╮(╯▽╰)╭,tanA/2=根号5,则(1-sinA-cosA)/(1-sinA+cosA)=?还有一题~sin6°cos24°sin78°cos48°的值为?
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/03 12:11:15
![╮(╯▽╰)╭,tanA/2=根号5,则(1-sinA-cosA)/(1-sinA+cosA)=?还有一题~sin6°cos24°sin78°cos48°的值为?](/uploads/image/z/7172595-27-5.jpg?t=%E2%95%AE%28%E2%95%AF%E2%96%BD%E2%95%B0%29%E2%95%AD%2CtanA%2F2%3D%E6%A0%B9%E5%8F%B75%2C%E5%88%99%EF%BC%881-sinA-cosA%EF%BC%89%2F%EF%BC%881-sinA%2BcosA%EF%BC%89%3D%3F%E8%BF%98%E6%9C%89%E4%B8%80%E9%A2%98%7Esin6%C2%B0cos24%C2%B0sin78%C2%B0cos48%C2%B0%E7%9A%84%E5%80%BC%E4%B8%BA%EF%BC%9F)
╮(╯▽╰)╭,tanA/2=根号5,则(1-sinA-cosA)/(1-sinA+cosA)=?还有一题~sin6°cos24°sin78°cos48°的值为?
╮(╯▽╰)╭,
tanA/2=根号5,则(1-sinA-cosA)/(1-sinA+cosA)=?
还有一题~
sin6°cos24°sin78°cos48°的值为?
╮(╯▽╰)╭,tanA/2=根号5,则(1-sinA-cosA)/(1-sinA+cosA)=?还有一题~sin6°cos24°sin78°cos48°的值为?
sinA=2sin(A/2)cos(A/2);1-cosA=2sin ² (A/2);1+cosA=2 cos ² (A/2)
又:tan(A/2)=√5,故:sin(A/2)cos(A/2)≠0
故:(1-sinA-cosA)/(1-sinA+cosA)
=[2sin ² (A/2)- 2sin(A/2)cos(A/2)]/[ 2 cos ² (A/2) - 2sin(A/2)cos(A/2)][分子、分母同时除以cos ² (A/2)]
=[2 tan²(A/2)- 2 tan(A/2)]/[ 2 - 2 tan(A/2)]
=(10- 2 √5)/(2 - 2 √5)
=-√5
sin6°cos24°sin78°cos48°
= sin6°sin78°cos24°cos48°
= sin6°cos12°cos24°cos48°
=16 sin6°cos6°cos12°cos24°cos48°/(16cos6°)
=8sin12° cos12°cos24°cos48°/(16cos6°)
=4sin24°cos24°cos48°/(16cos6°)
=2sin48°cos48°/(16cos6°)
=sin96°/(16cos6°)
= cos6°/(16cos6°)
=1/16
令t=tan(A/2)=√5,sinA=2t/(1+t²).cosA=(1-t²)/(1+t²).
(1-sinA-cosA)/(1-sinA+cosA)[代入计算]
=(1+t²-2t-1-t²)/(1+t²-2t+1+t²)=-t=-√5.
令t=tan(A/2)=√5,sinA=2t/(1+t²).cosA=(1-t²)/(1+t²).