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標題:
f.3 math (10)~~~~~~~
發問:
最佳解答:
cos 27*tan27-cos 63 = cos27*(sin27/cos27)-cos63 (tan x=sin x/cos x) = sin27-cos63 = sin27-sin(90-63) (cos x= sin(90-x)) = sin27-sin27 = 0 2009-04-23 23:14:51 補充： sin^2 42 + cos^2 48 = sin^2 42+ cos^2 42 (sin^2 x=cos^2(90-x)) =1 (sin^2 x+ cos^2 x=1)
其他解答:
1) (cos 27)(tan27) - (cos 63) = (cos 27)(sin27/cos27) - (cos 63) = sin27 - cos27 2) (sin42)^2 + (sin48) = (sin42)^2 + [sin(90-42)]^2 = (sin42)^2 + (cos42)^2 = 1 3) 4[cos(90-x)][tan(90-X)] = 4[cos(90-x)][sin(90-X)/cos(90-x)] = 4sin(90-x) = 4cosx
f.3 math (10)~~~~~~~
發問:
此文章來自奇摩知識+如有不便請留言告知
cos 27*tan27-cos 63 sin^2 42+sin^2 48 4cos(90-x)*tan(90-X)最佳解答:
cos 27*tan27-cos 63 = cos27*(sin27/cos27)-cos63 (tan x=sin x/cos x) = sin27-cos63 = sin27-sin(90-63) (cos x= sin(90-x)) = sin27-sin27 = 0 2009-04-23 23:14:51 補充： sin^2 42 + cos^2 48 = sin^2 42+ cos^2 42 (sin^2 x=cos^2(90-x)) =1 (sin^2 x+ cos^2 x=1)
其他解答:
1) (cos 27)(tan27) - (cos 63) = (cos 27)(sin27/cos27) - (cos 63) = sin27 - cos27 2) (sin42)^2 + (sin48) = (sin42)^2 + [sin(90-42)]^2 = (sin42)^2 + (cos42)^2 = 1 3) 4[cos(90-x)][tan(90-X)] = 4[cos(90-x)][sin(90-X)/cos(90-x)] = 4sin(90-x) = 4cosx
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