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

lim_(x→0) ((cos 2x−cos 6x)/(1−cos 3x cos 5x)) =?

$$\:\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\mathrm{2x}−\mathrm{cos}\:\mathrm{6x}}{\mathrm{1}−\mathrm{cos}\:\mathrm{3x}\:\mathrm{cos}\:\mathrm{5x}}\:=? \\ $$$$\:\:\:\: \\ $$

Answered by oubiji last updated on 20/Jan/25

lim_(x→0) ((cos 2x−cos 6x)/(1−cos 3x cos 5x)) =lim_(x→0) ((((cos 2x−1)/x^2 )+((1−cos 6x)/x^2 ))/(((1−cos 3x)/x^2 )cos 5x+((1−cos 5x)/x^2 ))) = ((−2+18)/((9/2)×1+((25)/2)))=((16)/(17))

$$\:\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\mathrm{2x}−\mathrm{cos}\:\mathrm{6x}}{\mathrm{1}−\mathrm{cos}\:\mathrm{3x}\:\mathrm{cos}\:\mathrm{5x}}\: \\ $$$$\:\:\:\:\:\:\:\:={li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\frac{\frac{\mathrm{cos}\:\mathrm{2}{x}−\mathrm{1}}{{x}^{\mathrm{2}} }+\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{6}{x}}{{x}^{\mathrm{2}} }}{\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{3}{x}}{{x}^{\mathrm{2}} }\mathrm{cos}\:\mathrm{5}{x}+\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{5}{x}}{{x}^{\mathrm{2}} }} \\ $$$$\:\:\:\:\:\:\:\:= \\ $$$$\frac{−\mathrm{2}+\mathrm{18}}{\frac{\mathrm{9}}{\mathrm{2}}×\mathrm{1}+\frac{\mathrm{25}}{\mathrm{2}}}=\frac{\mathrm{16}}{\mathrm{17}} \\ $$

Commented by MathematicalUser2357 last updated on 21/Jan/25

In second line, lim_(x→0) ((((cos 2x−1)/x^2 )+((1−cos 6x)/x^2 ))/(((1−cos 3x)/x^2 )cos 5x+((1−cos 5x)/x^2 )))=((−2+18)/((9/2)×1+((25)/(12))))=(−2+18)÷((9/2)×1+((25)/(12)))=((16)/(17))

$${In}\:{second}\:{line},\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\mathrm{cos}\:\mathrm{2}{x}−\mathrm{1}}{{x}^{\mathrm{2}} }+\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{6}{x}}{{x}^{\mathrm{2}} }}{\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{3}{x}}{{x}^{\mathrm{2}} }\mathrm{cos}\:\mathrm{5}{x}+\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{5}{x}}{{x}^{\mathrm{2}} }}=\frac{−\mathrm{2}+\mathrm{18}}{\frac{\mathrm{9}}{\mathrm{2}}×\mathrm{1}+\frac{\mathrm{25}}{\mathrm{12}}}=\left(−\mathrm{2}+\mathrm{18}\right)\boldsymbol{\div}\left(\frac{\mathrm{9}}{\mathrm{2}}×\mathrm{1}+\frac{\mathrm{25}}{\mathrm{12}}\right)=\frac{\mathrm{16}}{\mathrm{17}} \\ $$

Commented by khorshidi17 last updated on 22/Jan/25

solve K=sin^2 x×sin^2 2x×sin^2 4x max K=?

$${solve} \\ $$$${K}={sin}^{\mathrm{2}} {x}×{sin}^{\mathrm{2}} \mathrm{2}{x}×{sin}^{\mathrm{2}} \mathrm{4}{x} \\ $$$${max}\:{K}=? \\ $$

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