Помогите, пожалуйста Найдите a, если cos15°-sin15°=[tex] \frac{a}{4cos15} [/tex]. Created by Olya20023027. algebra-ru

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Olya20023027 @Olya20023027

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Помогите, пожалуйста

Найдите a, если cos15°-sin15°=[tex] \frac{a}{4cos15} [/tex]

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$cos15-sin15= \frac{a}{4cos15}\\ \\\\cos15-sin15= \frac{\sqrt3+1}{2\sqrt2} -\frac{\sqrt3-1}{2\sqrt2}= \frac{\sqrt3+1-\sqrt3+1}{2\sqrt2}= \frac{2}{2\sqrt2}= \frac{1}{\sqrt2} \\\\ \frac{1}{\sqrt2}=\frac{a}{4cos15} \\\\ \frac{1}{\sqrt2}=\frac{a}{4\cdot \frac{\sqrt3+1}{2\sqrt2}}\\\\\frac{1}{\sqrt2}=\frac{a\sqrt2}{2(\sqrt3+1)} \\\\a= \frac{2(\sqrt3+1)}{\sqrt2\cdot \sqrt2}\\\\a=\sqrt3+1$

$P.S.\; \; \; cos15=cos(45-30)=cos45\cdot cos30+sin45\cdot sin30=\\\\= \frac{\sqrt2}{2}\cdot \frac{\sqrt3}{2}+ \frac{\sqrt2}{2}\cdot \frac{1}{2}=\frac{\sqrt3}{2\sqrt2}+\frac{1}{2\sqrt2}=\frac{\sqrt3+1}{2\sqrt2}\; ;\\\\sin15=sin(45-30)=sin45\cdot cos30-sin30\cdot cos45=\\\\= \frac{\sqrt2}{2}\cdot \frac{\sqrt3}{2}-\frac{1}{2}\cdot \frac{\sqrt2}{2}= \frac{\sqrt3}{2\sqrt2}-\frac{1}{2\sqrt2} =\frac{\sqrt3-1}{2\sqrt2}\; ;$

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