Jika A + B + C = π, maka 
adalah
izharulhaq78
![\displaystyle A+B+C=180\\B+C=180-A\\\\\frac{\sin\dfrac12A+\cos\left(\dfrac12[B+C]+A\right)}{\sin\left(B+C+\dfrac12A\right)}=\frac{\sin\dfrac12A+\cos\left(\dfrac12[180-A]+A\right)}{\sin\left(\pi-A+\dfrac12A\right)}\\\frac{\sin\dfrac12A+\cos\left(\dfrac12[B+C]+A\right)}{\sin\left(B+C+\dfrac12A\right)}=\frac{\sin\dfrac12A+\cos\left(90-\dfrac12A+A\right)}{\sin\left(\pi-\dfrac12A\right)}](https://tex.z-dn.net/?f=%5Cdisplaystyle+A%2BB%2BC%3D180%5C%5CB%2BC%3D180-A%5C%5C%5C%5C%5Cfrac%7B%5Csin%5Cdfrac12A%2B%5Ccos%5Cleft%28%5Cdfrac12%5BB%2BC%5D%2BA%5Cright%29%7D%7B%5Csin%5Cleft%28B%2BC%2B%5Cdfrac12A%5Cright%29%7D%3D%5Cfrac%7B%5Csin%5Cdfrac12A%2B%5Ccos%5Cleft%28%5Cdfrac12%5B180-A%5D%2BA%5Cright%29%7D%7B%5Csin%5Cleft%28%5Cpi-A%2B%5Cdfrac12A%5Cright%29%7D%5C%5C%5Cfrac%7B%5Csin%5Cdfrac12A%2B%5Ccos%5Cleft%28%5Cdfrac12%5BB%2BC%5D%2BA%5Cright%29%7D%7B%5Csin%5Cleft%28B%2BC%2B%5Cdfrac12A%5Cright%29%7D%3D%5Cfrac%7B%5Csin%5Cdfrac12A%2B%5Ccos%5Cleft%2890-%5Cdfrac12A%2BA%5Cright%29%7D%7B%5Csin%5Cleft%28%5Cpi-%5Cdfrac12A%5Cright%29%7D "\displaystyle A+B+C=180\\B+C=180-A\\\\\frac{\sin\dfrac12A+\cos\left(\dfrac12[B+C]+A\right)}{\sin\left(B+C+\dfrac12A\right)}=\frac{\sin\dfrac12A+\cos\left(\dfrac12[180-A]+A\right)}{\sin\left(\pi-A+\dfrac12A\right)}\\\frac{\sin\dfrac12A+\cos\left(\dfrac12[B+C]+A\right)}{\sin\left(B+C+\dfrac12A\right)}=\frac{\sin\dfrac12A+\cos\left(90-\dfrac12A+A\right)}{\sin\left(\pi-\dfrac12A\right)}")
![\displaystyle \frac{\sin\dfrac12A+\cos\left(\dfrac12[B+C]+A\right)}{\sin\left(B+C+\dfrac12A\right)}=\frac{\sin\dfrac12A+\cos\left(90+\dfrac12A\right)}{\sin\left(\pi-\dfrac12A\right)}\\\frac{\sin\dfrac12A+\cos\left(\dfrac12[B+C]+A\right)}{\sin\left(B+C+\dfrac12A\right)}=\frac{\sin\dfrac12A-\sin\dfrac12A}{\sin\left(\pi-\dfrac12A\right)}\\\frac{\sin\dfrac12A+\cos\left(\dfrac12[B+C]+A\right)}{\sin\left(B+C+\dfrac12A\right)}=\frac{0}{\sin\left(\pi-\dfrac12A\right)}](https://tex.z-dn.net/?f=%5Cdisplaystyle+%5Cfrac%7B%5Csin%5Cdfrac12A%2B%5Ccos%5Cleft%28%5Cdfrac12%5BB%2BC%5D%2BA%5Cright%29%7D%7B%5Csin%5Cleft%28B%2BC%2B%5Cdfrac12A%5Cright%29%7D%3D%5Cfrac%7B%5Csin%5Cdfrac12A%2B%5Ccos%5Cleft%2890%2B%5Cdfrac12A%5Cright%29%7D%7B%5Csin%5Cleft%28%5Cpi-%5Cdfrac12A%5Cright%29%7D%5C%5C%5Cfrac%7B%5Csin%5Cdfrac12A%2B%5Ccos%5Cleft%28%5Cdfrac12%5BB%2BC%5D%2BA%5Cright%29%7D%7B%5Csin%5Cleft%28B%2BC%2B%5Cdfrac12A%5Cright%29%7D%3D%5Cfrac%7B%5Csin%5Cdfrac12A-%5Csin%5Cdfrac12A%7D%7B%5Csin%5Cleft%28%5Cpi-%5Cdfrac12A%5Cright%29%7D%5C%5C%5Cfrac%7B%5Csin%5Cdfrac12A%2B%5Ccos%5Cleft%28%5Cdfrac12%5BB%2BC%5D%2BA%5Cright%29%7D%7B%5Csin%5Cleft%28B%2BC%2B%5Cdfrac12A%5Cright%29%7D%3D%5Cfrac%7B0%7D%7B%5Csin%5Cleft%28%5Cpi-%5Cdfrac12A%5Cright%29%7D "\displaystyle \frac{\sin\dfrac12A+\cos\left(\dfrac12[B+C]+A\right)}{\sin\left(B+C+\dfrac12A\right)}=\frac{\sin\dfrac12A+\cos\left(90+\dfrac12A\right)}{\sin\left(\pi-\dfrac12A\right)}\\\frac{\sin\dfrac12A+\cos\left(\dfrac12[B+C]+A\right)}{\sin\left(B+C+\dfrac12A\right)}=\frac{\sin\dfrac12A-\sin\dfrac12A}{\sin\left(\pi-\dfrac12A\right)}\\\frac{\sin\dfrac12A+\cos\left(\dfrac12[B+C]+A\right)}{\sin\left(B+C+\dfrac12A\right)}=\frac{0}{\sin\left(\pi-\dfrac12A\right)}")
![\displaystyle \boxed{\boxed{\frac{\sin\dfrac12A+\cos\left(\dfrac12[B+C]+A\right)}{\sin\left(B+C+\dfrac12A\right)}=0}}](https://tex.z-dn.net/?f=%5Cdisplaystyle+%5Cboxed%7B%5Cboxed%7B%5Cfrac%7B%5Csin%5Cdfrac12A%2B%5Ccos%5Cleft%28%5Cdfrac12%5BB%2BC%5D%2BA%5Cright%29%7D%7B%5Csin%5Cleft%28B%2BC%2B%5Cdfrac12A%5Cright%29%7D%3D0%7D%7D "\displaystyle \boxed{\boxed{\frac{\sin\dfrac12A+\cos\left(\dfrac12[B+C]+A\right)}{\sin\left(B+C+\dfrac12A\right)}=0}}")
2 votes
Thanks 8
puccii
bisa gak sih kalau A=B=C ?
izharulhaq78
selama gak ada keterangannya, gak bisa
Callahan46
soal pendek jawaban panjang hasilnya 0 nyesek banget