數學 三角函數問題

解:
tan2A+sec2A=cosA+sinA

圖片參考：http://f.imagehost.org/0673/Tri.gif


A+45o=45o,315o或405o
∴A=0o,270o或360o

sin 2A/cos 2A + 1/cos 2A = cos A + sin A
(1 + sin 2A)/cos 2A = cos A + sin A
1 + 2 sin A cos A = (cos A + sin A)(cos^2A - sin ^2A)
1 + 2 sin A cos A = (cos A + sin A)(cos A + sin A)( cos A - sin A)
sin ^2 A + 2 sin A cos A + cos ^2 A = ( cos A + sin A)^2(cos A - sin A)
(cos A + sin A)^2 - (cos A + sin A)^2(cos A - sin A) = 0
(cos A + sin A)^2(1 - cos A + sin A) = 0
that is
cos A + sin A = 0...............(1) and
cos A - sin A = 1..................(2)
From (1), tan A = - 1
A = 135 and 315.
From (2),
sqrt2 cos (A + 45) = 1
so A = arccos (1/sqrt2) - 45
A = 0, 270.
so A = 0, 135, 270 and 315.



2009-05-01 09:39:46 補充：
Remark: A = 135 and 315 are rejected, because tan 2A = tan 270 and tan 630 are undefined.