7/8兀的三角函数
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解决时间 2021-02-11 10:23
- 提问者网友：树红树绿
- 2021-02-10 13:58
7/8兀的三角函数
最佳答案
- 五星知识达人网友：鱼忧
- 2021-02-10 15:29
sinπ/8 = √{(1-cosπ/4)/2} = √{(1-√2/2)/2} = √(2-√2)/2
cosπ/8 = √{(1+cosπ/4)/2} = √{(1+√2/2)/2} = √(2+√2)/2
tanπ/8 = (1-cosπ/4)/(sinπ/4) = (1-√2/2)/(√2/2) = √2-1
sin7π/8 = sinπ/8 = √(2-√2)/2
c os7π/8 = -cosπ/8 = -√(2+√2)/2
t an7π/8 = -tanπ/8 = 1-√2
cosπ/8 = √{(1+cosπ/4)/2} = √{(1+√2/2)/2} = √(2+√2)/2
tanπ/8 = (1-cosπ/4)/(sinπ/4) = (1-√2/2)/(√2/2) = √2-1
sin7π/8 = sinπ/8 = √(2-√2)/2
c os7π/8 = -cosπ/8 = -√(2+√2)/2
t an7π/8 = -tanπ/8 = 1-√2
全部回答
- 1楼网友：傲气稳了全场
- 2021-02-10 16:09
tan7/8丌=-tan丌/8=-tan{(丌/4)/2},
由tan2a=2tana/{1-(tana)^2}知1=2tan丌/8/{1-（tan丌/8）^2},(tan丌/8)^2+2tan丌/8-1=0,tan丌/8=√2-1
tan7/8丌=-√2+1
(sin7/8丌)^2+(cos7丌/8)^2=1,且sin7/8丌/cos7/8丌=tan7/8丌=-√2+1,
解得sin7/8丌=√(2-√2)/2
cos7/8丌=-√（2+√2）/2
由tan2a=2tana/{1-(tana)^2}知1=2tan丌/8/{1-（tan丌/8）^2},(tan丌/8)^2+2tan丌/8-1=0,tan丌/8=√2-1
tan7/8丌=-√2+1
(sin7/8丌)^2+(cos7丌/8)^2=1,且sin7/8丌/cos7/8丌=tan7/8丌=-√2+1,
解得sin7/8丌=√(2-√2)/2
cos7/8丌=-√（2+√2）/2
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