高一数学二倍角的正弦余弦正切公式数学题
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高一数学二倍角的正弦余弦正切公式数学题
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(1)由倍角公式:
2cos²α/2-1=cosα,∴2cos²α/2=1+cosα.
f(x)=cos(x+2π/3)+cosx+1
=cosxcos(2π/3)-sinxsin(2π/3)+cosx+1
=-(1/2)cosx-sinxsin(2π/3)+cosx+1
=(1/2)cosx-sinxsin(2π/3)+1
=cosπ/3cosx-sinπ/3sinx+1(其中:sin2π/3=sinπ/3)
=cos(x+π/6)+1
∵|cos(x+π/6|≤1,
∴0≤f(x)≤2..
(2)由tanα=3,且α∈(π/4,π/2)
∴sinα=3√10/10,cosα=√10/10,cotα=1/3,
sin2α=2sinαcosα=2×(3√10/10)×(√10/10)=3/5,
cos2α=2cos²α-1=2×(√10/10)²-1=-4/5,
cot2α=cos2α/sin2α=-4/3.
2cos²α/2-1=cosα,∴2cos²α/2=1+cosα.
f(x)=cos(x+2π/3)+cosx+1
=cosxcos(2π/3)-sinxsin(2π/3)+cosx+1
=-(1/2)cosx-sinxsin(2π/3)+cosx+1
=(1/2)cosx-sinxsin(2π/3)+1
=cosπ/3cosx-sinπ/3sinx+1(其中:sin2π/3=sinπ/3)
=cos(x+π/6)+1
∵|cos(x+π/6|≤1,
∴0≤f(x)≤2..
(2)由tanα=3,且α∈(π/4,π/2)
∴sinα=3√10/10,cosα=√10/10,cotα=1/3,
sin2α=2sinαcosα=2×(3√10/10)×(√10/10)=3/5,
cos2α=2cos²α-1=2×(√10/10)²-1=-4/5,
cot2α=cos2α/sin2α=-4/3.