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Gviona @Gviona

August 2022 1 33 Report

Объясните,пожалуйста,как решать такие выражения:

а). sin ( arccos 3/5)

б). tg ( arccos ( - 5/13))

в). sin ( arccos ( -0,8))

г). ctg ( arccos 4/5)

Answers & Comments

Artem112

Verified answer

$sin(arccosA)=\sqrt{1-A^2} \\\ cos(arccosA)=A \\\ tg(arccosA)=\frac{sin(arccosA)}{cos(arccosA)}=\frac{\sqrt{1-A^2}}{A} \\\ ctg(arccosA)=\frac{cos(arccosA)}{sin(arccosA)}=\frac{A}{\sqrt{1-A^2}}$

$sin(arccos\frac{3}{5} )=\sqrt{1-(\frac{3}{5})^2}=\frac{4}{5} \\\ tg(arccos(\frac{5}{13}))=\frac{\sqrt{1-(\frac{5}{13})^2}}{\frac{5}{13}}=\frac{\frac{12}{13}}{\frac{5}{13}}=\frac{12}{5} \\\ sin(arccos(-0.8))=\sqrt{1-(-0.8)^2}=0.6 \\\ ctg(arccos\frac{4}{5})=\frac{\frac{4}{5}}{\sqrt{1-(\frac{4}{5})^2}}=\frac{\frac{4}{5}}{\frac{3}{5}}=\frac{4}{3}$

7 votes Thanks 36

Answers & Comments

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