For an acute angle θ with 0° < θ < 90°, consider the expression 4 sin^2 θ + 1. For every such θ, this expression is always greater than or equal to which of the following options?

Introduction / Context:
This question combines trigonometric expressions with inequality reasoning. We are given an expression involving sin^2 θ and asked which of the listed expressions is always less than or equal to it, for all acute angles θ. Such problems encourage you to manipulate expressions algebraically and use the fact that sine values lie between 0 and 1 for 0° < θ < 90°.

Given Data / Assumptions:

• 0° < θ < 90° so θ is acute.
• The expression of interest is 4 sin^2 θ + 1.
• We must find which option is always less than or equal to 4 sin^2 θ + 1 for all such θ.

Concept / Approach:
We test each option by forming an inequality of the form 4 sin^2 θ + 1 ≥ (candidate expression). If we can show that this inequality holds for every θ in (0°, 90°), then that candidate is valid. The cleanest candidate to test is 4 sin θ, because both sides are simple functions of sin θ. For that one, a simple algebraic manipulation will reveal a perfect square, which is always non negative. For other options, you can quickly see counterexamples by picking convenient angles like 0°, 30°, 60°, or 90°.

Step-by-Step Solution:
Let s = sin θ, where 0 < s < 1 for 0° < θ < 90°. Consider the inequality 4s^2 + 1 ≥ 4s. Rearrange to one side: 4s^2 - 4s + 1 ≥ 0. Factor the quadratic: 4s^2 - 4s + 1 = 4(s^2 - s + 1/4) = 4(s - 1/2)^2. Since a square is always ≥ 0, (s - 1/2)^2 ≥ 0 for all real s. Therefore, 4(s - 1/2)^2 ≥ 0, which implies 4s^2 + 1 ≥ 4s for all 0 < s < 1. So 4 sin^2 θ + 1 is always greater than or equal to 4 sin θ. Now briefly test other options to see that they fail for some angles. For option A (2): at very small θ, sin θ is near 0, so 4 sin^2 θ + 1 is close to 1, which is less than 2. So it is not always ≥ 2. For option C (4 cos θ): at θ = 0°, 4 sin^2 θ + 1 = 1, but 4 cos θ = 4, so the inequality fails. For option D (4 tan θ): near θ = 90°, tan θ becomes very large while 4 sin^2 θ + 1 stays bounded between 1 and 5, so it fails. For option E (sin θ + cos θ): a detailed analysis shows 4 sin^2 θ + 1 is not tightly bounded by this expression in the same guaranteed way, and option B is the clear algebraic match.

Verification / Alternative check:
We can verify option B numerically at a few angles. At θ = 30°, sin θ = 1/2, so 4 sin^2 θ + 1 = 4 * 1/4 + 1 = 2, and 4 sin θ = 2. They are equal. At θ = 60°, sin θ = sqrt(3) / 2, so 4 sin^2 θ + 1 = 4 * 3/4 + 1 = 4, and 4 sin θ = 2 sqrt(3) ≈ 3.464, which is less than 4. At θ close to 0°, sin θ is very small, so 4 sin^2 θ + 1 is slightly above 1, and 4 sin θ is near 0, so the inequality holds there too.

Why Other Options Are Wrong:
As seen, option A fails at small angles, C fails at 0°, D fails near 90°, and E does not have a guaranteed inequality of the same simple squared form. Only 4 sin θ is always less than or equal to 4 sin^2 θ + 1 for all acute θ.

Common Pitfalls:
A common mistake is to guess based on values at special angles like 45° without checking the whole range. Another is to assume the largest looking expression must be the answer. Always convert to a single variable and check if the difference simplifies to a square or obviously non negative form.

Final Answer:
The expression 4 sin^2 θ + 1 is always greater than or equal to 4 sin θ for 0° < θ < 90°.