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Question Number 215884 by golsendro last updated on 20/Jan/25

$$\underset{x\to 0}{\lim }\frac{\cos 2\mathrm{x}-\cos 6\mathrm{x}}{1-\cos 3\mathrm{x}\cos 5\mathrm{x}}=?$$$$$$

Answered by oubiji last updated on 20/Jan/25

$$\underset{x\to 0}{\lim }\frac{\cos 2\mathrm{x}-\cos 6\mathrm{x}}{1-\cos 3\mathrm{x}\cos 5\mathrm{x}}$$$$=li\underset{x\to 0}{m}\frac{\frac{\cos 2x-1}{x^2}+\frac{1-\cos 6x}{x^2}}{\frac{1-\cos 3x}{x^2}\cos 5x+\frac{1-\cos 5x}{x^2}}$$$$=$$$$\frac{-2+18}{\frac{9}{2}\times 1+\frac{25}{2}}=\frac{16}{17}$$

Commented by MathematicalUser2357 last updated on 21/Jan/25

$$Insecondline,\underset{x\to 0}{\lim }\frac{\frac{\cos 2x-1}{x^2}+\frac{1-\cos 6x}{x^2}}{\frac{1-\cos 3x}{x^2}\cos 5x+\frac{1-\cos 5x}{x^2}}=\frac{-2+18}{\frac{9}{2}\times 1+\frac{25}{12}}=(-2+18)\mathbf{÷}(\frac{9}{2}\times 1+\frac{25}{12})=\frac{16}{17}$$

Commented by khorshidi17 last updated on 22/Jan/25

$$solve$$$$K={sin}^{2}x\times {sin}^{2}2x\times {sin}^{2}4x$$$$maxK=?$$

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