Key Concepts and Formulas

• Exponential Inequality: If $a > 1$, then $a^m \le a^n$ if and only if $m \le n$.
• Completing the Square: The expression $x^2 + bx + c$ can be written as $(x + \frac{b}{2})^2 + (c - \frac{b^2}{4})$.
• Range of Sine Function: $-1 \le \sin x \le 1$ for all real numbers $x$, and consequently, $0 \le \sin^2 x \le 1$.

Step-by-Step Solution

Step 1: Rewrite the Inequality Using Exponent Properties

The given inequality is: $2^{\sqrt{\sin^2 x - 2\sin x + 5}} \cdot \frac{1}{4^{\sin^2 y}} \le 1$ We rewrite the term $4^{\sin^2 y}$ as $(2^2)^{\sin^2 y} = 2^{2\sin^2 y}$. The inequality becomes: $2^{\sqrt{\sin^2 x - 2\sin x + 5}} \cdot \frac{1}{2^{2\sin^2 y}} \le 1$ $2^{\sqrt{\sin^2 x - 2\sin x + 5}} \le 2^{2\sin^2 y}$ Why: We want to compare the exponents directly. Since the base is 2 (which is greater than 1), the inequality between the exponential expressions holds if and only if the inequality between the exponents holds in the same direction.

Therefore, $\sqrt{\sin^2 x - 2\sin x + 5} \le 2\sin^2 y$

Step 2: Simplify the Expression Under the Square Root by Completing the Square

We focus on the expression under the square root: $\sin^2 x - 2\sin x + 5$. We treat this as a quadratic in terms of $\sin x$ and complete the square. Why: Completing the square helps us determine the minimum value of the expression. $\sin^2 x - 2\sin x + 5 = (\sin^2 x - 2\sin x + 1) + 4 = (\sin x - 1)^2 + 4$ The inequality now reads: $\sqrt{(\sin x - 1)^2 + 4} \le 2\sin^2 y$

Step 3: Determine the Minimum Value of the Left-Hand Side (LHS)

Consider the term inside the square root: $(\sin x - 1)^2 + 4$. Why: Finding the minimum value of the LHS is essential to establish conditions for the inequality to hold. Since $(\sin x - 1)^2$ is a square of a real number, its minimum possible value is 0. This occurs when $\sin x - 1 = 0$, i.e., when $\sin x = 1$. Therefore, the minimum value of $(\sin x - 1)^2 + 4$ is $0 + 4 = 4$. Consequently, the minimum value of the LHS, $\sqrt{(\sin x - 1)^2 + 4}$, is $\sqrt{4} = 2$.

Step 4: Analyze the Right-Hand Side (RHS) and Derive Necessary Conditions

The RHS of the inequality is $2\sin^2 y$. Why: To satisfy the inequality LHS $\le$ RHS, we need to ensure that the RHS can be at least as large as the minimum value of the LHS. We know that $0 \le \sin^2 y \le 1$ for any real number $y$. Multiplying by 2, we get $0 \le 2\sin^2 y \le 2$. So, the maximum possible value of the RHS is 2.

For the inequality $\sqrt{(\sin x - 1)^2 + 4} \le 2\sin^2 y$ to hold, and knowing that the LHS is always $\ge 2$, the RHS must be at least 2. Thus: $2\sin^2 y \ge 2$ $\sin^2 y \ge 1$ Since we also know that $\sin^2 y \le 1$, the only way for $\sin^2 y \ge 1$ to be true is if: $\sin^2 y = 1$ This implies $\sin y = 1$ or $\sin y = -1$. In other words, $|\sin y| = 1$.

Step 5: Find the Precise Conditions for Equality

We now know that a necessary condition for the original inequality to hold is $\sin^2 y = 1$. Let's substitute this back into the inequality from Step 2: $\sqrt{(\sin x - 1)^2 + 4} \le 2\sin^2 y$ $\sqrt{(\sin x - 1)^2 + 4} \le 2(1)$ $\sqrt{(\sin x - 1)^2 + 4} \le 2$ Why: We are finding the specific value of $\sin x$ that satisfies the inequality when the condition on $\sin y$ is met. Squaring both sides (valid because both sides are non-negative): $(\sin x - 1)^2 + 4 \le 4$ $(\sin x - 1)^2 \le 0$ Since the square of any real number cannot be negative, the only possibility is: $(\sin x - 1)^2 = 0$ This implies $\sin x - 1 = 0$, which means: $\sin x = 1$

Step 6: Verify the Solution against Options

The inequality is satisfied if and only if BOTH conditions are met:

1. $\sin x = 1$
2. $|\sin y| = 1$ (which is equivalent to $\sin^2 y = 1$)

These two conditions together imply that $\sin x = 1$ and $|\sin y| = 1$. If $\sin x = 1$, then $\sin x = |\sin y|$ holds because $1 = |\sin y|$ means $\sin y = 1$ or $\sin y = -1$, consistent with $|\sin y| = 1$.

Let's check the options: (A) $\sin x = |\sin y|$ (This is what we derived) (B) $\sin x = 2\sin y$ (If $\sin x = 1$, then $\sin y = 1/2$, contradicting $|\sin y| = 1$) (C) $2\sin x = \sin y$ (If $\sin x = 1$, then $\sin y = 2$, impossible) (D) $2|\sin x| = 3\sin y$ (If $\sin x = 1$, then $\sin y = 2/3$, contradicting $|\sin y| = 1$)

Therefore, the only equation that all pairs $(x, y)$ satisfying the inequality also satisfy is $\sin x = |\sin y|$. This corresponds to option (A).

Common Mistakes & Tips

• Tip: When dealing with inequalities, consider extreme values and boundary conditions to simplify the problem.
• Tip: Completing the square is a useful technique for finding the minimum or maximum value of a quadratic expression.
• Common Mistake: Forgetting the range of trigonometric functions. $\sin x$ always lies between -1 and 1.

Summary

By using properties of exponents, completing the square, and analyzing the ranges of sine functions, we found that the given inequality is satisfied if and only if $\sin x = 1$ and $|\sin y| = 1$. These conditions lead directly to the equation $\sin x = |\sin y|$, which corresponds to option (A).

Final Answer

The final answer is \boxed{\sin x = |\sin y|}, which corresponds to option (A).