Question

chứng minh rằng tam giác ABC

cos \(\dfrac{3A+2B+C}{2}\)= -sin \(\left(A+\dfrac{B}{2}\right)\)

Answer 1

ΔABC có góc A+góc B+góc C=180 độ

=>3*góc A+2*góc B+góc C=180 độ+2*góc A+góc B

=>\(\dfrac{3A+2B+C}{2}=90^0+A+\dfrac{B}{2}\)

=>\(cos\left(\dfrac{3A+2B+C}{2}\right)=-sin\left(A+\dfrac{B}{2}\right)\)