> 数学 >
证明:(cos30º+i·sin30º)^n=cos(n·30º)+i·sin(n·30º)
人气:373 ℃ 时间:2020-04-15 21:47:45
解答
证明:
当n=1时,cos30°+i·sin30°=cos30°+sin30°,成立
假设当n=k-1时成立,即(cos30°+i·sin30°)^(k-1)=cos[(k-1)·30°]+i·sin[(k-1)·30°]
下面证明当n=k时成立:
(cos30°+i·sin30°)^k
=(cos30°+i·sin30°)^(k-1)×(cos30°+i·sin30°)
={cos[(k-1)·30°]+i·sin[(k-1)·30°]}×(cos30°+i·sin30°)
=cos[(k-1)·30°]cos30°-sin[(k-1)·30°]sin30°
+i·{sin[(k-1)·30°]cos30°+cos[(k-1)·30°]sin30°}
=cos[(k-1)·30°+30°]+i·sin[(k-1)·30°+30°]
=cos(k·30°)+i·sin(k·30°)
所以当n=k时成立.
∴(cos30º+i·sin30º)^n=cos(n·30º)+i·sin(n·30º)
推荐
猜你喜欢
© 2024 79432.Com All Rights Reserved.
电脑版|手机版