Trigonometry & Inverse Trigonometry Question - None

Question

152. The minimum value of the expression $\dfrac{\sin^3\alpha + 6\sin^2\alpha + \sin\alpha + 2\cos^2\alpha - 8}{\sin\alpha - 1}$ is equal to:

Accepted Answer

Step 1: Let t = sin α where t ∈ [-1, 1) and t ≠ 1 (denominator non-zero). Since cos²α = 1 - sin²α = 1 - t², rewrite the expression as:

f(t) = (t³ + 6t² + t + 2(1 - t²) - 8)/(t - 1) = (t³ + 6t² + t + 2 - 2t² - 8)/(t - 1)

Step 2: Simplify the numerator:

f(t) = (t³ + 4t² + t - 6)/(t - 1)

Step 3: Factor the numerator by checking t = 1 (even though excluded): t³ + 4t² + t - 6 = (t - 1)(t² + 5t + 6) = (t - 1)(t + 2)(t + 3)

Step 4: Cancel the (t - 1) factor:

f(t) = (t + 2)(t + 3) = t² + 5t + 6 for t ∈ [-1, 1)

Step 5: Find the minimum of g(t) = t² + 5t + 6 on [-1, 1).

g'(t) = 2t + 5 = 0 gives t = -5/2 (outside domain)

Step 6: Since the parabola vertex is at t = -5/2 < -1, the function is increasing on [-1, 1). Minimum occurs at t = -1:

g(-1) = 1 - 5 + 6 = 2

∴ Answer: B (minimum value = 2)

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