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ID:34402030
大小:3.35 MB
页数:429页
时间:2019-03-05
《数理方程习题全解new》由会员上传分享,免费在线阅读,更多相关内容在教育资源-天天文库。
1、习题11.1计算下列各式的值,并且写出相应的三角表达式和指数表达式444cosjsin22j11j44(1)255513j213jcosjsin2233cosjsin21j355cosjsin33222j1j3,2cosjsin,2e3。33551355(2)jcosjsincosjsin22333313j2
2、213jj,cosjsin,e3。22331322(3)13j2j2cosjsin2233222k2k332cosjsin222coskjsink33kk21cosj1sin33k211j3k1,02221j3,1j3;2291222jj2cosjsin,2cosjsin;2e3,2e3。33331311
3、4(4)422j422j28cosjsin224432k2k28cos4jsin4443kk28cosjsink3,2,1,0216216337728cosjsin,28cosjsin,161616163315159928cosjsin,28cosjsin;1616161633731539jjjj28e16,28e
4、16,28e16,28e16。1.2证明nknkkk(1)cosnjsinnCncossinjn3,2,1;k0nnknkk证cosnjsinncosjsinCncosjsink0nknkkkCncossinjk01sinnn12(2)cosk21k12sin2nnn1ejn11cosn1jsinn1jk证coskjsinkejk0k0k01e1cosjsin1cosn1jsinn1
5、1cosjsin21cos92n1coscosn1coscosn1sinsinn1coskk021cos1coscosn1sinsinn1cosn1224sin21cosn1cosn11cosncosn122224sin4sin222nnn12sinsinsinn122122224sin2sin22此式称为拉格朗日三角恒等式。从上式可以得到1sinnnn1
6、2coskcosk12k1k02sin21.3解方程3(1)z8j0;解z38j23j23cosjsin2222z2coskjsinkk2,1,0363631k0:z2cosjsin2j3j662222k1:z2cosjsin2cosjsin2j3636224477k2:z2cos
7、jsin2cosjsin363666312cosjsin2j3j66224(2)z40;解z4424124cosjsin932k12k1=2cosjsink3,2,1,04411k0:z12cosjsin2j1j44223311k1:z22
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