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\(P=\dfrac{2sin\alpha-3cos\alpha}{3sin\alpha+2cos\alpha}\\ =\dfrac{\dfrac{2sin\alpha}{cos\alpha}-\dfrac{3cos\alpha}{cos\alpha}}{\dfrac{3sin\alpha}{cos\alpha}+\dfrac{2cos\alpha}{cos\alpha}}\\ =\dfrac{2tan\alpha-3}{3tan\alpha+2}=\dfrac{2.3-3}{3.3+2}=\dfrac{3}{11}\)
Ta có: \(1 + {\tan ^2}\alpha = \frac{1}{{{{\cos }^2}\alpha }}\quad (\alpha \ne {90^o})\)
\( \Rightarrow \frac{1}{{{{\cos }^2}\alpha }} = 1 + {3^2} = 10\)
\( \Leftrightarrow {\cos ^2}\alpha = \frac{1}{{10}} \Leftrightarrow \cos \alpha = \pm \frac{{\sqrt {10} }}{{10}}\)
Vì \({0^o} < \alpha < {180^o}\) nên \(\sin \alpha > 0\).
Mà \(\tan \alpha = 3 > 0 \Rightarrow \cos \alpha > 0 \Rightarrow \cos \alpha = \frac{{\sqrt {10} }}{{10}}\)
Lại có: \(\sin \alpha = \cos \alpha .\tan \alpha = \frac{{\sqrt {10} }}{{10}}.3 = \frac{{3\sqrt {10} }}{{10}}.\)
\( \Rightarrow P = \dfrac{{2.\frac{{3\sqrt {10} }}{{10}} - 3.\frac{{\sqrt {10} }}{{10}}}}{{3.\frac{{3\sqrt {10} }}{{10}} + 2.\frac{{\sqrt {10} }}{{10}}}} = \dfrac{{\frac{{\sqrt {10} }}{{10}}\left( {2.3 - 3} \right)}}{{\frac{{\sqrt {10} }}{{10}}\left( {3.3 + 2} \right)}} = \dfrac{3}{{11}}.\)
\(A=\dfrac{\dfrac{3sina}{sina}-\dfrac{cosa}{sina}}{\dfrac{2sina}{sina}+\dfrac{cosa}{sina}}=\dfrac{3-cota}{2+cota}=\dfrac{3-3}{2+3}=0\)
\(B=\dfrac{\dfrac{sin^2a}{sin^2a}-\dfrac{3sina.cosa}{sin^2a}+\dfrac{2}{sin^2a}}{\dfrac{2sin^2a}{sin^2a}+\dfrac{sina.cosa}{sin^2a}+\dfrac{cos^2a}{sin^2a}}=\dfrac{1-3cota+2\left(1+cot^2a\right)}{2+cota+cot^2a}=\dfrac{1-3.3+2\left(1+3^2\right)}{2+3+3^2}=...\)
a. \(A=\dfrac{3sin\alpha-cos\alpha}{2sin\alpha+cos\alpha}=\dfrac{3\dfrac{sin\alpha}{cos\alpha}-1}{2\dfrac{sin\alpha}{cos\alpha}+1}=\dfrac{3.\dfrac{1}{3}-1}{2.\dfrac{1}{3}+1}=0\)
b.\(B=\dfrac{sin^2\alpha-3sin\alpha.cos\alpha+2}{2sin^2\alpha+sin\alpha.cos\alpha+cos^2\alpha}\)\(=\dfrac{1-\dfrac{3cos\alpha}{sin\alpha}+\dfrac{2}{sin^2\alpha}}{2+\dfrac{cos\alpha}{sin\alpha}+\dfrac{cos^2\alpha}{sin^2\alpha}}=\dfrac{1-3.3+\dfrac{2}{sin^2\alpha}}{2+3+3^2}\)
Mà \(\dfrac{cos\alpha}{sin\alpha}=3,cos^2\alpha+sin^2\alpha=1\Rightarrow sin^2\alpha=\dfrac{1}{10}\)
\(B=\dfrac{1-3.3+\dfrac{2}{\dfrac{1}{10}}}{2+3+3^2}=\dfrac{6}{7}\)
b) \(\sin x+\cos x=\dfrac{3}{2}\)
\(\left(\sin x+\cos x\right)^2=\dfrac{1}{4}\)
\(\sin^2x+\cos^2x+2\sin x\cos x=\dfrac{1}{4}\)
\(2\sin x\cos x=-\dfrac{3}{4}=\sin2x\)
Tham khảo:
a)
Gọi M(x;y) là điểm trên đường tròn đơn vị sao cho \(\widehat {xOM} = \alpha \). Gọi N, P tương ứng là hình chiếu vuông góc của M lên các trục Ox, Oy.
