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Cách 1:
Ta có: \(tan\alpha=\sqrt{2}\Rightarrow\left\{{}\begin{matrix}\dfrac{sin\alpha}{cos\alpha}=\sqrt{2}\\1+\left(\sqrt{2}\right)^2=\dfrac{1}{cos^2\alpha}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}sin\alpha=\sqrt{2}\cdot cos\alpha\\cos^2\alpha=\dfrac{1}{3}\end{matrix}\right.\)
\(P=\dfrac{sin\alpha-cos\alpha}{sin^3\alpha+3cos^3\alpha+2sin\alpha}\)
\(=\dfrac{\sqrt{2}\cdot cos\alpha-cos\alpha}{\left(\sqrt{2}\cdot cos\alpha\right)^3+3cos^3\alpha+2\cdot\sqrt{2}\cdot cos\alpha}\)
\(=\dfrac{cos\alpha\left(\sqrt{2}-1\right)}{2\sqrt{2}\cdot cos^3\alpha+3cos^3\alpha+2\sqrt{2}\cdot cos\alpha}\)
\(=\dfrac{cos\alpha\left(\sqrt{2}-1\right)}{cos\alpha\left(2\sqrt{2}\cdot cos^2\alpha+3cos^2\alpha+2\sqrt{2}\right)}\)
\(=\dfrac{\sqrt{2}-1}{2\sqrt{2}\cdot cos^2\alpha+3cos^2\alpha+2\sqrt{2}}\)
Thay \(cos^2\alpha=\dfrac{1}{3}\) vào \(P\) ta có:
\(P=\dfrac{\sqrt{2}-1}{2\sqrt{2}\cdot\dfrac{1}{3}+3\cdot\dfrac{1}{3}+2\sqrt{2}}=\dfrac{\sqrt{2}-1}{1+\dfrac{8}{3}\sqrt{2}}\)
\(=\dfrac{3\left(\sqrt{2}-1\right)}{3\left(1+\dfrac{8}{3}\sqrt{2}\right)}=\dfrac{3\left(\sqrt{2}-1\right)}{3+8\sqrt{2}}\)
\(=\dfrac{3\left(\sqrt{2}-1\right)}{3+2^3\sqrt{2}}=\dfrac{a\left(\sqrt{b}-1\right)}{a+b^3\sqrt{b}}\)
\(\Rightarrow\left\{{}\begin{matrix}a=3\\b=2\end{matrix}\right.\Rightarrow a+b=5\)
Chọn đáp án A.
Cách 2:
\(P=\dfrac{sin\alpha-cos\alpha}{sin^3\alpha+3cos^3\alpha+2sin\alpha}=\dfrac{\left(sin\alpha-cos\alpha\right)\div cos^3\alpha}{\left(sin^3\alpha+3cos^3\alpha+2sin\alpha\right)\div cos^3\alpha}\)
\(=\dfrac{\dfrac{sin\alpha}{cos^3\alpha}-\dfrac{1}{cos^2\alpha}}{\dfrac{sin^3\alpha}{cos^3\alpha}+3+2\cdot\dfrac{sin\alpha}{cos^3\alpha}}=\dfrac{\dfrac{sin\alpha}{cos\alpha}\cdot\dfrac{1}{cos^2\alpha}-\dfrac{1}{cos^2\alpha}}{tan^3\alpha+3+2\cdot\dfrac{sin\alpha}{cos\alpha}\cdot\dfrac{1}{cos^2\alpha}}\)
\(=\dfrac{tan\alpha\cdot\left(1+tan^2\alpha\right)-\left(1+tan^2\alpha\right)}{tan^3\alpha+3+2tan\alpha\cdot\left(1+tan^2\alpha\right)}\)
Thay \(tan\alpha=\sqrt{2}\) vào ta có:
\(P=\dfrac{\sqrt{2}\cdot\left[1+\left(\sqrt{2}\right)^2\right]-\left[1+\left(\sqrt{2}\right)^2\right]}{\left(\sqrt{2}\right)^3+3+2\sqrt{2}\cdot\left[1+\left(\sqrt{2}\right)^2\right]}=\dfrac{3\sqrt{2}-3}{2\sqrt{2}+3+6\sqrt{2}}\)
\(=\dfrac{3\left(\sqrt{2}-1\right)}{3+8\sqrt{2}}=\dfrac{3\left(\sqrt{2}-1\right)}{3+2^3\sqrt{2}}=\dfrac{a\left(\sqrt{b}-1\right)}{a+b^3\sqrt{b}}\)
