Solving for the Trigonometric Expression Value

We are given two equations:

• $m = a \cos^3\beta$
• $n = b \sin^3\beta$

Our goal is to find the value of the expression $(\frac{m}{a})^{\frac{2}{3}} + (\frac{n}{b})^{\frac{2}{3}}$.

Step-by-Step Algebraic Manipulation

First, let's rearrange the given equations to express $\cos\beta$ and $\sin\beta$ in terms of $m, n, a,$ and $b$.

1. From the first equation, $m = a \cos^3\beta$, we can isolate $\cos^3\beta$: $$ \frac{m}{a} = \cos^3\beta $$ Now, take the cube root of both sides: $$ (\frac{m}{a})^{\frac{1}{3}} = \cos\beta $$
2. Similarly, from the second equation, $n = b \sin^3\beta$, we isolate $\sin^3\beta$: $$ \frac{n}{b} = \sin^3\beta $$ Take the cube root of both sides: $$ (\frac{n}{b})^{\frac{1}{3}} = \sin\beta $$

Evaluating the Target Expression

Now, let's look at the expression we need to evaluate: $(\frac{m}{a})^{\frac{2}{3}} + (\frac{n}{b})^{\frac{2}{3}}$.

We can rewrite the terms using exponent rules, specifically $(x^p)^q = x^{pq}$.

1. The first term is $(\frac{m}{a})^{\frac{2}{3}}$. We know that $(\frac{m}{a})^{\frac{1}{3}} = \cos\beta$. So, $$ (\frac{m}{a})^{\frac{2}{3}} = \left((\frac{m}{a})^{\frac{1}{3}}\right)^2 = (\cos\beta)^2 = \cos^2\beta $$
2. The second term is $(\frac{n}{b})^{\frac{2}{3}}$. We know that $(\frac{n}{b})^{\frac{1}{3}} = \sin\beta$. So, $$ (\frac{n}{b})^{\frac{2}{3}} = \left((\frac{n}{b})^{\frac{1}{3}}\right)^2 = (\sin\beta)^2 = \sin^2\beta $$

Substitute these simplified terms back into the expression:

Applying the Trigonometric Identity

We use the fundamental Pythagorean trigonometric identity, which states that for any angle $\beta$:

Therefore, the value of the given expression is 1.