(2011•渭南三模)设函数f(x)=cos(2x−π3)−cos2x,x∈R.
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(2011•渭南三模)设函数f(x)=cos(2x−
)−cos2x
π |
3 |
(I)f(x)=cos(2x−
π
3)−cos2x=cos2xcos
π
3+sin2xsin
π
3−cos2x
=
3
2sin2x−
1
2cos2x=sin(2x−
π
6).∵x∈(0,
π
2),∴2x−
π
6∈(−
π
6,
5π
6),
∴sin(2x−
π
6)∈(−
1
2,1],即f(x)在(0,
π
2)的值域为(−
1
2,1].
(II)由(I)可知,f(A)=sin(2A−
π
6),∴sin(2A−
π
6)=1.
∵0<A<π,∴−
π
6<2A−
π
6<
11π
6,∴2A−
π
6=
π
2,A=
π
3.
∵a2=b2+c2-2bccosA,把a=
7,b=3代入,得到c2-3c+2=0,∴c=1或c=2.
π
3)−cos2x=cos2xcos
π
3+sin2xsin
π
3−cos2x
=
3
2sin2x−
1
2cos2x=sin(2x−
π
6).∵x∈(0,
π
2),∴2x−
π
6∈(−
π
6,
5π
6),
∴sin(2x−
π
6)∈(−
1
2,1],即f(x)在(0,
π
2)的值域为(−
1
2,1].
(II)由(I)可知,f(A)=sin(2A−
π
6),∴sin(2A−
π
6)=1.
∵0<A<π,∴−
π
6<2A−
π
6<
11π
6,∴2A−
π
6=
π
2,A=
π
3.
∵a2=b2+c2-2bccosA,把a=
7,b=3代入,得到c2-3c+2=0,∴c=1或c=2.
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