## [問題]三角函數

### [問題]三角函數

f(n)=(sinx)^n+(cosx)^n

qeypour

### Re: [問題]三角函數

qeypour 寫到:f(n)=(sinx)^n+(cosx)^n

☆ ~ 幻 星 ~ ☆

### Re: [問題]三角函數

☆ ~ 幻 星 ~ ☆ 寫到:
qeypour 寫到:f(n)=(sinx)^n+(cosx)^n

qeypour

[f(5)-f(3)]/f(1)
= [(sinx)^5+(cosx)^5-(sinx)^3-(cosx)^3]/ [sinx+cosx]
={(sinx)^3[(sinx)^2-1]+(cosx)^3[ (cosx)^2-1]}/ [sinx+cosx]
={(sinx)^3[-(cosx)^2]+(cosx)^3[-(sinx)^2]}/ [sinx+cosx]
=-(sinx)^2*(cosx)^2*(sinx+cosx)/ [sinx+cosx]
=-(sinx)^2*(cosx)^2

[f(7)-f(5)]/f(3)
= [(sinx)^7+(cosx)^7-(sinx)^5-(cosx)^5]/ [(sinx)^3+(cosx)^3]
={(sinx)^5[(sinx)^2-1]+(cosx)^5[ (cosx)^2-1]}/ [(sinx)^3+(cosx)^3]
={(sinx)^5[-(cosx)^2]+(cosx)^5[-(sinx)^2]}/ [(sinx)^3+(cosx)^3]
=-(sinx)^2*(cosx)^2*[(sinx)^3+(cosx)^3]/ [(sinx)^3+(cosx)^3]
=-(sinx)^2*(cosx)^2

[f(5)-f(3)]/f(1)= [f(2k+3)-f(2k+1)]/f(2k-1), k=1,2,3,4,---

[f(5)-f(3)]/f(1)= [f(2k+4)-f(2k+2)]/f(2k), k=1,2,3,4,---

[f(5)-f(3)]/f(1)= [f(m+4)-f(m+2)]/f(m), m=0,1,2,3,4,---

p

p 寫到:另解
[f(5)-f(3)]/f(1)
= [(sinx)^5+(cosx)^5-(sinx)^3-(cosx)^3]/ [sinx+cosx]
={(sinx)^3[(sinx)^2-1]+(cosx)^3[ (cosx)^2-1]}/ [sinx+cosx]
={(sinx)^3[-(cosx)^2]+(cosx)^3[-(sinx)^2]}/ [sinx+cosx]
=-(sinx)^2*(cosx)^2*(sinx+cosx)/ [sinx+cosx]
=-(sinx)^2*(cosx)^2

[f(7)-f(5)]/f(3)
= [(sinx)^7+(cosx)^7-(sinx)^5-(cosx)^5]/ [(sinx)^3+(cosx)^3]
={(sinx)^5[(sinx)^2-1]+(cosx)^5[ (cosx)^2-1]}/ [(sinx)^3+(cosx)^3]
={(sinx)^5[-(cosx)^2]+(cosx)^5[-(sinx)^2]}/ [(sinx)^3+(cosx)^3]
=-(sinx)^2*(cosx)^2*[(sinx)^3+(cosx)^3]/ [(sinx)^3+(cosx)^3]
=-(sinx)^2*(cosx)^2

[f(5)-f(3)]/f(1)= [f(2k+3)-f(2k+1)]/f(2k-1), k=1,2,3,4,---

[f(5)-f(3)]/f(1)= [f(2k+4)-f(2k+2)]/f(2k), k=1,2,3,4,---

[f(5)-f(3)]/f(1)= [f(m+4)-f(m+2)]/f(m), m=0,1,2,3,4,---

3Q~~~~~~~~~~~~~~~

qeypour