Question

Пожалуйста решите максимально подробно:

Answer 1

$\cos2x= \frac{ctg^2x-1}{ctg^2x+1}$. Пользуясь этой формулой, получим $\cos(2arcctg \frac{1}{4} )= \dfrac{ctg^2(arcctg\frac{1}{4} )-1}{ctg^2(arcctg\frac{1}{4} )+1} = \dfrac{(\frac{1}{4} )^2-1}{(\frac{1}{4} )^2+1} =- \dfrac{15}{17}$

Answer 2

$cos(2arcctg\frac{1}{4})=cos2a\; ,\; \; a=arcctg\frac{1}{4}\\\\cos2a=cos^2a-sin^2a=cos^2(arcctg\frac{1}{4})-sin^2(arcctg\frac{1}{4})\\\\1+tg^2a=\frac{1}{cos^2a}\; \; \Rightarrow \; \; cos^2a=\frac{1}{1+tg^2a}= \frac{1}{1+\frac{1}{ctg^2a}}} = \frac{ctg^2a}{1+ctg^2a}$

$ctga=ctg(arcctg\frac{1}{4})=\frac{1}{4}\\\\cos^2a=cos^2(arcctg\frac{1}{4})= \frac{ctg^2(arcctg\frac{1}{4})}{1+ctg^2(arcctg\frac{1}{4})} = \frac{(\frac{1}{4})^2}{1+(\frac{1}{4})^2} = \frac{\frac{1}{16}}{1+\frac{1}{16}} = \frac{1}{17}$

$1+ctg^2a= \frac{1}{sin^2a} \; \; \Rightarrow \; \; sin^2a= \frac{1}{1+ctg^2a}\\\\sin^2a=sin^2(arcctg\frac{1}{4})= \frac{1}{1+(\frac{1}{4})^2} = \frac{1}{1+\frac{1}{16}} = \frac{16}{17} \\\\\\cos2a=cos(2arcctg\frac{1}{4})=cos^2(arcctg\frac{1}{4})-sin^2(arcctg\frac{1}{4})=\\\\= \frac{1}{17}- \frac{16}{17}=- \frac{15}{17}$

Answer 3

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