Question

Trong tam giacsbABC thỏa các đẳng thức sau. Tìm các hệ số a;b;c

1) sinA + sinB + sinC = a+bcosA/2.cosB/2.cosC/2.

2) sin4A+sin4B+sin4C = a+bsin2A.sin2B.sin2C.

3) cos4A+cos4B+cos4c = a+bcos2A.cos2B.cos2C.

4) cos²A +cos²B+cos²C = a+bcosA.cosB.cosC

Answer 1

3/

\(cos4A+cos4B+cos4C=2cos\left(2A+2B\right).cos\left(2A-2B\right)+2cos^22C-1\)

\(=2cos2C.cos\left(2A-2B\right)+2cos^22C-1\)

\(=2cos2C\left(cos\left(2A-2B\right)+cos2C\right)-1\)

\(=2cos2C\left(cos\left(2A-2B\right)+cos\left(2A+2B\right)\right)-1\)

\(=4cos2A.cos2B.cos2C-1\Rightarrow\left\{{}\begin{matrix}a=-1\\b=4\end{matrix}\right.\)

4/

\(cos^2A+cos^2B+cos^2C=\frac{1}{2}+\frac{1}{2}cos2A+\frac{1}{2}+\frac{1}{2}cos2B+\frac{1}{2}+\frac{1}{2}cos2C\)

\(=\frac{3}{2}+\frac{1}{2}\left(cos2A+cos2B+cos2C\right)\)

\(=\frac{3}{2}+\frac{1}{2}\left[2cos\left(A+B\right).cos\left(A-B\right)+2cos^2C-1\right]\)

\(=1+\frac{1}{2}\left(-2cosC.cos\left(A-B\right)+2cos^2C\right)\)

\(=1-cosC\left(cos\left(A-B\right)-cosC\right)\)

\(=1-cosC\left(cos\left(A-B\right)+cos\left(A+B\right)\right)\)

\(=1-2cosA.cosB.cosC\) \(\Rightarrow\left\{{}\begin{matrix}a=1\\b=-2\end{matrix}\right.\)

Answer 2

1/ \(sinA+sinB+sin2\frac{C}{2}=2sin\frac{A+B}{2}cos\frac{A-B}{2}+2sin\frac{C}{2}cos\frac{C}{2}\)

\(=2cos\frac{C}{2}.cos\frac{A-B}{2}+2cos\frac{A+B}{2}.cos\frac{C}{2}=2cos\frac{C}{2}\left(cos\frac{A-B}{2}+cos\frac{A+B}{2}\right)\)

\(=4cos\frac{A}{2}cos\frac{B}{2}cos\frac{C}{2}\Rightarrow\left\{{}\begin{matrix}a=0\\b=4\end{matrix}\right.\)

2/ \(sin4A+sin4B+sin4C=2sin\left(2A+2B\right)cos\left(2A-2B\right)+2sin2C.cos2C\)

\(=-2sin2C.cos\left(2A-2B\right)+2sin2C.cos2C\)

\(\)\(=2sin2C\left(cos2C-cos\left(2A-2B\right)\right)\)

\(=-4sin2C.sin\left(C+A-B\right)sin\left(C-A+B\right)\)

\(=-4sin2A.sin2B.sin2C\Rightarrow\left\{{}\begin{matrix}a=0\\b=-4\end{matrix}\right.\)