So we seek $ \max(\sin 3x \sin x) $. Use identity:

["# Maximizing the Product: Understanding $ \max(\sin 3x \sin x) $ Using Trigonometric Identities", "When analyzing functions involving products of trigonometric terms, one powerful common approach is to use trigonometric identities to simplify and maximize expressions. A classic example is finding the maximum value of $ \sin 3x \cdot \sin x $. This article explores how to compute this maximum using a key identity, and the mathematical techniques behind it — all optimized with SEO for search visibility.", "---", "## Why Maximize $ \sin 3x \sin x $?", "Maximizing $ \sin 3x \sin x $ is not just an academic exercise—it appears in signal processing, wave mechanics, and optimization problems. By identifying its maximum value and the angle(s) $ x $ that achieve it, we unlock deeper insight into periodic functions and enable applications like pulse width modulation or harmonic analysis.", "---", "## The Trigonometric Identity That Simplifies the Problem", "To simplify $ \sin 3x \sin x $, we apply a powerful product-to-sum identity:", "$$\n\sin A \sin B = \frac{1}{2} [\cos(A - B) - \cos(A + B)]\n$$", "Let $ A = 3x $, $ 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 we rewrite our target function:", "$$\n\sin 3x \sin x = \frac{1}{2} [\cos 2x - \cos 4x]\n$$", "Our goal is to maximize this expression over real numbers $ x $:", "$$\nf(x) = \sin 3x \sin x = \frac{1}{2} [\cos 2x - \cos 4x]\n$$", "---", "## Finding the Maximum of $ f(x) = \frac{1}{2} [\cos 2x - \cos 4x] $", "We now aim to find:", "$$\n\max_x \left( \frac{1}{2} [\cos 2x - \cos 4x] \right)\n$$", "Factor out the constant:", "$$\nf(x) = \frac{1}{2} \cos 2x - \frac{1}{2} \cos 4x\n$$", "To maximize $ f(x) $, note that $ \cos 2x $ and $ \cos 4x $ are bounded between $[-1,1]$. However, their values depend on $ x $, and we cannot independently maximize both terms. Instead, we use a double-angle identity to write everything in terms of $ \cos 2x $:", "Recall:\n$$\n\cos 4x = 2\cos^2 2x - 1\n$$", "Substitute:", "$$\nf(x) = \frac{1}{2} \cos 2x - \frac{1}{2} (2\cos^2 2x - 1) = \frac{1}{2} \cos 2x - \cos^2 2x + \frac{1}{2}\n$$", "Let $ u = \cos 2x $. Since $ \cos 2x \in [-1, 1] $, we define:", "$$\nf(u) = \frac{1}{2} u - u^2 + \frac{1}{2}, \quad u \in [-1, 1]\n$$", "Now, maximize this quadratic function. Write:", "$$\nf(u) = -u^2 + \frac{1}{2}u + \frac{1}{2}\n$$", "This is a downward-opening parabola. Its maximum occurs at the vertex:", "$$\nu = -\frac{b}{2a} = -\frac{1/2}{2(-1)} = \frac{1}{4}\n$$", "Check that $ u = \frac{1}{4} \in [-1,1] $ — yes, it is valid.", "Now compute $ f\left(\frac{1}{4}\right) $:", "$$\nf\left(\frac{1}{4}\right) = -\left(\frac{1}{4}\right)^2 + \frac{1}{2} \cdot \frac{1}{4} + \frac{1}{2} = -\frac{1}{16} + \frac{1}{8} + \frac{1}{2}\n$$", "Convert to sixteenths:", "$$\n= -\frac{1}{16} + \frac{2}{16} + \frac{8}{16} = \frac{9}{16}\n$$", "So:", "$$\n\max(\sin 3x \sin x) = \frac{1}{2} \cdot \frac{9}{16} = \frac{9}{32}\n$$", "Wait — correction: recall $ f(x) = \frac{1}{2}(\cos 2x - \cos 4x) = \frac{1}{2} \left( u - (2u^2 - 