Proszę o pomoc
ania: Zamień na postać iloczynową sinα+sinβ+sinγ, gdzie α,β,γ−kąty trójkąta
29 lis 19:07
M:
1 mar 06:03
Mei Lin:
α+β+γ=180
o
y=180
o−(α+β)
sinα+sinβ+sinγ=sinα+sinβ+sin(180
o−(α+β)=(sinα+sinβ)+sin(α+β)=
| | α+β | | α−β) | | α+β | | α+β | |
=2sin( |
| )*cos( |
| )+2sin( |
| )*cos( |
| )= |
| | 2 | | 2 | | 2 | | 2 | |
| | α+β | | α−β | | α+β | |
=2sin( |
| )[cos( |
| )+cos( |
| )]= |
| | 2 | | 2 | | 2 | |
| | α+β | | α | | β | | γ | | α | | β | |
=4sin( |
| )*cos |
| *cos |
| =4cos |
| *cos |
| *cos |
| |
| | 2 | | 2 | | 2 | | 2 | | 2 | | 2 | |
1 mar 17:25