But where: $ f(x) = 1 - \sin 3x \sin x $, so minimum of $ f $ is when $ \sin 3x \sin x $ is **maximized**, which is $ rac{9}{16} $, so:
["Understanding the Minimum of ( f(x) = 1 - \sin 3x \sin x ): When Does the Function Reach Its Lowest Value?", "In the study of periodic functions, identifying minima and maxima is essential for understanding their behavior. Consider the function:", "[\nf(x) = 1 - \sin 3x \sin x\n]", "Since ( f(x) ) is defined as 1 minus the product of two sine functions, its minimum value occurs when the product ( \sin 3x \sin x ) is maximized—a crucial insight for simplifying optimization.", "---", "### Step 1: Maximize ( \sin 3x \sin x )", "We analyze ( g(x) = \sin 3x \sin x ), seeking its maximum value. Using a well-known trigonometric identity:", "[\n\sin A \sin B = \frac{1}{2} [\cos(A - B) - \cos(A + B)]\n]", "Set ( A = 3x ) and ( B = x ):", "[\n\sin 3x \sin x = \frac{1}{2} \left[ \cos(2x) - \cos(4x) \right]\n]", "Thus,", "[\ng(x) = \frac{1}{2} \left[ \cos 2x - \cos 4x \right]\n]", "To maximize ( g(x) ), we analyze ( \cos 2x - \cos 4x ). Recall the double-angle identity:", "[\n\cos 4x = 2\cos^2 2x - 1\n]", "Substitute:", "[\ng(x) = \frac{1}{2} \left[ \cos 2x - (2\cos^2 2x - 1) \right] = \frac{1}{2} \left[ \cos 2x - 2\cos^2 2x + 1 \right]\n]", "Let ( u = \cos 2x ), with ( -1 \le u \le 1 ). Then:", "[\ng(x) = \frac{1}{2} \left( -2u^2 + u + 1 \right) = -u^2 + \frac{u}{2} + \frac{1}{2}\n]", "This is a quadratic in ( u ), opening downward. Its maximum occurs at the vertex:", "[\nu = -\frac{b}{2a} = -\frac{1/2}{2 \cdot (-1)} = \frac{1}{4}\n]", "Now compute the maximum value:", "[\ng_{\ ext{max}} = -\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} = \frac{-1 + 2 + 8}{16} = \frac{9}{16}\n]", "Thus,", "[\n\sin 3x \sin x \le \frac{9}{16}\n]", "with equality when ( \cos 2x = \frac{1}{4} ) and parameters satisfy the identity.", "---", "### Step 2: Find When the Minimum of ( f(x) ) Occurs", "Since ( f(x) = 1 - \sin 3x \sin x ), the minimum of ( f ) is:", "[\nf_{\ ext{min}} = 1 - \max(\sin 3x \sin x) = 1 - \frac{9}{16} = \frac{7}{16}\n]", "This minimum occurs when ( \sin 3x \sin x = \frac{9}{16} ), which happens when ( \cos 2x = \frac{1}{4} ).", "---", "### Step 3: Solve for ( x ) Values That Maximize the Product", "We solve:", "[\n\cos 2x = \frac{1}{4}\n]", "Then:", "[\n2x = \pm \cos^{-1}\left( \frac{1}{4} \right) + 2k\pi \quad \Rightarrow \quad x = \frac{1}{2} \cos^{-1}\left( \frac{1}{4} \right) + k\pi \quad \ ext{or} \quad x = -\frac{1}{2} \cos^{-1}\left( \frac{1}{4} \right) + k\pi\n]", "These values of ( x ) are where ( f(x) ) reaches its minimum of ( \frac{7}{16} ).", "---", "### Conclusion", "The minimum value of ( f(x) = 1 - \sin 3x \sin x ) is achieved precisely when ( \sin 3x \sin x ) reaches its maximum of ( \frac{9}{16} ), occurring at:", "[\n\sin 2x = \frac{1}{4}\n]", "Understanding when the product reaches its peak allows exact determination of the global minimum of ( f(x) ), showcasing the power of trigonometric identities in optimization.", "---", "Keywords:\n( f(x) = 1 - \sin 3x \sin x ), minimum value, maximize ( \sin 3x \sin x ), ( \cos 2x = \frac{1}{4} ), trigonometric optimization, maximum value ( \frac{9}{16} ), calculus and trigonometry, periodic functions, algebra of trig functions.", "---", "Meta Description:\nDiscover when ( f(x) = 1 - \sin 3x \sin x ) reaches its minimum value. Learn how maximizing ( \sin 3x \sin x ) (max ( \frac{9}{16} )) unlocks the minimum of ( f(x) ), with exact conditions and applications in trigonometric modeling."]
