CHỈ CÓ CÔNG THỨC :\(\cos\left(\beta-\alpha\right)=\cos\beta\cos\alpha+\sin\beta\sin\alpha\)

viết cái đề mà cũng không đúng thì người khác trả lời làm sao hiểu nổi

a,

\(\begin{array}{l}\cos \left( {\alpha  - b} \right) + \cos \left( {\alpha  + \beta } \right)\\ = \cos \alpha \cos \beta  + \sin \alpha sin\beta  + \cos \alpha \cos \beta  - \sin \alpha sin\beta \\ = 2\cos \alpha \cos \beta \end{array}\)

\(\begin{array}{l}\cos \left( {\alpha  - b} \right) - \cos \left( {\alpha  + \beta } \right)\\ = \cos \alpha \cos \beta  + \sin \alpha sin\beta  - \cos \alpha \cos \beta  + \sin \alpha sin\beta \\ = 2\sin \alpha sin\beta \end{array}\)

b,

\(\begin{array}{l}\sin \left( {\alpha  - \beta } \right) - \sin \left( {\alpha  + \beta } \right)\\ = \sin \alpha \cos \beta  - \cos \alpha sin\beta  - \sin \alpha \cos \beta  - \cos \alpha sin\beta \\ =  - 2\cos \alpha sin\beta \end{array}\)

\(\begin{array}{l}\sin \left( {\alpha  - \beta } \right) + \sin \left( {\alpha  + \beta } \right)\\ = \sin \alpha \cos \beta  - \cos \alpha sin\beta  + \sin \alpha \cos \beta  + \cos \alpha sin\beta \\ = 2\sin \alpha \cos \beta \end{array}\)

2.

ĐK: \(2x-y\ge0;y\ge0;y-x-1\ge0;y-3x+5\ge0\)

\(\left\{{}\begin{matrix}xy-2y-3=\sqrt{y-x-1}+\sqrt{y-3x+5}\left(1\right)\\\left(1-y\right)\sqrt{2x-y}+2\left(x-1\right)=\left(2x-y-1\right)\sqrt{y}\left(2\right)\end{matrix}\right.\)

\(\left(2\right)\Leftrightarrow\left(1-y\right)\sqrt{2x-y}+y-1+2x-y-1-\left(2x-y-1\right)\sqrt{y}=0\)

\(\Leftrightarrow\left(1-y\right)\left(\sqrt{2x-y}-1\right)+\left(2x-y-1\right)\left(1-\sqrt{y}\right)=0\)

\(\Leftrightarrow\left(1-\sqrt{y}\right)\left(\sqrt{2x-y}-1\right)\left(1+\sqrt{y}\right)+\left(\sqrt{2x-y}-1\right)\left(1-\sqrt{y}\right)\left(\sqrt{2x-y}+1\right)=0\)

\(\Leftrightarrow\left(1-\sqrt{y}\right)\left(\sqrt{2x-y}-1\right)\left(\sqrt{y}+\sqrt{2x-y}+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=1\\y=2x-1\end{matrix}\right.\) (Vì \(\sqrt{y}+\sqrt{2x-y}+2>0\))

Nếu \(y=1\), khi đó:

\(\left(1\right)\Leftrightarrow x-5=\sqrt{-x}+\sqrt{-3x+6}\)

Phương trình này vô nghiệm

Nếu \(y=2x-1\), khi đó:

\(\left(1\right)\Leftrightarrow2x^2-5x-1=\sqrt{x-2}+\sqrt{4-x}\) (Điều kiện: \(2\le x\le4\))

\(\Leftrightarrow2x\left(x-3\right)+x-3+1-\sqrt{x-2}+1-\sqrt{4-x}=0\)

\(\Leftrightarrow\left(x-3\right)\left(\dfrac{1}{1+\sqrt{4-x}}-\dfrac{1}{1+\sqrt{x-2}}+2x+1\right)=0\)

Ta thấy: \(1+\sqrt{x-2}\ge1\Rightarrow-\dfrac{1}{1+\sqrt{x-2}}\ge-1\Rightarrow1-\dfrac{1}{1+\sqrt{x-2}}\ge0\)

Lại có: \(\dfrac{1}{1+\sqrt{4-x}}>0\); \(2x>0\)

\(\Rightarrow\dfrac{1}{1+\sqrt{4-x}}-\dfrac{1}{1+\sqrt{x-2}}+2x+1>0\)

Nên phương trình \(\left(1\right)\) tương đương \(x-3=0\Leftrightarrow x=3\Rightarrow y=5\)

Ta thấy \(\left(x;y\right)=\left(3;5\right)\) thỏa mãn điều kiện ban đầu.

Vậy hệ phương trình đã cho có nghiệm \(\left(x;y\right)=\left(3;5\right)\)

\(A=cos^2a+cos^2b+2cosa.cosb+sin^2a+sin^2b+2sina.sinb\)

\(=2+2\left(cosa.cosb+sina.sinb\right)\)

\(=2+2.cos\left(a-b\right)=2+2.cos\frac{\pi}{3}=3\)

\(B=cos^2a+sin^2b+2cosa.sinb+cos^2b+sin^2a-2sina.cosb\)

\(=2-2\left(sina.cosb-cosa.sinb\right)\)

\(=2-2sin\left(a-b\right)=2-2sin\frac{\pi}{3}=2-\sqrt{3}\)

\(K=\frac{2sin\left(\frac{a+b}{2}\right).cos\left(\frac{a+b}{2}\right)+2sin\left(\frac{a+b}{2}\right).cos\left(\frac{a-b}{2}\right)}{2cos^2\left(\frac{a+b}{2}\right)-1+2cos\left(\frac{a+b}{2}\right).cos\left(\frac{a-b}{2}\right)+1}\)

\(K=\frac{sin\left(\frac{a+b}{2}\right)\left[cos\left(\frac{a+b}{2}\right)+cos\left(\frac{a-b}{2}\right)\right]}{cos\left(\frac{a+b}{2}\right)\left[cos\left(\frac{a+b}{2}\right)+cos\left(\frac{a-b}{2}\right)\right]}\)

\(K=\frac{sin\left(\frac{a+b}{2}\right)}{cos\left(\frac{a+b}{2}\right)}=tan\left(\frac{a+b}{2}\right)\)