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\(a,\sqrt{2}sin\left(\alpha+\dfrac{\pi}{4}\right)-cos\alpha\\ =\sqrt{2}\left(sin\alpha cos\dfrac{\pi}{4}+cos\alpha sin\dfrac{\pi}{4}\right)-cos\alpha\\ =\sqrt{2}\left(sin\alpha\cdot\dfrac{\sqrt{2}}{2}+cos\alpha\cdot\dfrac{\sqrt{2}}{2}\right)-cos\alpha\\ =\sqrt{2}\cdot sin\alpha\cdot\dfrac{\sqrt{2}}{2}+\sqrt{2}\cdot cos\alpha\cdot\dfrac{\sqrt{2}}{2}-cos\alpha\\ =sin\alpha+cos\alpha-cos\alpha\\ =sin\alpha\)
\(b,\left(cos\alpha+sin\alpha\right)^2-sin2\alpha\\ =cos^2\alpha+sin^2\alpha=2cos\alpha sin\alpha-2sin\alpha cos\alpha\\ =sin^2\alpha+cos^2\alpha\\ =1\)
Ta có:
\(\begin{array}{l}\sin \left( { - \frac{{15\pi }}{2} - \alpha } \right) - \cos \left( {13\pi + \alpha } \right) = \sin \left( { -\frac{{16\pi }}{2} +\frac{{\pi }}{2} + \alpha } \right) - \cos \left( {12\pi + \pi + \alpha } \right) = \sin \left( {-8\pi + \frac{\pi }{2} - \alpha } \right) - \cos \left( { \pi + \alpha } \right) \\ = \sin \left( {\frac{\pi }{2} - \alpha } \right) + \cos \left( \alpha \right) = \cos \left( \alpha \right) + \cos \left( \alpha \right) = 2\cos \left( \alpha \right) = 2.\left( { - \frac{5}{{13}}} \right) = \frac{{ - 10}}{{13}}\end{array}\)
\(a,\dfrac{1}{tan\alpha+1}+\dfrac{1}{cot\alpha+1}\\ =\dfrac{cot\alpha+1+tan\alpha+1}{\left(tan\alpha+1\right)\left(cot\alpha+1\right)}\\ =\dfrac{tan\alpha+cot\alpha+2}{tan\alpha\cdot cot\alpha+tan\alpha+cot\alpha+1}\\ =\dfrac{tan\alpha+cot\alpha+2}{tan\alpha+cot\alpha+2}\\ =1\)
\(b,cos\left(\dfrac{\pi}{2}-\alpha\right)-sin\left(\pi+\alpha\right)\\ =sin\alpha+sin\alpha\\ =2sin\alpha\)
\(c,sin\left(\alpha-\dfrac{\pi}{2}\right)+cos\left(-\alpha+6\pi\right)-tan\left(\alpha+\pi\right)cot\left(3\pi-\alpha\right)\\ =-sin\left(\dfrac{\pi}{2}-\alpha\right)+cos\left(\alpha\right)-tan\left(\alpha\right)cot\left(\pi-\alpha\right)\\ =-cos\left(\alpha\right)+cos\left(\alpha\right)+tan\left(\alpha\right)\cdot cot\left(\alpha\right)\\ =1\)
\(\cos \alpha = - \sqrt {1 - {{\left( { - \frac{5}{{13}}} \right)}^2}} = - \frac{{12}}{{13}}\) (vì \(\pi < \alpha < \frac{{3\pi }}{2}\))
\(\sin \left( {\alpha + \frac{\pi }{6}} \right) = \sin \alpha \cos \frac{\pi }{6} + \cos \alpha sin\frac{\pi }{6} = \frac{{ - 12 + 5\sqrt 3 }}{{26}}\)
\(\cos \left( {\frac{\pi }{4} - \alpha } \right) = \cos \frac{\pi }{4}\cos \alpha + \sin \frac{\pi }{4}sin\alpha = \frac{{ - 17\sqrt 2 }}{{26}}\)
Ta có:
a) \(\sin \left( {\alpha + \frac{\pi }{6}} \right) = \sin \alpha \cos \frac{\pi }{6} + \cos \alpha \sin \frac{\pi }{6} = \frac{{\sqrt 6 }}{3}.\frac{{\sqrt 3 }}{2} + \left( { - \frac{1}{{\sqrt 3 }}} \right).\frac{1}{2} = \frac{{ - \sqrt 3 + 3\sqrt 2 }}{6}\)
b) \(\cos \left( {\alpha + \frac{\pi }{6}} \right) = \cos \alpha .\cos \frac{\pi }{6} - \sin \alpha \sin \frac{\pi }{6} = \left( { - \frac{1}{{\sqrt 3 }}} \right).\frac{{\sqrt 3 }}{2} - \frac{{\sqrt 6 }}{3}.\frac{1}{2} = - \frac{{3 + \sqrt 6 }}{6}\)
c) \(\sin \left( {\alpha - \frac{\pi }{3}} \right) = \sin \alpha \cos \frac{\pi }{3} - \cos \alpha \sin \frac{\pi }{3} = \frac{{\sqrt 6 }}{3}.\frac{1}{2} - \left( { - \frac{1}{{\sqrt 3 }}} \right).\frac{{\sqrt 3 }}{2} = \frac{{3 + \sqrt 6 }}{6}\)
d) \(\cos \left( {\alpha - \frac{\pi }{6}} \right) = \cos \alpha \cos \frac{\pi }{6} + \sin \alpha \sin \frac{\pi }{6} = \left( { - \frac{1}{{\sqrt 3 }}} \right).\frac{{\sqrt 3 }}{2} + \frac{{\sqrt 6 }}{3}.\frac{1}{2} = \frac{{ - 3 + \sqrt 6 }}{6}\)
a, Ta có: \({\sin ^2}x + co{s^2}x = 1\)
\(\begin{array}{l} \Leftrightarrow {\sin ^2}\alpha + {\left( {\frac{1}{3}} \right)^2} = 1\\ \Leftrightarrow \sin \alpha = \pm \sqrt {1 - {{\left( {\frac{1}{3}} \right)}^2}} = \pm \frac{{2\sqrt 2 }}{3}\end{array}\)
Vì \( - \frac{\pi }{2} < \alpha < 0\) nên \(sin\alpha < 0 \Rightarrow \sin \alpha = - \frac{{2\sqrt 2 }}{3}\).
\(b)\;\,sin2\alpha = 2sin\alpha .cos\alpha = 2.\left( { - \frac{{2\sqrt 2 }}{3}} \right).\frac{1}{3} = - \frac{{4\sqrt 2 }}{9}\)
\(c)\;cos(\alpha + \frac{\pi }{3}) = cos\alpha .cos\frac{\pi }{3} - sin\alpha .sin\frac{\pi }{3}\)\( = \frac{1}{3}.\frac{1}{2} - \left( { - \frac{{2\sqrt 2 }}{3}} \right).\frac{{\sqrt 3 }}{2} = \frac{{2\sqrt 6 + 1}}{6}\).
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 \(\left(x;y\right)=\left(3;5\right)\)
Ta có: \(\cos \left( {\pi - \alpha } \right) = - \cos \alpha \)
Vậy ta chọn đáp án B