高中三角函数题目

Question

已知sinx+siny=4/5,codx+cosy=3/5 求cos(x-y)
计算 [sin50+cos40*(1+tan60*tan10)]/cos^2 20
化简 (2cos^2 a-1)/[2tan*(45-a)*sin^2(45+a)] （我帮弧度制换成角度值了）
已知cos(a+b)=1/3 cos(a-b)=1/6 求tana*tanb 的值
在△ABC中 求证 tanA+tanB+tanC=tanA*tanB*tanC,
tan(A/2)*tan(B/2)+tan(B/2)*tan(C/2)+tan(C/2)*tan(A/2)=1
建议把过程写清楚点 适合在试卷上用的语言

Answer 1

1.两个式子分别平方： (sinx+siny)^2 =(sinx)^2 + 2sinxsiny + (siny)^2=16/25
(cosx+cosy)^2 =(cosx)^2 + 2cosxcosy + (cosy)^2=9/25
两式相加：(sinx)^2 + (cosx)^2 + 2sinxsiny + 2cosxcosy + (siny)^2 + (cosy)^2=16/25 + 9/25
1 + 2sinxsiny + 2cosxcosy + 1 =1
2(sinxsiny + cosxcosy)=-1
cos(x-y)=-1/2

2. [sin50+cos40*(1+tan60*tan10)]/(cos20)^2
={sin50 + cos40*[1 + (sin60sin10/cos60cos10)]} / (cos20)^2
={sin50 + cos40*[(cos60cos10/cos60cos10) + (sin60sin10/cos60cos10)]} / (cos20)^2
={sin50 + cos40*[(cos60cos10+sin60sin10)/(cos60cos10)]} / (cos20)^2
={sin50 + cos40*[cos(60-10) / (cos60cos10)]} / (cos20)^2
=[sin50 + (cos40cos50)/(cos60cos10)] / (cos20)^2
={sin50 + [cos40cos(90-40) / (1/2)cos10]} / (cos20)^2
=[sin(90-40) / (cos20)^2] + [2cos40sin40 / cos10*(cos20)^2]
=[(cos40) / (1+cos40)/2] + [(sin80) / cos10*(cos20)^2]
=[2cos40 / (1+cos40)] + [sin(90-10) / cos10*(cos20)^2]
=[2cos40 / (1+cos40)] + {(cos10) / cos10*[(1+cos40)/2]}
=(2cos40 + 2)/(1+cos40)
=2

3. (2cos^2 a-1)/[2tan*(45-a)*sin^2(45+a)]
= (2cos^2 a-1)/{2tan*(45-a)*[sin^2(90-(45-a)]}
= (2cos^2 a-1)/[2tan*(45-a)*cos^2(45-a)]
=(cos2a)/[2sin(45-a)cos(45-a)]
=(cos2a)/[sin(90-2a)]
=(cos2a)/(cos2a)=1

4.cos(a+b)=cosacosb - sinasinb=1/3
cos(a-b)=cosacosb + sinasinb=1/6
两式相加：2cosacosb=1/2
两式相减：2sinasinb=-1/6
tana*tanb=(sinasinb)/(cosacosb)=-1/3

5.在三角形中：tanC=tan[π-(A+B)]=-tan(A+B)
根据正切的两角和公式，变形后得：tanA+tanB={[tan(A+B)]*(1-tanAtanB)}
展开：tanA+tanB=tan(A+B) - tan(A+B)tanAtanB
将tanC=-tan(A+B)代入上式：tanA+tanB = -tanC + tanCtanAtanB
整理后： tanA+tanB+tanC = tanAtanBtanC

在三角形中：A+B+C=π ，则：A/2 + B/2 + C/2 =π/2
tan(A/2)*tan(B/2) + tan(B/2)*tan(C/2) + tan(C/2)*tan(A/2)
=tan(A/2)[tan(B/2)+tan(C/2)] + tan(B/2)*tan(C/2)
={tan[π/2 - (B/2 + C/2)]}*{[tan(B/2 + C/2)][1 - tan(B/2)*tan(C/2)]} + tan(B/2)*tan(C/2)
=cot(B/2 + C/2)*[tan(B/2 + C/2)][1 - tan(B/2)*tan(C/2)] + tan(B/2)*tan(C/2)
=1[1 - tan(B/2)*tan(C/2)] + tan(B/2)*tan(C/2)
=1 - tan(B/2)tan(C/2) + tan(B/2)*tan(C/2)
=1

Answer 2

（1）将两个等式分别平方后相加得:
sinx的平方+cosx的平方+siny的平方+cosy的平方+2（sinxsiny+cosxcosy）=7/5
cos(x-y)=sinxsiny+cosxcosy
所以cos(x-y)=-3/10
（4）cosacosb-sinasinb=1/3
cosacosb+sinasinb=1/6
两式相加得cosacosb=1/4
两式相减得sinasinb=-1/12
所以tanatanb=-1/3

Answer 3

1. sinx+siny=4/5 (1)
cosx+cosy=3/5(2) (1)^2+(2)^2 = sin^2x+sin^2y+2sinxsiny+cos^2x+cos^2y+2cosxcosy=16/25+9/25=1
所以cosxcosy+sinxsiny=cos(x-y)=-1/2
2. cos40*(1+tan60*tan10)= cos40*(cos60*cos10+sin60*sin10)/(cos60cos10) = cos40*cos50/(cos60cos10) = sin10*sin50*cos50/(sin10cos10cos60)=sin10*sin100/sin20cos60=sin10*sin80/(sin20*cos60)=sin10*cos10/(sin20*cos60)=sin20/(2sin20cos60)=1
所以 [sin50+cos40*(1+tan60*tan10)]/cos^2 20 = (cos40+1)/[(1+cos40)/2]=2
3.原式=(1+cos2a-1)/(2ctg(45+a)sin^2(45+a))= cos2a/[2sin(45+a)cos(45+a)]=cos2a/sin(90+2a)=cos2a/cos2a= 1
4.cos(a+b)=cosacosb-sinasinb=1/3 cos(a-b)=cosacosb+sinasinb=1/6 两个方程可得 cosacosb=1/4 sinasinb=-1/12 tana*tanb = sinasinb/cosacosb=-1/3
5.(1) tan(A+B)= (tanA+tanB)/(1-tanAtanB) 所以 tanA+tanB= tan(A+B) - tanAtanBtan(A+B)
而tan(A+B)=tan(180-C)=-tanC所以 tanA+tanB= -tanC+tanAtanBtanC 即
tanA+tanB+tanC=tanA*tanB*tanC
(2)tan(A/2+B/2)=(tan(A/2)+tan(B/2))/(1-tanA/2tanB/2) 所以1-tanA/2tanB/2 = (tan(A/2)+tan(B/2))/tan((A+B)/2) = (tan(A/2)+tan(B/2))/tan((180-C)/2)=(tan(A/2)+tan(B/2))/tan(90-C/2)=(tan(A/2)+tan(B/2))/ctan(C/2)=(tan(A/2)+tan(B/2))tan(C/2)
即 1-tanA/2tanB/2 = (tan(A/2)+tan(B/2))tan(C/2) 展开即得
tan(A/2)*tan(B/2)+tan(B/2)*tan(C/2)+tan(C/2)*tan(A/2)=1

Answer 4

答：（1)因为（sinx+siny)^2+(cosx+cosy)^2=（sinx^2+siny^2+2*sin*cosy)+(cosx^2+cosy^2+2*cosx*cosy)=2+2*cos(x-y)=(3/5)^2+(4/5)^2=1
所以可以知道cos(x-y)=-1/2