Saved Bookmarks
| 1. |
\frac { \operatorname { sin } A + \operatorname { sin } 3 A + \operatorname { sin } 5 A + \operatorname { sin } 7 A } { \operatorname { cos } A + \operatorname { cos } 3 A + \operatorname { cos } 5 A + \operatorname { cos } 7 A } = \operatorname { tan } 4 A |
|
Answer» SinA+sin3A+sin5A+sin7A/cosA+cos3A+cos5A+cos7A=(sinA+sin7A)+(sin3A+sin5A)/(cosA+cos7A)+(cos3A+cos5A)=(2sin4Acos3A+2sin4AcosA)/(2cos4Acos3A+2cos4AcosA)=2sin4A(cos3A+cosA)/2cos4A(cos3A+cosA)=sin4A/cos4A=tan4A (Proved) thanks you |
|