Question

高一数学，如何化简三角函数，急！～

一共10小题，要详细的解答，不要只给公式，全答出来会加赏20分。急，大家帮忙啊

Answer 1

解：
1、原式=6√5((√3/2)sinx+(1/2)cosx)=6√5（sinx cos π/6+cosx sinπ/6）=6√5sin(x+π/6)
2、原式= √3((√3/2)cosx-(1/2)sinx)=√3(sinπ/3cosx-cos π/3 sinx)=√3sin(π/3-x)
3、原式=2(√3/2sin(x/2)+1/2cos(x/2)）
=2(sinx cos π/6+cosx sinπ/6)
=2sin(x/2+π/6)
4、原式=√2/4sin(π/4-x)+√6/4cos(π/4-x)
=(√2/2)((1/2)sin(π/4-x)+(√3/2)cos(π/4-x)
=(√2/2)((sinπ/6sin(π/4-x)+cosπ/6cos(π/4-x)
=(√2/2)((sinπ/6sin(π/4-x)+cosπ/6cos(π/4-x)
=(√2/2)(cos(π/6-π/4+x))
=(√2/2)(cos(x-π/12))
5、原式=sin(360°-13°)cos(180°-32°)+sin(90°-13°)cos(90°-32°)
=(-sin13°)(-cos32°)+cos13°sin32°
=sin13°cos32°+cos13°sin32°
=sin(13°+32°)=sin45°=√2/2
6、原式=sin(180°-16°)sin(180°+44°)+sin(270°-16°)sin(270°+44°)
=-sin16°sin44°+cos16°cos44°
=cos(16°+44°)=cos60°=1/2
7、原式=sin(α+β)cos(γ-β)-cos(α+β)sin(-(γ-β))
=sin(α+β)cos(γ-β)+cos(α+β)sin(γ-β)
=sin(α+β+γ-β)= sin(α+γ)
8、原式=sin(α-β)sin(β-γ)-cos(α-β)cos(-(β-γ))
=sin(α-β)sin(β-γ)-cos(α-β)cos(β-γ)
=-(cos(α-β)cos(β-γ)-sin(α-β)sin(β-γ))
=-cos(α-β+β-γ)=-cos(α-γ)
9、原式=(tan(π+π/4)+tan(5π/12))/(1-tan(5π/12)tan(π/4))
=(tanπ/4+tan(5π/12))/(1-tan(5π/12)tan(π/4))
= tan(π/4+5π/12)
= tan(3π/12+5π/12)
= tan(8π/12)
= tan(2π/3)=tan（π-π/3）＝-tanπ/3=-√3
10、(sin(α+β)-2sinαcosβ)/(2sinαsinβ+cos(α+β))
=(sin(α+β)-sin(α+β)+sin(α-β))/(cos(α-β)-cos(α+β)+cos(α+β))
=sin(α-β)/cos(α-β) =tan(α-β)