Ta có: \(\left\{ \begin{array}{l}x = \cos \alpha \\y = \sin \alpha \end{array} \right. \Rightarrow \left\{ \begin{array}{l}{\cos ^2}\alpha = {x^2}\\{\sin ^2}\alpha = {y^2}\end{array} \right.\)(1)
Mà \(\left\{ \begin{array}{l}\left| x \right| = ON\\\left| y \right| = OP = MN\end{array} \right. \Rightarrow \left\{ \begin{array}{l}{x^2} = {\left| x \right|^2} = O{N^2}\\{y^2} = {\left| y \right|^2} = M{N^2}\end{array} \right.\)(2)
Từ (1) và (2) suy ra \({\sin ^2}\alpha + {\cos ^2}\alpha = O{N^2} + M{N^2} = O{M^2}\) (do \(\Delta OMN\) vuông tại N)
\( \Rightarrow {\sin ^2}\alpha + {\cos ^2}\alpha = 1\) (vì OM =1). (đpcm)
b)
Ta có: \(\tan \alpha = \frac{{\sin \alpha }}{{\cos \alpha }}\;\;(\alpha \ne {90^o})\)
\( \Rightarrow 1 + {\tan ^2}\alpha = 1 + \frac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }} = \frac{{{{\cos }^2}\alpha }}{{{{\cos }^2}\alpha }} + \frac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }} = \frac{{{{\sin }^2}\alpha + {{\cos }^2}\alpha }}{{{{\cos }^2}\alpha }}\)
Mà theo ý a) ta có \({\sin ^2}\alpha + {\cos ^2}\alpha = 1\) với mọi góc \(\alpha \)
\( \Rightarrow 1 + {\tan ^2}\alpha = \frac{1}{{{{\cos }^2}\alpha }}\) (đpcm)
c)
Ta có: \(\cot \alpha = \frac{{\cos \alpha }}{{\sin \alpha }}\;\;\;({0^o} < \alpha < {180^o})\)
\( \Rightarrow 1 + {\cot ^2}\alpha = 1 + \frac{{{{\cos }^2}\alpha }}{{{{\sin }^2}\alpha }} = \frac{{{{\sin }^2}\alpha }}{{{{\sin }^2}\alpha }} + \frac{{{{\cos }^2}\alpha }}{{{{\sin }^2}\alpha }} = \frac{{{{\sin }^2}\alpha + {{\cos }^2}\alpha }}{{{{\sin }^2}\alpha }}\)
Mà theo ý a) ta có \({\sin ^2}\alpha + {\cos ^2}\alpha = 1\) với mọi góc \(\alpha \)
\( \Rightarrow 1 + {\cot ^2}\alpha = \frac{1}{{{{\sin }^2}\alpha }}\) (đpcm)
Bạn xem lại biểu thức A. Biểu thức $A$ sau khi rút gọn thì \(A=\frac{-2\sin ^2a}{3\cos 2a}\) vẫn phụ thuộc vào $a$
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Sử dụng công thức: \(\sin (90-a)=\cos a; \cot (90-a)=\tan a\), ta có:
\(B=\tan ^260(\sin ^8a-\cos ^8a)+4\cos 60(\cos ^6a-\sin ^6a)-\cos ^6a(\tan ^2a-1)^3\)
\(=3(\sin ^8a-\cos ^8a)+2(\cos ^6a-\sin ^6a)-\cos ^6a\left(\frac{\sin ^2a}{\cos ^2a}-1\right)^3\)
\(=3(\sin ^8a-\cos ^8a)+2(\cos ^6a-\sin ^6a)-(\sin ^2a-\cos ^2a)^3\)
\(=3(\sin ^2a-\cos ^2a)(\sin ^2a+\cos ^2a)(\sin ^4a+\cos ^4a)+2(\cos ^2a-\sin ^2a)(\cos ^4a+\sin ^2a\cos ^2a+\sin ^4a)-(\sin ^2a-\cos ^2a)^3\)
\(=3(\sin ^2-\cos ^2a)(\sin ^4a+\cos ^4a)-2(\sin ^2a-\cos ^2a)(\cos ^4a+\sin ^2a\cos ^2a+\sin ^4a)-(\sin ^2a-\cos ^2a)^3\)
\(=(\sin ^2a-\cos ^2a)[3(\sin ^4a+\cos ^4a)-2(\cos ^4a+\sin ^2a\cos ^2a+\sin ^4a)-(\sin ^2a-\cos ^2a)^2]\)
\(=(\sin ^2a-\cos ^2a).0=0\). Do đó giá trị của biểu thức không phụ thuộc vào $a$
a/\(sina-1=2sin\dfrac{a}{2}.cos\dfrac{a}{2}-sin^2\dfrac{a}{2}-cos^2\dfrac{a}{2}=-\left(sin\dfrac{a}{2}-cos\dfrac{a}{2}\right)^2\)
