tgα=1/ctgα=-5/12
т.к. α∈2 четверти, то cosα=-√(1/(1+tg²α))=
-√(1/(1+(25/144))=-√(144/169)=-12/13
sinα=√(1-cos²α)=√(1-(144/169)=5/13
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tgα=1/ctgα=-5/12
т.к. α∈2 четверти, то cosα=-√(1/(1+tg²α))=
-√(1/(1+(25/144))=-√(144/169)=-12/13
sinα=√(1-cos²α)=√(1-(144/169)=5/13