Jika A+B+C=
, maka 
A. -2
B. -1
C. 0
D. 1
E. 2
A. -2
B. -1
C. 0
D. 1
E. 2
izharulhaq78
![\displaystyle \text{ubah }\pi\text{ menjadi }180^\circ\text{ agar lebih mudah (kalau tidak mau tak apa)}\\A+B+C=\pi\\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(180-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(180-\dfrac12A\right)}](https://tex.z-dn.net/?f=%5Cdisplaystyle+%5Ctext%7Bubah+%7D%5Cpi%5Ctext%7B+menjadi+%7D180%5E%5Ccirc%5Ctext%7B+agar+lebih+mudah+%28kalau+tidak+mau+tak+apa%29%7D%5C%5CA%2BB%2BC%3D%5Cpi%5C%5CA%2BB%2BC%3D180%5C%5CB%2BC%3D180-A%5C%5C%5C%5C%5Cfrac%7B%5Csin%5Cdfrac12A%2B%5Ccos%5Cleft%5B%5Cdfrac12%28B%2BC%29%2BA%5Cright%5D%7D%7B%5Csin%5Cleft%28B%2BC%2B%5Cdfrac12A%5Cright%29%7D%3D%5Cfrac%7B%5Csin%5Cdfrac12A%2B%5Ccos%5Cleft%5B%5Cdfrac12%28180-A%29%2BA%5Cright%5D%7D%7B%5Csin%5Cleft%28180-A%2B%5Cdfrac12A%5Cright%29%7D%5C%5C%5Cfrac%7B%5Csin%5Cdfrac12A%2B%5Ccos%5Cleft%5B%5Cdfrac12%28B%2BC%29%2BA%5Cright%5D%7D%7B%5Csin%5Cleft%28B%2BC%2B%5Cdfrac12A%5Cright%29%7D%3D%5Cfrac%7B%5Csin%5Cdfrac12A%2B%5Ccos%5Cleft%5B90-%5Cdfrac12A%2BA%5Cright%5D%7D%7B%5Csin%5Cleft%28180-%5Cdfrac12A%5Cright%29%7D "\displaystyle \text{ubah }\pi\text{ menjadi }180^\circ\text{ agar lebih mudah (kalau tidak mau tak apa)}\\A+B+C=\pi\\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(180-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(180-\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(180-\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(180-\dfrac12A\right)}\\\frac{\sin\dfrac12A+\cos\left[\dfrac12(B+C)+A\right]}{\sin\left(B+C+\dfrac12A\right)}=\frac{0}{\sin\left(180-\dfrac12A\right)}](https://tex.z-dn.net/?f=%5Cdisplaystyle+%5Cfrac%7B%5Csin%5Cdfrac12A%2B%5Ccos%5Cleft%5B%5Cdfrac12%28B%2BC%29%2BA%5Cright%5D%7D%7B%5Csin%5Cleft%28B%2BC%2B%5Cdfrac12A%5Cright%29%7D%3D%5Cfrac%7B%5Csin%5Cdfrac12A%2B%5Ccos%5Cleft%5B90%2B%5Cdfrac12A%5Cright%5D%7D%7B%5Csin%5Cleft%28180-%5Cdfrac12A%5Cright%29%7D%5C%5C%5Cfrac%7B%5Csin%5Cdfrac12A%2B%5Ccos%5Cleft%5B%5Cdfrac12%28B%2BC%29%2BA%5Cright%5D%7D%7B%5Csin%5Cleft%28B%2BC%2B%5Cdfrac12A%5Cright%29%7D%3D%5Cfrac%7B%5Csin%5Cdfrac12A-%5Csin%5Cdfrac12A%7D%7B%5Csin%5Cleft%28180-%5Cdfrac12A%5Cright%29%7D%5C%5C%5Cfrac%7B%5Csin%5Cdfrac12A%2B%5Ccos%5Cleft%5B%5Cdfrac12%28B%2BC%29%2BA%5Cright%5D%7D%7B%5Csin%5Cleft%28B%2BC%2B%5Cdfrac12A%5Cright%29%7D%3D%5Cfrac%7B0%7D%7B%5Csin%5Cleft%28180-%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(180-\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(180-\dfrac12A\right)}\\\frac{\sin\dfrac12A+\cos\left[\dfrac12(B+C)+A\right]}{\sin\left(B+C+\dfrac12A\right)}=\frac{0}{\sin\left(180-\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%5B%5Cdfrac12%28B%2BC%29%2BA%5Cright%5D%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}}")
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