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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 \(\left(x;y\right)=\left(3;5\right)\)
Theo Viet: \(\left\{{}\begin{matrix}tana+tanb=p\\tana.tanb=q\end{matrix}\right.\)
\(\Rightarrow tan\left(a+b\right)=\frac{tana+tanb}{1-tana.tanb}=\frac{p}{1-q}\)
\(\Rightarrow A=cos^2\left(a+b\right)+psin\left(a+b\right)+q.sin^2\left(a+b\right)\)
\(=\frac{1}{cos^2\left(a+b\right)}\left(1+p.\frac{sin\left(a+b\right)}{cos\left(a+b\right)}+q.\frac{sin^2\left(a+b\right)}{cos^2\left(a+b\right)}\right)\)
\(=\left[1+tan^2\left(a+b\right)\right]\left[1+p.tan\left(a+b\right)+q.tan^2\left(a+b\right)\right]\)
\(=\left[1+\frac{p^2}{\left(1-q\right)^2}\right]\left[1+\frac{p^2}{1-q}+\frac{p^2q}{\left(1-q\right)^2}\right]\)
\(=\left[1+\frac{p^2}{\left(1-q\right)^2}\right]\left[1+\frac{p^2}{\left(1-q\right)^2}\right]=\left[1+\frac{p^2}{\left(1-q\right)^2}\right]^2\)
+) Xét \(\beta = - \alpha \), khi đó:
\(\begin{array}{l}cos\beta = cos\left( {-{\rm{ }}\alpha } \right) = cos\alpha ;\\sin\beta = sin\left( {-{\rm{ }}\alpha } \right) = -sin\alpha \Leftrightarrow sin\alpha = -sin\beta .\end{array}\)
Do đó A thỏa mãn.
Đáp án: A
\(a)\;sin(\alpha + \beta ).sin(\alpha - \beta ) = \;\frac{1}{2}.\left[ {cos\left( {\alpha + \beta - \alpha + \beta } \right) - cos\left( {\alpha + \beta + \alpha - \beta } \right)} \right]\)
\(\begin{array}{l} = \;\frac{1}{2}.(cos2\beta - cos2\alpha ) = \;\frac{1}{2}.(1 - 2si{n^2}\beta - 1 + 2si{n^2}\alpha )\\ = si{n^2}\alpha - si{n^2}\beta \end{array}\)
\(\begin{array}{l}b)\;co{s^4}\alpha - co{s^4}\left( {\alpha - \frac{\pi }{2}} \right) = \;co{s^4}\alpha - si{n^4}\alpha \\ = \;(co{s^2}\alpha + si{n^2}\alpha )(co{s^2}\alpha - si{n^2}\alpha )\\ = \;co{s^2}\alpha -si{n^2}\alpha = cos2\alpha .\end{array}\)
1.a) \(4cos\dfrac{\alpha}{2}.cos\dfrac{\beta}{2}.cos\dfrac{f}{2}\)
\(=\dfrac{1}{2}.4\left[cos\left(\dfrac{\alpha-\beta}{2}\right)+cos\left(\dfrac{\alpha+\beta}{2}\right)\right].cos\dfrac{f}{2}\)
\(=2.cos\left(\dfrac{\alpha-\beta}{2}\right)cos\dfrac{f}{2}+2.cos\left(\dfrac{\alpha+\beta}{2}\right).cos\dfrac{f}{2}\)
\(=cos\left(\dfrac{\alpha-\left(\beta+f\right)}{2}\right)+cos\left(\dfrac{\alpha-\beta+f}{2}\right)+cos\left(\dfrac{\alpha+\beta-f}{2}\right)+cos\left(\dfrac{\alpha+\beta+f}{2}\right)\)
\(=cos\left(\dfrac{2\alpha-\pi}{2}\right)+cos\left(\dfrac{\pi-2\beta}{2}\right)+cos\left(\dfrac{\pi-2f}{2}\right)+cos\left(\dfrac{\pi}{2}\right)\)
\(=cos\left(-\dfrac{\pi}{2}+\alpha\right)+cos\left(\dfrac{\pi}{2}-\beta\right)+cos\left(\dfrac{\pi}{2}-f\right)\)
\(=sin\alpha+sin\beta+sinf\) (đpcm)
a2) \(1+4sin\dfrac{\alpha}{2}.sin\dfrac{\beta}{2}.sin\dfrac{f}{2}\)
\(=1+2\left[cos\left(\dfrac{\alpha-\beta}{2}\right)-cos\left(\dfrac{\alpha+\beta}{2}\right)\right].sin\dfrac{f}{2}\)
\(=1+2.cos\left(\dfrac{\alpha-\beta}{2}\right).sin\dfrac{f}{2}-2.cos\left(\dfrac{\alpha+\beta}{2}\right).sin\dfrac{f}{2}\)
\(=1+sin\left(\dfrac{f-\alpha+\beta}{2}\right)+sin\left(\dfrac{a-\beta+f}{2}\right)-sin\left(\dfrac{f-\left(\alpha+\beta\right)}{2}\right)-sin\left(\dfrac{\alpha+\beta+f}{2}\right)\)
\(=1+sin\left(\dfrac{\pi-2\alpha}{2}\right)+sin\left(\dfrac{\pi-2\beta}{2}\right)-sin\left(\dfrac{2f-\pi}{2}\right)-sin\left(\dfrac{\pi}{2}\right)\)
\(=sin\left(\dfrac{\pi}{2}-\alpha\right)+sin\left(\dfrac{\pi}{2}-\beta\right)+sin\left(\dfrac{\pi}{2}-f\right)\)
\(=cos\alpha+cos\beta+cosf\) (đpcm)