●Aが鋭角から、cosA>0なので、
sinA=1/5のとき、
cosA=√{1-sin²A}=√{1-(1/5)²}=√{1-(1/25)}=√{24/25}=2√6/5
●余角の三角比を利用し
(1)cos(90-A)=sinA=1/5
(2)sin(90-A)=cosA=2√6/5
●Aが鋭角から、cosA>0なので、
sinA=1/5のとき、
cosA=√{1-sin²A}=√{1-(1/5)²}=√{1-(1/25)}=√{24/25}=2√6/5
●余角の三角比を利用し
(1)cos(90-A)=sinA=1/5
(2)sin(90-A)=cosA=2√6/5
Users viewing this question
are also looking at these questions 😉