LG a
Cho \(z = c{\rm{os}}\varphi {\rm{ + }}i\sin \varphi \left( {\varphi \in R} \right)\). Chứng minh rằng với mọi số nguyên \(n \ge 1\), ta có
\({z^n} + {1 \over {{z^n}}} = 2\cos n\varphi ,{z^n} - {1 \over {{z^n}}} = 2i\sin n\varphi \)
Giải chi tiết:
\({z^n} = \cos n\varphi + i\sin n\varphi ,{1 \over {{z^n}}} = \cos n\varphi - i\sin n\varphi \) nên
\({z^n} + {1 \over {{z^n}}} = 2\cos n\varphi ,{z^n} - {1 \over {{z^n}}} = 2i\sin n\varphi \)
(Đặc biệt \({z} + {1 \over z} = 2\cos \varphi ,z - {1 \over z} = 2i\sin \varphi \)).
LG b
Từ câu a), chứng minh rằng
\(c{\rm{o}}{{\rm{s}}^4}\varphi = {1 \over 8}\left( {{\rm{cos4}}\varphi + 4\cos 2\varphi + 3} \right)\)
\({\sin ^5}\varphi = {1 \over {16}}\left( {\sin 5\varphi - 5\sin 3\varphi + 10\sin \varphi } \right)\)
Giải chi tiết:
\(c{\rm{o}}{{\rm{s}}^4}\varphi = {\left[ {{1 \over 2}\left( {z + {1 \over z}} \right)} \right]^{ - 4}} \)
\(= {1 \over {{2^4}}}\left[ {{z^4} + {1 \over {{z^4}}} + C_4^1\left( {{z^2} + {1 \over {{z^2}}}} \right) + C_4^2} \right]\)
\( = {1 \over {{2^4}}}\left( {2\cos 4\varphi + 4.2cos2\varphi + 6} \right) \)
\(= {1 \over 8}\left( {\cos 4\varphi + 4cos2\varphi + 3} \right)\)
\({\sin ^5}\varphi = {\left[ {{1 \over {2i}}\left( {z - {1 \over z}} \right)} \right]^5}\)
\( = {1 \over {{2^5}i}}\left[ {\left( {{z^5} - {1 \over {{z^5}}}} \right) - C_5^1\left( {{z^3} - {1 \over {{z^3}}}} \right) + C_5^2\left( {z - {1 \over z}} \right)} \right]\)
\( = {1 \over {{2^5}}}\left( {2\sin 5\varphi - 2C_5^1\sin 3\varphi + 2C_5^2\sin \varphi } \right)\)
\(={1 \over {16}}\left( {\sin 5\varphi - 5\sin 3\varphi + 10\sin \varphi } \right)\).