\(\Rightarrow\left\{{}\begin{matrix}a=3\\b=2\end{matrix}\right.\Rightarrow a+b=3+2=5\)
Chọn đáp án A
a: \(VT=\dfrac{\left(sina+cosa\right)^3-3\cdot sina\cdot cosa\left(sina+cosa\right)}{sina+cosa}\)
=(sina+cosa)^2-3*sina*cosa
=sin^2a+cos^2a-sina*cosa
=1-sina*cosa=VP
c: VT=(sin^2a+cos^2a)^2-2*sin^2a*cos^2a-(sin^2a+cos^2a)^3+3*sin^2a*cos^2a*(sin^2a+cos^2a)
=1-2sin^2a*cos^2a-1+3*sin^2a*cos^2a
=sin^2a*cos^2a=VP
a) \({\cos ^2}\alpha + {\sin ^2}\alpha = 1\)
b) \(\tan \alpha .\cot \alpha = \frac{{\sin \alpha }}{{\cos \alpha }}.\frac{{\cos \alpha }}{{\sin \alpha }} = 1\)
c) \(\frac{{{{\sin }^2}\alpha + {{\cos }^2}\alpha }}{{{{\cos }^2}\alpha }} = \frac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }} + \frac{{{{\cos }^2}\alpha }}{{{{\cos }^2}\alpha }} = {\tan ^2}\alpha + 1\)
d) \(\frac{1}{{{{\sin }^2}\alpha }} = \frac{{{{\sin }^2}\alpha + {{\cos }^2}\alpha }}{{{{\sin }^2}\alpha }} = \frac{{{{\sin }^2}\alpha }}{{{{\sin }^2}\alpha }} + \frac{{{{\cos }^2}\alpha }}{{{{\sin }^2}\alpha }} = 1 + {\cot ^2}\alpha \)
a) Ta có:
\(\begin{array}{l}{\sin ^4}\alpha - {\cos ^4}\alpha = 1 - 2{\cos ^2}\alpha \\ \Leftrightarrow \left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right)\left( {{{\sin }^2}\alpha - {{\cos }^2}\alpha } \right) = 1 - 2{\cos ^2}\alpha \\ \Leftrightarrow {\sin ^2}\alpha - {\cos ^2}\alpha - 1 + 2{\cos ^2}\alpha = 0\\ \Leftrightarrow {\sin ^2}\alpha + {\cos ^2}\alpha - 1 = 0\\ \Leftrightarrow 1 - 1 = 0\\ \Leftrightarrow 0 = 0\end{array}\)
Đẳng thức luôn đúng
b) Ta có:
\(\begin{array}{l}\tan \alpha + \cot \alpha = \frac{1}{{\sin \alpha .\cos \alpha }}\\ \Leftrightarrow \frac{{\sin \alpha }}{{\cos \alpha }} + \frac{{\cos \alpha }}{{\sin \alpha }} = \frac{1}{{\sin \alpha .\cos \alpha }}\\ \Leftrightarrow \frac{{{{\sin }^2}\alpha + {{\cos }^2}\alpha }}{{\cos \alpha .\sin \alpha }} = \frac{1}{{\sin \alpha .\cos \alpha }}\\ \Leftrightarrow \frac{1}{{\sin \alpha .\cos \alpha }} = \frac{1}{{\sin \alpha .\cos \alpha }}\end{array}\)
Đẳng thức luôn đúng
a, Ta có: \(sin^2\alpha+cos^2\alpha=1\Leftrightarrow\left(\dfrac{3}{5}\right)^2+cos^2\alpha=1\Leftrightarrow cos\alpha=\pm\dfrac{4}{5}\)
Vậy đẳng thức có thể đồng thời xảy ra.
b, Ta có: \(1+cot^2\alpha=\dfrac{1}{sin^2\alpha}\Rightarrow1+cot^2\alpha=\dfrac{1}{\left(\dfrac{1}{3}\right)^2}\Rightarrow cot\alpha=\pm2\sqrt{2}\)
Hai đẳng thức không thể đồng thời xảy ra.
c, Ta có: \(tan\alpha\cdot cot\alpha=1\Rightarrow3\cdot cot\alpha=1\Rightarrow cot\alpha=\dfrac{1}{3}\)
Đẳng thức có thể đồng thời xảy ra.
