诱导公式 200分1.化简:sin(30°+α)×tan(45°+α)×tan(45°-α) sec(60°-α)2.求值:sin²1°+sin²2°+sin²3°+.+sin²89°3.已知θ=(0,2π),且sinθ、COSθ是方程X²-KX+K-1=0的根 求K和θ的值4.已知SINα+SINβ+SINr+0,且cosα+cosβ+cosr=0求证：cos(α-β)= 负的二分之一

1,sec(60°-α)=1/cos(60°-α)=1/sin[90-(60°-α)]=1/sin(30°+α)
而tan(45°-α)=1/tan(45°+α)
这样原式=1
2sin²89°=cos²1°……前后相加得44+sin²45°=44.5
3，sinθ+COSθ=k sinθCOSθ=k-1
而sin²θ+cos²θ=1
所以k²=sin²θ+cos²θ+2sinθCOSθ=k²-2k+2
所以k=1
sinθCOSθ=0 而θ=(0,2π),所以θ=π/2或=π
4，sin²r+cos²r=1
(SINα+SINβ)²+(cosα+cosβ)²=1
有2+2(SINαSINβ+cosαcosβ)=2+2cos(α-β)= 1
所以cos(α-β)= 负的二分之一

1、sin(30°+α)×tan(45°+α)×tan(45°-α) sec(60°-α)
=sin(30°+α)×tan(45°+α)×tan(45°-α)/ cos(60°-α)
=[sin(30°+α)/ cos(60°-α)]×tan(45°+α)×tan(45°-α)
=[sin(30°+α)/sin(30°+α)]×tan(45°+α)×tan(45°-α)
=tan(45°+α)×tan(90°-45°-α)
=tan(45°+α)×cot(45°+α)
=1
2、sin²1°+sin²2°+sin²3°+….+sin²89°
= (sin²1°+ sin²89°)+ (sin²2°+ sin²88°) + (sin²3°+ sin²87°)+…+ (sin²44°+ sin²46°)+ sin²45°
= (sin²1°+ cos²1°)+ (sin²2°+ cos²2°) + (sin²3°+ cos²3°)+…+ (sin²44°+ cos²44°)+ sin²45°
=1×44+1/2
=44.5
3、由韦达定理得
sinθ+cosθ=k ............ ①
sinθ*cosθ=k-1
①式两边平方
（sinθ+cosθ）²=k²
1+2sinθ*cosθ=k²
将sinθ*cosθ=k-1带入
1+2*（k-1）=k²
解得k=1
4、sinθ+cosθ=1
sinθ*cosθ=0
已知θ属于（0，2π），
所以θ=π/2
∵sina+sinb+sinr=0
∴sinr=-(sina+sinb)........ ①
∵cosa+cosb+cosr=0
∴cosr=-(cosa+cosb)......... ②
①式平方+②式平方,得
1=(sina+sinb)^2+(cosa+cosb)^2
=2+2(sinasinb+cosacosb)
=2+2cos(a-b)
∴cos(a-b)=-1/2
不准抄袭~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

1,sec(60°-α)=1/cos(60°-α)=1/sin[90-(60°-α)]=1/sin(30°+α)
而tan(45°-α)=1/tan(45°+α)
这样原式=1
2,前后相加
sin²1°+sin²2°+sin²3°+….+sin²89°
= (sin²1°+ sin²89°)+ (sin²2°+ sin²88°) + (sin²3°+ sin²87°)+…+ (sin²44°+ sin²46°)+ sin²45°
= (sin²1°+ cos²1°)+ (sin²2°+ cos²2°) + (sin²3°+ cos²3°)+…+ (sin²44°+ cos²44°)+ sin²45°
=1×44+1/2
=44.5
3，转换
sinθ+COSθ=k,sinθ×COSθ=k-1
sin²θ+COS²θ+2sinθ×COSθ=k²
1+2（k-1)=k²
1+2k-2=k²
k²-2k+1=0
(k-1)²=0
k=1
θ=(0,2π), θ=π/2
4，证明：
cos(α-β)= cosαcosβ+sinαsinβ
SINα+SINβ=-SINr(1); cosα+cosβ=-cosr（2）
把（1）两边平方得：sin²α+2sinαsinβ+sin²β=sin²r (3)
把（2）两边平方得: cos²α+2cosαcosβ+cos²β=cos²r (4)
(3)+(4) 得 1+2(sinαsinβ+cosαcosβ)+1=1
2(sinαsinβ+cosαcosβ)=-1
cosαcosβ+sinαsinβ=-1/2
cos(α-β)= -1/2