b/\(P=\dfrac{cosa+cos5a+2cos3a}{sina+sin5a+2sin3a}=\dfrac{2cos3a.cos2a+2cos3a}{2sin3a.cos2a+2sin3a}=\dfrac{2cos3a\left(cos2a+1\right)}{2sin3a\left(cos2a+1\right)}=cot3a\)
c/\(P=sin\left(30+60\right)=sin90=1\)
d/
\(A=cos\dfrac{2\pi}{7}+cos\dfrac{6\pi}{7}+cos\dfrac{4\pi}{7}\Rightarrow A.sin\dfrac{\pi}{7}=sin\dfrac{\pi}{7}.cos\dfrac{2\pi}{7}+sin\dfrac{\pi}{7}cos\dfrac{4\pi}{7}+sin\dfrac{\pi}{7}.cos\dfrac{6\pi}{7}\)
\(=\dfrac{1}{2}sin\dfrac{3\pi}{7}-\dfrac{1}{2}sin\dfrac{\pi}{7}+\dfrac{1}{2}sin\dfrac{5\pi}{7}-\dfrac{1}{2}sin\dfrac{3\pi}{7}+\dfrac{1}{2}sin\dfrac{7\pi}{7}-\dfrac{1}{2}sin\dfrac{5\pi}{7}\)
\(=-\dfrac{1}{2}sin\dfrac{\pi}{7}\Rightarrow A=-\dfrac{1}{2}\)
e/
\(tan\dfrac{\pi}{24}+tan\dfrac{7\pi}{24}=\dfrac{sin\dfrac{\pi}{24}}{cos\dfrac{\pi}{24}}+\dfrac{sin\dfrac{7\pi}{24}}{cos\dfrac{7\pi}{24}}=\dfrac{sin\dfrac{\pi}{24}cos\dfrac{7\pi}{24}+sin\dfrac{7\pi}{24}cos\dfrac{\pi}{24}}{cos\dfrac{\pi}{24}.cos\dfrac{7\pi}{24}}\)
\(=\dfrac{sin\left(\dfrac{\pi}{24}+\dfrac{7\pi}{24}\right)}{\dfrac{1}{2}cos\dfrac{\pi}{4}+\dfrac{1}{2}cos\dfrac{\pi}{3}}=\dfrac{2sin\dfrac{\pi}{3}}{cos\dfrac{\pi}{4}+cos\dfrac{\pi}{3}}=\dfrac{\sqrt{3}}{\dfrac{\sqrt{2}}{2}+\dfrac{1}{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}+1}\)
a)
Trên nửa đường tròn đơn vị, lấy điểm M sao cho \(\widehat {xOM} = \alpha \)
Gọi H, K lần lượt là các hình chiếu vuông góc của M trên Ox, Oy.
Ta có: tam giác vuông OHM vuông tại H và \(\alpha = \widehat {xOM}\)
Do đó: \(\sin \alpha = \frac{{MH}}{{OM}} = MH;\;\cos \alpha = \frac{{OH}}{{OM}} = OH.\)
\( \Rightarrow {\cos ^2}\alpha + {\sin ^2}\alpha = O{H^2} + M{H^2} = O{M^2} = 1\)
b) Ta có:
\(\begin{array}{l}\;\tan \alpha = \frac{{\sin \alpha }}{{\cos \alpha }};\;\cot \alpha = \frac{{\cos \alpha }}{{\sin \alpha }}.\\ \Rightarrow \;\tan \alpha .\cot \alpha = \frac{{\sin \alpha }}{{\cos \alpha }}.\frac{{\cos \alpha }}{{\sin \alpha }} = 1\end{array}\)
c) Với \(\alpha \ne {90^o}\) ta có:
\(\begin{array}{l}\;\tan \alpha = \frac{{\sin \alpha }}{{\cos \alpha }};\;\\ \Rightarrow \;1 + {\tan ^2}\alpha = 1 + \frac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }} = \frac{{{{\sin }^2}\alpha + {{\cos }^2}\alpha }}{{{{\cos }^2}\alpha }} = \frac{1}{{{{\cos }^2}\alpha }}\;\end{array}\)
d) Ta có:
\(\begin{array}{l}\cot \alpha = \frac{{\cos \alpha }}{{\sin \alpha }};\;\\ \Rightarrow \;1 + {\cot ^2}\alpha = 1 + \frac{{{{\cos }^2}\alpha }}{{{{\sin }^2}\alpha }} = \frac{{{{\sin }^2}\alpha + {{\cos }^2}\alpha }}{{{{\sin }^2}\alpha }} = \frac{1}{{{{\sin }^2}\alpha }}\;\end{array}\)
\(1+\tan^2a=\dfrac{1}{\sin^2a}=1+\dfrac{1}{16}=\dfrac{17}{16}\)
\(\Leftrightarrow\sin^2a=\dfrac{16}{17}\)
\(\Leftrightarrow\cos^2a=\dfrac{1}{17}\)
\(A=2\cdot\sin^2a+\cos^2a=2\cdot\dfrac{16}{17}+\dfrac{1}{17}=\dfrac{33}{17}\)