中四數學M2題, 急!

第一題
lim x->π cos(π/2)除(x-π)
=lim x->π 1除1 (By L'Hospital Rule)
=1如果題目是 lim x->π cos(x/2)除(x-π)
lim x->π cos(x/2)除(x-π)
=lim x->π (-sin(x/2))*(1/2)除(1) (By L'Hospital Rule)
=(-sin(π/2))*(1/2)
=-1/2第二題
2^x^3=5^x^2 ?是2乘以(x的3次方)=5乘以(x的2次方)的話,
2*x^3=5*x^2
(x^2)(2*x-5)=0
x^2=0 or 2*x-5=0
x=0 or x=5/2是2的(x^3)次方=5的(x^2)次方的話,
2^(x^3)=5^(x^2)
log(2^(x^3))=log(5^(x^2))
(x^3)*log(2)=(x^2)*log(5)
(x^2)*(x*log(2)-log(5))=0
x^2=0 or x*log(2)-log(5)=0
x=0 or x=(log(5))/(log(2))第三題
LHS
=cos(A+B)cos(A-B)
=(cos(A)cos(B)-sin(A)sin(B))(cos(A)cos(B)+sin(A)sin(B))
=((cos(A))^2)((cos(B))^2)-((sin(A))^2)((sin(B))^2)
=(1-(sin(A))^2)((cos(B))^2)-((sin(A))^2)((sin(B))^2)
=((cos(B))^2)-((sin(A))^2)((cos(B))^2)-((sin(A))^2)((sin(B))^2)
=((cos(B))^2)-((sin(A))^2)((cos(B))^2+(sin(B))^2)
=(cos(B))^2-(sin(A))^2
=(cos(B))^2-(1-(cos(A))^2)
=(cos(A))^2+(cos(B))^2-1
=RHS

2011-07-03 08:50:48 補充：
我在微分時犯了錯誤

糾正:
第一題
lim x->π cos(π/2)除(x-π)
=lim x->π 0除1 (By L'Hospital Rule)
=0

謝謝你的指出

1) lim x-> π_Cos(π/2)/(x－π)
Sol
Cos(π/2)=0
x－π->0，x－π->0，x－π<>0forever
So
limx-> π_Cos(π/2)/(x－π)=0解答錯誤

1A ) lim x-> π_Cos(x/2)/(x－π)
Sol
lim x-> π_Cos(x/2)/(x－π) 0/0 type
= lim x-> π_－(1/2)Sin(x/2)/1
=－1/2

2) 2^(x^3) = 5^(x^2) 求x
Sol
2^(x^3)=5^(x^2)
log[2^(x^3)]=log[5^(x^2)]
x^3log2=x^2log5
x^2(xlog2－log5)=0
x=0or x=log5/log2

3) Cos(A+B)Cos(A－B)=Cos^ 2 A +Cos^2 B－1
Sol
Cos(A+B)Cos(A－B)
=[CosACosB－SinASinB](CosACosb+SinASinB)
=Cos^2 ACos^2 B－Sin^2 ASin^2B
=Cos^2 ACos^2 B－(1－Cos^ 2 A )(1－Cos^2 B)
=Cos^2 ACos^2 B－(1－Cos^2 B－Cos^ 2 A +Cos^2 ACos^2 B)
=Cos^ 2 A +Cos^2 B－1