Question

高一数学 倍角公式

tan2Θ=-2√2,π<2Θ<2π，求2cos^2Θ/2-sinΘ-1 / √2sin(Θ+π/4)

求过程，谢谢

Answer 1

2cos^2Θ/2-sinΘ-1 / √2sin(Θ+π/4) =(cosΘ-sinΘ) /(cosΘ+sinΘ) {在分子和分母上同时除以cosΘ}
=(1-tanΘ) / (1+tanΘ)

tan2Θ=-2√2 =2tanΘ / (1-tan^2 Θ) 解得 tanΘ =√2或-√2/2
tanΘ =√2时 原式 = 2√2 -3
tanΘ =-√2/2时 原式 = 3 + 2√2

Answer 2

cosπ/12 x cos5π/12=1/2(cos(π/12+5π/12)+cos(π/12-5π/12))
=1/2(cos(π/2+cos(-π/3))
=1/2cosπ/3
=1/4

Answer 3

cosπ/12 x cos5π/12
=1/2(cos(π/12+5π/12)+cos(π/12-5π/12))
=1/2(cos(π/2+cos(-π/3))
=1/2cosπ/3
=1/2*1/2=1/4