sin(30°+α)×tan(45°+α)×tan(45°-α) sec(60°-α)
=sin(30°+α)×tan(45°+α)×tan(45°-α)/ cos(60°-α)
=[sin(30°+α)/ cos(60°-α)]×tan(45°+α)×tan(45°-α)
=[sin(30°+α)/sin(30°+α)]×tan(45°+α)×tan(45°-α)
=tan(45°+α)×tan(90°-45°-α)
=tan(45°+α)×cot(45°+α)
=1
sin²1°+sin²2°+sin²3°+….+sin²89°
= (sin²1°+ sin²89°)+ (sin²2°+ sin²88°) + (sin²3°+ sin²87°)+…+ (sin²44°+ sin²46°)+ sin²45°
= (sin²1°+ cos²1°)+ (sin²2°+ cos²2°) + (sin²3°+ cos²3°)+…+ (sin²44°+ cos²44°)+ sin²45°
=1×44+1/2
=44.5
由韦达定理得
sinθ+cosθ=k ............ ①
sinθ*cosθ=k-1
①式两边平方
（sinθ+cosθ）²=k²
1+2sinθ*cosθ=k²
将sinθ*cosθ=k-1带入
1+2*（k-1）=k²
解得k=1
sinθ+cosθ=1
sinθ*cosθ=0
已知θ属于（0，2π），
所以θ=π/2
∵sina+sinb+sinr=0
∴sinr=-(sina+sinb)........ ①
∵cosa+cosb+cosr=0
∴cosr=-(cosa+cosb)......... ②
①式平方+②式平方,得
1=(sina+sinb)^2+(cosa+cosb)^2
=2+2(sinasinb+cosacosb)
=2+2cos(a-b)
∴cos(a-b)=-1/2

1.化简:sin(30°+α)×tan(45°+α)×tan(45°-α) sec(60°-α) =sin(30°+α)×tan(45°+α)×tan(45°-α)/ cos(60°-α) =sin(30°+α)/[cos(90°-30°-α]×tan(45°+α)×tan(90°-45°-α) =s...

2.求值:sin²1°+sin²2°+sin²3°+。。。。+sin²89° sin(π/2-α)=cosα
sin²89°=cos²1°，
=(sin²1°+sin²89°)+(sin²2°+sin²88°）+....+（sin²44°+sin²46°)+sin²45°+....
=1+1+...1+1/2=89/2
1.化简:sin(30°+α)×tan(45°+α)×tan(45°-α) sec(60°-α)
=sin(30°+α)×tan(45°+α)×tan(45°-α)/ cos(60°-α)
=sin(30°+α)/[cos(90°-30°-α]×tan(45°+α)×tan(90°-45°-α)
=sin(30°+α)/cos[90°-(30°+α)]×tan(45°+α)×tan[90°-(45+α)]
=sin(30°+α)/sin(30°+α)×tan(45°+α)×cot(45+α)]
=1
3.已知θ=(0,2π),且sinθ、COSθ是方程X²-KX+K-1=0的根 求K和θ的值
sinθ+COSθ=k,sinθ×COSθ=k-1
sin²θ+COS²θ+2sinθ×COSθ=k²
1+2（k-1)=k²
1+2k-2=k²
k²-2k+1=0
(k-1)²=0
k=1
θ=(0,2π), θ=π/2
4.已知SINα+SINβ+SINr=0,且cosα+cosβ+cosr=0
求证：cos(α-β)= 负的二分之一

cos(α-β)= cosαcosβ+sinαsinβ
SINα+SINβ=-SINr(1); cosα+cosβ=-cosr（2）
把（1）两边平方得：sin²α+2sinαsinβ+sin²β=sin²r (3)
把（2）两边平方得: cos²α+2cosαcosβ+cos²β=cos²r (4)
(3)+(4) 得 1+2(sinαsinβ+cosαcosβ)+1=1
2(sinαsinβ+cosαcosβ)=-1
cosαcosβ+sinαsinβ=-1/2
cos(α-β)= -1/2