Wait — we want to **minimize** $ f(x) = 1 - \sin 3x \sin x $, so we want the **maximum** of $ \sin 3x \sin x $.

["Title: Maximize ( \sin 3x \sin x ) to Minimize ( f(x) = 1 - \sin 3x \sin x ): A Step-by-Step Guide", "---", "In optimization problems involving trigonometric functions, understanding the relationship between complementary expressions is key. Here, our goal is to minimize the function:", "[\nf(x) = 1 - \sin 3x \sin x\n]", "This is equivalent to maximizing the product ( \sin 3x \sin x ), since:", "[\n\ ext{Maximize } \sin 3x \sin x \quad \Rightarrow \quad f(x) \ ext{ reaches its minimum at } f(x) = 1 - \max(\sin 3x \sin x)\n]", "In this article, we’ll explore how to simplify, analyze, and maximize ( \sin 3x \sin x ) — a crucial step in solving the optimization problem.", "---", "### Step 1: Use Trigonometric Identities to Simplify ( \sin 3x \sin x )", "We begin by expressing the product ( \sin 3x \sin x ) using a known trigonometric identity. Recall the product-to-sum formula:", "[\n\sin A \sin B = \frac{1}{2} [\cos(A - B) - \cos(A + B)]\n]", "Let ( A = 3x ) and ( B = x ). Then:", "[\n\sin 3x \sin x = \frac{1}{2} [\cos(3x - x) - \cos(3x + x)] = \frac{1}{2} [\cos 2x - \cos 4x]\n]", "So the expression becomes:", "[\n\sin 3x \sin x = \frac{1}{2} (\cos 2x - \cos 4x)\n]", "---", "### Step 2: Maximize ( \sin 3x \sin x = \frac{1}{2} (\cos 2x - \cos 4x) )", "Let ( g(x) = \sin 3x \sin x = \frac{1}{2} (\cos 2x - \cos 4x) ). To maximize ( g(x) ), we analyze its maximum value over its domain.", "Note that both ( \cos 2x ) and ( \cos 4x ) oscillate between (-1) and (1), but their difference is constrained by trigonometric bounds. Since ( | \cos \ heta | \leq 1 ), we estimate the maximum possible value of ( \cos 2x - \cos 4x ).", "The maximum occurs when ( \cos 2x = 1 ) and ( \cos 4x = -1 ), since these set the difference ( \cos 2x - \cos 4x = 1 - (-1) = 2 ), the maximum possible.", "Does such an ( x ) exist? Let’s solve:", "- ( \cos 2x = 1 \Rightarrow 2x = 2n\pi \Rightarrow x = n\pi )\n- Try ( x = \pi ): then ( \cos 4x = \cos(4\pi) = 1 ), not (-1)", "Not simultaneously satisfied. We need values where ( \cos 2x \approx 1 ) and ( \cos 4x \approx -1 ), close to the theoretical max.", "Let’s define:", "[\n\cos 2x - \cos 4x = \cos 2x - (2\cos^2 2x - 1) = -2\cos^2 2x + \cos 2x + 1\n]", "Let ( u = \cos 2x ), ( u \in [-1, 1] ), then:", "[\ng(x) = \frac{1}{2} (-2u^2 + u + 1)\n]", "Now maximize ( h(u) = -2u^2 + u + 1 ) over ( u \in [-1,1] )", "This is a quadratic in ( u ), opening downward. Vertex at:", "[\nu_{\ ext{max}} = -\frac{b}{2a} = -\frac{1}{2(-2)} = \frac{1}{4}\n]", "Evaluate:", "[\nh\left(\frac{1}{4}\right) = -2\left(\frac{1}{16}\right) + \frac{1}{4} + 1 = -\frac{1}{8} + \frac{2}{8} + \frac{8}{8} = \frac{9}{8}\n]", "Then:", "[\ng(x)_{\ ext{max}} = \frac{1}{2} \cdot \frac{9}{8} = \frac{9}{16}\n]", "---", "### Step 3: Confirm Feasibility — Does Maximum Actualize?", "We found ( h(u) ) achieves ( 9/8 ) at ( u = 1/4 ), but this requires:", "- ( \cos 2x = 1/4 )\n- ( \cos 4x = -1 )", "Check if consistent.", "From ( \cos 4x = -1 ), ( 4x = \pi + 2n\pi \Rightarrow x = \frac{\pi}{4} + \frac{n\pi}{2} )", "Try ( x = \pi/4 ):", "- ( 2x = \pi/2 \Rightarrow \cos 2x = 0 <br/>\ne 1/4 )", "Try ( x ) near where ( \cos 2x = 1/4 ), e.g., ( 2x \approx \cos^{-1}(1/4) \approx 1.318 ) rad ⇒ ( x \approx 0.659 )", "Then compute ( 4x \approx 2.636 ), ( \cos(2.636) \approx \cos(151^\circ) \approx -0.9 ) — close to (-1), but slightly off.", "Thus, exact simultaneous max is not achieved, but ( h(u) ) gives the supremum of ( \sin 3x \sin x ), which is ( \frac{9}{16} ).", "In real analysis, continuous functions attaining maxima on compact intervals achieve their maximum, but here the oscillatory nature means the global maximum of ( g(x) ) is approached but not necessarily reached over ( \mathbb{R} ), though local maxima converge densely.", "For most practical purposes — and especially in optimization over intervals — we consider the maximum value as:", "[\n\max (\sin 3x \sin x) = \frac{9}{16}\n]", "---", "### Step 4: Compute the Minimum of ( f(x) = 1 - \sin 3x \sin x )", "Now substitute:", "[\n\min f(x) = 1 - \max(\sin 3x \sin x) = 1 - \frac{9}{16} = \frac{7}{16}\n]", "---", "### Step 5: Find When Maximum Occurs — Critical Points", "To locate when ( \sin 3x \sin x = \frac{9}{16} ), return to:", "[\ng(x) = \frac{1}{2} (\cos 2x - \cos 4x)\n]", "Set ( \cos 2x = 1 ), ( \cos 4x = -1 ) as candidates (though not simultaneously satisfied exactly). The function reaches high values near where ( 2x ) is small and ( 4x ) near ( \pi ).", "But due to periodicity, maxima occur periodically. Since the expression is periodic, the minimum of ( f(x) ) occurs periodically at points where ( \sin 3x \sin x ) reaches ( \frac{9}{16} ).", "---", "### Conclusion: Key Takeaway", "Minimizing ( f(x) = 1 - \sin 3x \sin x ) this way — maximizing the product via trigonometric identities and optimization — reveals:", "- Use identities to simplify products.\n- Reduce to single-variable functions.\n- Maximize using calculus or algebraic maxima of composite expressions.\n- Evaluate maximum at critical points or via function bounds.", "This method exemplifies applying trigonometric identities and optimization strategies effectively in real-world calculus problems.", "---", "Keywords: ( \sin 3x \sin x ), maximize ( \sin 3x \sin x ), minimize ( 1 - \sin 3x \sin x ), trigonometric optimization, use identities, calculus and trigonometry, maximum value of trig expression, f(x) minimum.", "Meta Description: Learn how to minimize ( f(x) = 1 - \sin 3x \sin x ) by maximizing ( \sin 3x \sin x ) using trigonometric identities and calculus-based optimization — a clear guide for algebra and calculus students.", "---", "Note: For graphing or numerical maxima, use software to plot ( \sin 3x \sin x ) over one period (e.g., ( 0 \leq x \leq 2\pi )) and verify the maximum is near ( 9/16 ), confirming our analytical result."]