Question

高中数学 设cosa =-根号5/5,tanb=1/3,π<a<3π/2,0<b<π/2,求a-b的值

cosa=-√5/5,π<α<3π/2
sina=-2√5/5

tanβ=1/3,,0<β<π/2
sinβ=√10/10
cosβ=3√10/10

0<β<π/2
-π/2<-β<0
π/2<α-β<3π/2

sin(α-β)
=sinacosβ-cosasinβ
=-2√5/5*3√10/10-(-√5/5)*√10/10
=-6√50/50+√50/50
=-5√50/50
=-√50/10
=-5√2/10
=-√2/2

π/2<α-β<3π/2
π<α-β<3π/2

所以sin(α-β)=-√2/2
即α-β=5π/4 答案是这样吗？？

Answer 1

结果是对的
你做得太麻烦啦；用正切比较简单：
tanα=sinα/cosα=2,
tanβ=1/3;
tan(α-β)=(tanα-tanβ)/[1+tanαtanβ]=(5/3)/(5/3)=1;
又: π<α<3π/2; 0＜β<π/2
所以：π/2<α-β<3π/2;
;所以：α-β=5π/4

Answer 2

因为π＜a＜3π/2,
cosa=-√5/5令x=-√5 r=5 所以y= - 2√5
tana=y/x=2
tan(a-b)=(2-1/3)/[1+2*(1/3)]=3*(2-1/3)/3*[1+2*(1/3)]=(6-1)/(3+2)=1
π ＜ a ≤ 3π/2
0<b<π/2 ==> -π/2 < -b < 0
...........................................................两式相加得：
π/2＜a-b＜3π/2
所以a-b=5π/4

Answer 3

非常正确！

Answer 4

是

Answer 5

对