问题
解答题
已知-
(1)求sinx-cosx的值; (2)求tan2x; (3)求3sin2
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答案
(1)∵-
<x<0,sinx=-π 2 3 5
∴cosx=
=1-sin2x 4 5
∴sinx-cosx=-
-3 5
=-4 5 7 5
(2)tanx=
=-- 3 5 4 5
则tan2x=3 4
=2tanx 1-tan2x
=-2×(-
)3 4 1-(-
)23 4 24 7
(3)∵tanx=-
=3 4
,-2tan x 2 1-tan2 x 2
<x<0π 2
∴tan
=-x 2 1 3
则3sin2
-2sinx 2
cosx 2
+cos2x 2
=x 2
=3sin2
-2sinx 2
cosx 2
+cos2x 2 x 2 sin2
+cos2x 2 x 2
=3tan2
-2tanx 2
+1x 2 tan2
+1x 2
=
+1 3
+12 3
+11 9 9 5
∴3sin2
-2sinx 2
cosx 2
+cos2x 2
的值为x 2 9 5