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三角恒等变换题1.已知函数f﹙x﹚=a﹙cos²x+sinxcosx﹚+b(1)当a>0时,求f﹙x﹚的单调增

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三角恒等变换题
1.已知函数f﹙x﹚=a﹙cos²x+sinxcosx﹚+b
(1)当a>0时,求f﹙x﹚的单调增区间
(2)当a<0且x∈[0,兀/2]时,f﹙x﹚的值域是[3,4]求a,b的值.
f﹙x﹚=a﹙cos²x+sinxcosx﹚+b
=1/2a﹙2cos²x+2sinxcosx﹚+b
=1/2a(1+cos2x+sin2x)+b
=√2/2asin(2x+π/4)+1/2a+b
2x+π/4x∈[2kπ-π/2,2kπ+π/2]单调增区间
x∈[kπ-3π/8,kπ+π/8]单调增区间
当a<0且x∈[0,π/2]时f﹙x﹚的值域是[3,4]
f﹙x﹚=√2/2asin(2x+π/4)+1/2a+b
=√2/2asin[π/2+(2x-π/4)]+1/2a+b
= -√2/2acos(2x-π/4)+1/2a+b
当f﹙0﹚时有最大值=4
f﹙0﹚=-√2/2acos(-π/4)+1/2a+b=-1/2a1/2a+b=4
b=4
当f﹙[π/2﹚时有最小值=3
f﹙π/2﹚== -√2/2acos(π-π/4)+1/2a+b=3
a+b=3
a=-1
a=-1,b=4