a)
Ta có:
\({\cos ^4}\alpha {\sin ^4}\alpha = \left( {{{\cos }^2}\alpha - {{\sin }^2}\alpha } \right)\left( {{{\cos }^2}\alpha + {{\sin }^2}\alpha } \right) \\= {\cos ^2}\alpha - {\sin ^2}\alpha = {\cos ^2}\alpha - (1 - {\cos ^2}\alpha ) \\= {\cos ^2}\alpha - 1 + {\cos ^2}\alpha = 2{\cos ^2}\alpha - 1\)
(đpcm)
b)
Ta có:
\(\frac{{{{\cos }^2}\alpha + {{\tan }^2}\alpha - 1}}{{{{\sin }^2}\alpha }} = \frac{{{{\cos }^2}\alpha \; + {{\tan }^2}\alpha - {{\sin }^2}\alpha - {{\cos }^2}\alpha }}{{{{\sin }^2}\alpha }} \\= \frac{{{{\tan }^2}\alpha - {{\sin }^2}\alpha }}{{{{\sin }^2}\alpha }} = \frac{{\frac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }} - {{\sin }^2}\alpha }}{{{{\sin }^2}\alpha }} \\= \frac{1}{{{{\cos }^2}\alpha }} - 1 = {\tan ^2}\alpha \)
(đpcm)
a) Ta có: \({\left( {\sin \alpha + \cos \alpha } \right)^2} = {\sin ^2}\alpha + 2\sin \alpha \cos \alpha + {\cos ^2}\alpha = 1 + \sin 2\alpha \;\)
b) \({\cos ^4}\alpha - {\sin ^4}\alpha = \left( {{{\cos }^2}\alpha - {{\sin }^2}\alpha } \right)\left( {{{\cos }^2}\alpha + {{\sin }^2}\alpha } \right) = \cos 2\alpha \;\)
a) Trong Hình 5, M là điểm biểu diễn của góc lượng giác \(\alpha \) trên đường tròn lượng giác. Ta có:
OK = MH = \(\sin \alpha \)
OH = KM = \(\cos \alpha \)
\(\begin{array}{l}O{M^2} = O{H^2} + M{H^2}\\ \Rightarrow 1 = {\sin ^2}\alpha + {\cos ^2}\alpha \end{array}\)
b) \(1 + {\tan ^2}\alpha = \frac{{{{\cos }^2}\alpha }}{{{{\cos }^2}\alpha }} + \frac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }} = \frac{1}{{{{\cos }^2}\alpha }}\)
1) \(cot\alpha=\sqrt[]{5}\Rightarrow tan\alpha=\dfrac{1}{\sqrt[]{5}}\)
\(C=sin^2\alpha-sin\alpha.cos\alpha+cos^2\alpha\)
\(\Leftrightarrow C=\dfrac{1}{cos^2\alpha}\left(tan^2\alpha-tan\alpha+1\right)\)
\(\Leftrightarrow C=\left(1+tan^2\alpha\right)\left(tan^2\alpha-tan\alpha+1\right)\)
\(\Leftrightarrow C=\left(1+\dfrac{1}{5}\right)\left(\dfrac{1}{5}-\dfrac{1}{\sqrt[]{5}}+1\right)\)
\(\Leftrightarrow C=\dfrac{6}{5}\left(\dfrac{6}{5}-\dfrac{\sqrt[]{5}}{5}\right)=\dfrac{6}{25}\left(6-\sqrt[]{5}\right)\)
1: \(cota=\sqrt{5}\)
=>\(cosa=\sqrt{5}\cdot sina\)
\(1+cot^2a=\dfrac{1}{sin^2a}\)
=>\(\dfrac{1}{sin^2a}=1+5=6\)
=>\(sin^2a=\dfrac{1}{6}\)
\(C=sin^2a-sina\cdot\sqrt{5}\cdot sina+\left(\sqrt{5}\cdot sina\right)^2\)
\(=sin^2a\left(1-\sqrt{5}+5\right)=\dfrac{1}{6}\cdot\left(6-\sqrt{5}\right)\)
2: tan a=3
=>sin a=3*cosa
\(1+tan^2a=\dfrac{1}{cos^2a}\)
=>\(\dfrac{1}{cos^2a}=1+9=10\)
=>\(cos^2a=\dfrac{1}{10}\)
\(B=\dfrac{3\cdot cosa-cosa}{27\cdot cos^3a+3\cdot cos^3a+2\cdot3\cdot cosa}\)
\(=\dfrac{2\cdot cosa}{30cos^3a+6cosa}=\dfrac{2}{30cos^2a+6}\)
\(=\dfrac{2}{3+6}=\dfrac{2}{9}\)