已知tan(α+π⼀4)=3求sin(2α+π⼀3)

2025-06-27 03:01:16
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回答1:

解答:
∵tan(α+π/4)=3
∴(tanα+1)/(1-tanα)=3
∴tanα+1=3-3tanα
∴ 4tanα=2
∴ tanα=1/2
∴ sin2α=2sinαcosα=2sinαcosα/(sin²α+cos²α)
分子分母同时除以cos²α
sin2α=2tanα/(1+tan²α)=1/(1+1/4)=4/5
∴ cos2α=cos²α-sin²α=(cos²α-sin²α)/(cos²α+sin²α)
分子分母同时除以cos²α
cos2α=(1-tan²α/(1+tan²α)=(1-1/4)/(1+1/4)=3/5
∴ sin(2α+π/3)
=sin(2α)cos(π/3)+cos(2α)sin(π/3)
=(4/5)*(1/2)+(3/5)*(√3/2)
=(4+3√3)/10