Question

如图，已知△ABC是直角三角形，∠C=90°。 (1)利用锐角的正弦和锐角的余弦定义，证明：si

n^2A+cos^2A=1；(2)利用这一关系式解答下列问题：①已知sinA=3/7，求cosA；②已知sinA-cosA=1/5，求sinA+cosA的值。

Answer 1

sin^2A+cos^2A=(a/c)^2+(b/c)^2=(a^2+b^2)/c^2=1, (勾股定哩),
(2) cosA=根号(1-sin^2A)=根号[1-(3/7)^2]=根号[1-9/49]=根号(40/49)=2(根号10)/7. (sinA-cosA)^2=sin^2A+cos^2A-2sinAconA=1-2sinAconA=(1/5)^2, 2sinAconA=1-1/25=24/25,
( sinA+conA)^2=sin^2A+con^2A+2sinAconA=1+24/25=49/25, sinA+conA=根号(49/25)=7/5.

追问

能不能有点条理，就是排版比较好，要不然太乱，看不懂

追答

sin^2A+cos^2A=(a/c)^2+(b/c)^2=(a^2+b^2)/c^2=1, (勾股定哩),
(2)cosA=根号(1-sin^2A)=根号[1-(3/7)^2]=根号[1-9/49]=根号(40/49)=2(根号10)/7.
(sinA-cosA)^2=sin^2A+cos^2A-2sinAconA=1-2sinAconA=(1/5)^2,
2sinAconA=1-1/25=24/25,
( sinA+conA)^2=sin^2A+con^2A+2sinAconA=1+24/25=49/25,
sinA+conA=根号(49/25)=7/5.