1) \right) = \frac{1}{2} ( -2u^2 + u + 1 ) = -u^2 + \frac{1}{2}u + \frac{1}{2} $", "We already calculated $ f(1/4) = \frac{9}{16} $? But let's double-check:", "$$\n-\left(\frac{1}{4}\right)^2 = -\frac{1}{16},\quad \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{8} = \frac{2}{16},\quad \frac{1}{2} = \frac{8}{16} \\n\Rightarrow -\frac{1}{16} + \frac{2}{16} + \frac{8}{16} = \frac{9}{16}\n$$", "Yes, $ f(x) = \frac{9}{16} $ when $ \cos 2x = \frac{1}{4} $, and since this $ u $ is attainable (e.g., $ 2x = \cos^{-1}(1/4) $), the maximum is valid.", "But wait — is this the maximum of $ \sin 3x \sin x $? Let’s confirm using an alternative approach.", "---", "## Alternative Verification: Using $ \sin 3x \sin x $ Directly", "Recall the identity:", "$$\n\sin 3x = 3\sin x - 4\sin^3 x\n$$", "So:", "$$\n\sin 3x \sin x = (3\sin x - 4\sin^3 x)\sin x = 3\sin^2 x - 4\sin^4 x\n$$", "Let $ y = \sin^2 x $, $ y \in [0,1] $. Then:", "$$\nf(y) = 3y - 4y^2\n$$", "This is a quadratic. Maximum at $ y = -\frac{3}{2(-4)} = \frac{3}{8} $", "Value:", "$$\nf\left(\frac{3}{8}\right) = 3 \cdot \frac{3}{8} - 4 \cdot \left(\frac{3}{8}\right)^2 = \frac{9}{8} - 4 \cdot \frac{9}{64} = \frac{9}{8} - \frac{36}{64} = \frac{9}{8} - \frac{9}{16} = \frac{18 - 9}{16} = \frac{9}{16}\n$$", "Same result! So:", "$$\n\max(\sin 3x \sin x) = \frac{9}{16}\n$$", "---", "## Final Thoughts and Applications", "The maximum value of $ \sin 3x \sin x $ is $ \frac{9}{16} $, achieved when $ \sin^2 x = \frac{3}{8} $. This result illustrates how trigonometric identities transform complex products into manageable algebraic forms, enabling precise optimization.", "This kind of analysis is essential in fields like:", "- Signal processing (product of carrier waves)\n- Mechanical systems (harmonic motion products)\n- Engineering design (amplitude modulation)\n- Mathematical olympiads and advanced calculus", "---", "## Summary", "- Use the identity $ \sin A \sin B = \frac{1}{2}[\cos(A-B) - \cos(A+B)] $ to rewrite the product.\n- Simplify to $ f(x) = \frac{1}{2}[\cos 2x - \cos 4x] $.\n- Further reduce using $ \cos 4x = 2\cos^2 2x - 1 $.\n- Maximize the resulting quadratic in $ \cos 2x $.\n- Verify maximum numerically and analytically yields $ \frac{9}{16} $.", "---", "## SEO Keywords & Phrases (Optimized)", "- $ \max(\sin 3x \sin x) $\n- Maximize $ \sin 3x \sin x $\n- Trigonometric identity simplification\n- Product-to-sum identity\n- $ \sin A \sin B $ formula\n- Maximum of $ \cos 2x - \cos 4x $\n- Analytical optimization of trig functions\n- Trigonometric optimization techniques\n- $ \cos 4x $ in harmonic analysis\n- $ \sin^2 x $ and $ \sin 3x $ product maximum", "---", "## Related Reading", "- How to maximize trigonometric products\n- Using double-angle identities in calculus\n- Applications of product-to-sum formulae\n- Finding extrema of periodic functions", "---", "Maximizing $ \sin 3x \sin x $ showcases the elegance of trigonometric identities — transforming complexity into clarity. Whether for math students, engineers, or researchers, mastering these techniques unlocks powerful tools for solving real-world problems."]