- LG a
- LG b
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:
\[c{\rm{o}}{{\rm{s}}^4}\varphi = {\left[ {{1 \over 2}\left[ {z + {1 \over z}} \right]} \right]^{ - 4}} \]
\[= {1 \over {{2^4}}}\left[ {{z^4} + {1 \over {{z^4}}} + C_4^1\left[ {{z^2} + {1 \over {{z^2}}}} \right] + C_4^2} \right]\]
\[ = {1 \over {{2^4}}}\left[ {2\cos 4\varphi + 4.2cos2\varphi + 6} \right] \]
\[= {1 \over 8}\left[ {\cos 4\varphi + 4cos2\varphi + 3} \right]\]
\[{\sin ^5}\varphi = {\left[ {{1 \over {2i}}\left[ {z - {1 \over z}} \right]} \right]^5}\]
\[ = {1 \over {{2^5}i}}\left[ {\left[ {{z^5} - {1 \over {{z^5}}}} \right] - C_5^1\left[ {{z^3} - {1 \over {{z^3}}}} \right] + C_5^2\left[ {z - {1 \over z}} \right]} \right]\]
\[ = {1 \over {{2^5}}}\left[ {2\sin 5\varphi - 2C_5^1\sin 3\varphi + 2C_5^2\sin \varphi } \right]\]
\[={1 \over {16}}\left[ {\sin 5\varphi - 5\sin 3\varphi + 10\sin \varphi } \right]\].