Trigonometric Expression Inequality Explained

The question asks for the inequality that must hold true for all possible values of the angle $\theta$ concerning the given expression:

\( P = \sin^{20}\theta + \cos^{48}\theta \)

Our goal is to determine the valid range for the value of P.

Analyzing the Trigonometric Expression Components

Let's break down the expression P:

• The first term is $\sin^{20}\theta$.
• The second term is $\cos^{48}\theta$.

We know the fundamental properties of sine and cosine functions:

• The value of $\sin\theta$ ranges from -1 to 1 (inclusive).
• The value of $\cos\theta$ ranges from -1 to 1 (inclusive).

When we raise these values to an even power (like 20 or 48), the results are always non-negative:

• \( 0 \le \sin^2\theta \le 1 \)
• \( 0 \le \cos^2\theta \le 1 \)

Simplifying with Substitution

To simplify the analysis, let's use a substitution. Let $x = \sin^2\theta$.

Based on the properties above, the range for $x$ is:

\( 0 \le x \le 1 \)

Using the trigonometric identity $\sin^2\theta + \cos^2\theta = 1$, we can express $\cos^2\theta$ in terms of $x$:

\( \cos^2\theta = 1 - \sin^2\theta = 1 - x \)

Now, we can rewrite the expression P using $x$:

• \( \sin^{20}\theta = (\sin^2\theta)^{10} = x^{10} \)
• \( \cos^{48}\theta = (\cos^2\theta)^{24} = (1-x)^{24} \)

Substituting these back into the expression for P gives us a function of $x$:

\( P(x) = x^{10} + (1-x)^{24} \)

We need to find the range of this function $P(x)$ for $x$ values between 0 and 1.

Evaluating P at Boundary Values (x=0 and x=1)

Let's check the value of P at the extremes of the possible range for $x$:

• Case 1: $x=0$

  This corresponds to $\sin^2\theta = 0$, which means $\sin\theta = 0$ (e.g., $\theta = 0^\circ$ or $\theta = 180^\circ$).

  Calculation: \( P(0) = 0^{10} + (1-0)^{24} = 0 + 1^{24} = 1 \)

• Case 2: $x=1$

  This corresponds to $\sin^2\theta = 1$, which means $\sin\theta = \pm 1$ (e.g., $\theta = 90^\circ$ or $\theta = 270^\circ$).

  Calculation: \( P(1) = 1^{10} + (1-1)^{24} = 1 + 0^{24} = 1 \)

At both boundaries ($x=0$ and $x=1$), the value of P is 1.

Analyzing P for Intermediate Values (0 < x < 1)

Now, consider the case when $x$ is strictly between 0 and 1 (i.e., $0 < x < 1$).

Key property: For any number $y$ such that $0 < y < 1$, raising $y$ to a larger positive exponent results in a smaller value. For instance, $y^3 < y^2$.

Applying this property:

• Since $0 < x < 1$, we have \( x^{10} \le x^2 \). (The equality holds only if x=0 or x=1, but we are considering $0 < x < 1$).
• Since $0 < 1-x < 1$, we have \( (1-x)^{24} \le (1-x)^2 \).

Therefore, we can bound P as follows:

\( P(x) = x^{10} + (1-x)^{24} \le x^2 + (1-x)^2 \)

Let's analyze the upper bound function $g(x) = x^2 + (1-x)^2$ on the interval $[0, 1]$:

\( g(x) = x^2 + (1 - 2x + x^2) = 2x^2 - 2x + 1 \)

This is a quadratic function. To find its maximum value on $[0, 1]$, we check the endpoints:

• \( g(0) = 2(0)^2 - 2(0) + 1 = 1 \)
• \( g(1) = 2(1)^2 - 2(1) + 1 = 1 \)

The maximum value of $g(x)$ on the interval $[0, 1]$ is 1.

Since \( P(x) \le g(x) \) and $g(x) \le 1$, we can conclude that:

\( P(x) \le 1 \)

Determining the Lower Bound for P

We need to determine if P can ever be equal to 0.

The expression is \( P = \sin^{20}\theta + \cos^{48}\theta \).

Since $\sin^{20}\theta$ and $\cos^{48}\theta$ are powers of real numbers with even exponents, they are always greater than or equal to 0.

• \( \sin^{20}\theta \ge 0 \)
• \( \cos^{48}\theta \ge 0 \)

For P to equal 0, both $\sin^{20}\theta$ and $\cos^{48}\theta$ must simultaneously be 0.

This requires $\sin\theta = 0$ AND $\cos\theta = 0$.

However, $\sin\theta$ and $\cos\theta$ cannot both be 0 for the same angle $\theta$, because $\sin^2\theta + \cos^2\theta = 1$. If $\sin\theta = 0$, then $\cos\theta = \pm 1$, and if $\cos\theta = 0$, then $\sin\theta = \pm 1$.

Therefore, P can never be exactly 0. It must always be strictly positive.

\( P > 0 \)

Final Inequality Conclusion

By combining our findings:

• We found that \( P \le 1 \).
• We found that \( P > 0 \).

Thus, the inequality that holds for all values of $\theta$ is:

\( 0 < P \le 1 \)

Comparing with Given Options

Let's match our result with the provided options:

• Option 1: \( P \ge 1 \) - Incorrect, P can be less than 1.
• Option 2: \( 0 < P \le 1 \) - Correct.
• Option 3: \( 1 < P < 3 \) - Incorrect, P is never greater than 1.
• Option 4: \( 0 \le P \le 1 \) - Incorrect, P can never be exactly 0.