 | ■No8629に返信(堅持さんの記事)
> △ABCのとき次の証明をせよ。
>
> 1,sinA+sinB+sinC=4cos(A/2)cos(B/2)cos(C/2)
>
> 2,tanA+tanB+tanC=tanAtanBtanC
>
> すいません証明ばかりで、よろしくお願いします。
1,A+B+C=πより，C=π-(A+B)
sinA+sinB+sinC=sinA+sinB+sin{π-(A+B)}=sinA+sinB+sin(A+B)
=sinA+sinB+sinAcosB+cosAsinB
=sinA(1+cosB)+sinB(1+cosA)
=sinA*2cos^2 (B/2)+sinB*2cos^2 (A/2)
=4sin(A/2)cos(A/2)cos^2 (B/2)+4sin(B/2)cos(B/2)cos^2 (A/2)
=4cos(A/2)cos(B/2){sin(A/2)cos(B/2)+cos(A/2)+sin(B/2)}
=4cos(A/2)cos(B/2)sin{(A+B)/2}
=4cos(A/2)cos(B/2)sin{(π/2)+C/2)
=4cos(A/2)cos(B/2)cos(C/2)■
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