已知f(x)=sin2x+cos2x(I)求f(π12)的值(II)设A为三角形ABC的内角,f(A2)=22,求tanA的值.-数学

题目简介

已知f(x)=sin2x+cos2x(I)求f(π12)的值(II)设A为三角形ABC的内角,f(A2)=22,求tanA的值.-数学

题目详情

已知f(x)=sin2x+cos2x
(I)求f(
π
12
)的值
(II)设A为三角形ABC的内角,f(
A
2
)=
2
2
,求tanA的值.
题型:解答题难度:中档来源:不详

答案

(I)∵f(x)=sin2x+cos2x=
2
sin(2x+class="stub"π
4
),
∴f(class="stub"π
12
)=
2
sin(2×class="stub"π
12
+class="stub"π
4

=
2
(sinclass="stub"π
6
cosclass="stub"π
4
+cosclass="stub"π
6
sinclass="stub"π
4

=
2
2
4
+
6
4

=
1+
3
2

(II)∵f(class="stub"A
2
)=
2
2

∴f(class="stub"A
2
)=
2
sin(2×class="stub"A
2
+class="stub"π
4
)=
2
2

∴sin(A+class="stub"π
4
)=class="stub"1
2

∵A为三角形ABC的内角,
∴A+class="stub"π
4
=class="stub"5π
6

∴tanA=tan(class="stub"5π
6
-class="stub"π
4
)=
tanclass="stub"5π
6
-tanclass="stub"π
4
1+tanclass="stub"5π
6
tanclass="stub"π
4
=-class="stub"3
5
-class="stub"4
5
2

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