ $AB P^2 + B^2 P^2 + A^2 Q^2 + AB Q^2 = AB P^2 + 2AB PQ + AB Q^2$ Subtract $AB P^2 + AB Q^2$ from both sides: $B^2 P^2 + A^2 Q^2 = 2AB PQ$ Rearrange the terms: $B^2 P^2 - 2AB PQ + A^2 Q^2 = 0$ This is a perfect square: $(BP - AQ)^2 = 0$ This implies $BP - AQ = 0$, so $BP = AQ$. Dividing by $AB$ (assuming $A, B \neq 0$):

$$\frac{P}{A} = \frac{Q}{B}$$

Applying this identity to our problem with $P=\sin^2\alpha$, $Q=\cos^2\alpha$, $A=a$, $B=b$: From $\frac{(\sin^2\alpha)^2}{a} + \frac{(\cos^2\alpha)^2}{b} = \frac{(\sin^2\alpha+\cos^2\alpha)^2}{a+b}$, it follows that:

$$\frac{\sin^2\alpha}{a} = \frac{\cos^2\alpha}{b}$$

Let this common ratio be $k$. So, $\sin^2\alpha = ak$ and $\cos^2\alpha = bk$. Using the identity $\sin^2\alpha + \cos^2\alpha = 1$: $ak + bk = 1$ $k(a+b) = 1$ $k = \frac{1}{a+b}$ Therefore, we have:

$$\sin^2\alpha = \frac{a}{a+b}$$ $$\cos^2\alpha = \frac{b}{a+b}$$

Now, we need to prove the relation $\frac{\sin^8\alpha}{a^3}+\frac{\cos^8\alpha}{b^3}=\frac{1}{(a+b)^3}$. Substitute the values of $\sin^2\alpha$ and $\cos^2\alpha$ into the left-hand side (LHS) of the relation to be proven:

$$\text{LHS} = \frac{(\sin^2\alpha)^4}{a^3}+\frac{(\cos^2\alpha)^4}{b^3}$$ $$\text{LHS} = \frac{\left(\frac{a}{a+b}\right)^4}{a^3}+\frac{\left(\frac{b}{a+b}\right)^4}{b^3}$$ $$\text{LHS} = \frac{\frac{a^4}{(a+b)^4}}{a^3}+\frac{\frac{b^4}{(a+b)^4}}{b^3}$$ $$\text{LHS} = \frac{a^4}{a^3(a+b)^4}+\frac{b^4}{b^3(a+b)^4}$$ $$\text{LHS} = \frac{a}{(a+b)^4}+\frac{b}{(a+b)^4}$$ $$\text{LHS} = \frac{a+b}{(a+b)^4}$$ $$\text{LHS} = \frac{1}{(a+b)^3}$$

This is equal to the right-hand side (RHS) of the relation. Hence, the relation is proven.