已知cosx2+2sinx2=0,(Ⅰ)求tanx的值;(Ⅱ)求cos2x2sin(π4+x)•cosx的值.-数学

题目简介

已知cosx2+2sinx2=0,(Ⅰ)求tanx的值;(Ⅱ)求cos2x2sin(π4+x)•cosx的值.-数学

题目详情

已知cos
x
2
+2sin
x
2
=0

(Ⅰ)求tanx的值;
(Ⅱ)求
cos2x
2
sin(
π
4
+x)•cosx
的值.
题型:解答题难度:中档来源:不详

答案

cosclass="stub"x
2
+2sinclass="stub"x
2
=0
解得tanclass="stub"x
2
=-class="stub"1
2

(1)tanx=
2tanclass="stub"x
2
1-(tanclass="stub"x
2
)
2
=
-class="stub"1
2
×2
1-class="stub"1
4
=-class="stub"4
3

(2)class="stub"cos2x
2
sin(class="stub"π
4
+x)•cosx
=
cos2x-sin2x
sinxcosx+cos2x
=
1-tan2x
tanx+1
=
1+class="stub"4
3
-class="stub"4
3
+1
=-7

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