Cho $\cos \alpha {\rm{ = }}\dfrac{3}{4};\sin \alpha > 0$ . Tính \(\cos 2\alpha ,\sin \alpha \)
$\begin{array}{l}\cos \alpha =\dfrac{3}{4};\sin \alpha > 0 \Rightarrow {\sin ^2}\alpha = 1 - \dfrac{9}{{16}} = \dfrac{7}{{16}} \Rightarrow \sin \alpha = \dfrac{{\sqrt 7 }}{4}\\{\cos 2\alpha} = 1 - 2{\sin ^2}\alpha = 1 - 2.\dfrac{7}{{16}} = \dfrac{1}{8}\end{array}$
Cho ${\rm{cos}}\alpha {\rm{ = }}\dfrac{3}{4};\sin \alpha > 0$; ${\rm{sin}}\beta {\rm{ = }}\dfrac{3}{4};cos\beta < 0$ Tính $\cos \left( {\alpha + \beta } \right)$
${\rm{cos}}\alpha {\rm{ = }}\dfrac{3}{4};\sin \alpha > 0 \Rightarrow {\sin ^2}\alpha = 1 - \dfrac{9}{{16}} = \dfrac{7}{{16}} \Rightarrow \sin \alpha = \dfrac{{\sqrt 7 }}{4}$
${\rm{sin}}\beta {\rm{ = }}\dfrac{3}{4};cos\beta < 0 \Rightarrow co{s^2}\beta = 1 - \dfrac{9}{{16}} = \dfrac{7}{{16}} \Rightarrow cos\beta = - \dfrac{{\sqrt 7 }}{4}$
$\cos \left( {\alpha + \beta } \right) = \cos \alpha \cos \beta - \sin \alpha \sin \beta = \dfrac{3}{4}.\left( { - \dfrac{{\sqrt 7 }}{4}} \right) - \dfrac{3}{4}.\left( {\dfrac{{\sqrt 7 }}{4}} \right) = - \dfrac{{3\sqrt 7 }}{8}$
Cho \(\cos \alpha = m\) . Tính \({\sin ^2}\dfrac{\alpha }{2}\)
\({\sin ^2}\dfrac{\alpha }{2} = \dfrac{{1 - \cos \alpha }}{2} = \dfrac{{1 - m}}{2}\)
Thu gọn biểu thức \(\dfrac{{\sin \alpha + \sin 2\alpha }}{{1 + \cos \alpha + \cos 2\alpha }}\) ta được kết quả:
\(\dfrac{{\sin \alpha + \sin 2\alpha }}{{1 + \cos \alpha + \cos 2\alpha }} = \dfrac{{\sin \alpha + 2\sin \alpha \cos \alpha }}{{1 + \cos \alpha + 2{{\cos }^2}\alpha - 1}} = \dfrac{{\sin \alpha \left( {1 + 2\cos \alpha } \right)}}{{\cos \alpha \left( {1 + 2\cos \alpha } \right)}} = \tan \alpha \)
Thu gọn \(A = {\sin ^2}\alpha + {\sin ^2}\beta + 2\sin \alpha \sin \beta .\cos \left( {\alpha + \beta } \right)\) ta được:
$A = {\sin ^2}\alpha + {\sin ^2}\beta $ $+ 2\sin \alpha \sin \beta .\cos \left( {\alpha + \beta } \right)$
$= {\sin ^2}\alpha + {\sin ^2}\beta $ $+ 2\sin \alpha \sin \beta .\left( {\cos \alpha .\cos\beta - \sin \alpha .\sin\beta } \right)$
$= {\sin ^2}\alpha + {\sin ^2}\beta - 2{\sin ^2}\alpha {\sin ^2}\beta $ $ + 2\sin \alpha \sin \beta \cos \alpha .\cos\beta$
$= {\sin ^2}\alpha \left( {1 - {{\sin }^2}\beta } \right) + {\sin ^2}\beta \left( {1 - {{\sin }^2}\alpha } \right) + 2\sin \alpha \sin \beta \cos \alpha .\cos\beta$
$= {\sin ^2}\alpha {\cos ^2}\beta + {\sin ^2}\beta {\cos ^2}\alpha $ $+ 2\sin \alpha \sin \beta \cos \alpha .\cos\beta$
$= {\left( {\sin \alpha \cos \beta + \sin \beta \cos \alpha } \right)^2} = {\sin ^2}\left( {\alpha + \beta } \right)$
Biết $\cos \left( {\alpha + \beta } \right) = 0$ thì $\sin \left( {\alpha + 2\beta } \right)$ bằng:
Ta có:
$\sin\left( {\alpha + 2\beta } \right) = \sin \alpha .\cos 2\beta + \cos \alpha .\sin 2\beta$
$= \sin \alpha .\left( {1 - 2{{\sin }^2}\beta } \right) + 2\cos \alpha .\sin \beta \cos \beta$
$= \sin \alpha $ $ + 2\sin\beta \left( {\cos \alpha .\cos\beta - \sin \alpha .\sin\beta } \right)$
$= \sin \alpha + 2\sin\beta \cos \left( {\alpha + \beta } \right) $
$= \sin \alpha$
Tính \(\dfrac{{2\sin \alpha + 3\cos \alpha }}{{4\sin \alpha - 5\cos \alpha }}\) biết \(\tan \alpha = 3\).
Ta có:
\(\dfrac{{2\sin \alpha + 3\cos \alpha }}{{4\sin \alpha - 5\cos \alpha }} = \dfrac{{2\tan \alpha + 3}}{{4\tan \alpha - 5}} = \dfrac{{2.3 + 3}}{{4.3 - 5}} = \dfrac{9}{7}\)
Tính \(\dfrac{{\sin \alpha + \sin \beta c{\rm{os}}\left( {\alpha + \beta } \right)}}{{\cos \alpha - \sin \beta \sin \left( {\alpha + \beta } \right)}}\)
Ta có
\(\dfrac{{\sin \alpha + \sin \beta \cos \left( {\alpha + \beta } \right)}}{{\cos \alpha - \sin \beta \sin \left( {\alpha + \beta } \right)}}\)\( = \dfrac{{\sin \alpha + \dfrac{1}{2}\left[ {\sin \left( {\alpha + 2\beta } \right) + \sin \left( { - \alpha } \right)} \right]}}{{\cos \alpha + \dfrac{1}{2}\left[ {\cos \left( {\alpha + 2\beta } \right) - \cos \left( { - \alpha } \right)} \right]}}\) \( = \dfrac{{\sin \alpha + \dfrac{1}{2}\left[ {\sin \left( {\alpha + 2\beta } \right) - \sin \alpha } \right]}}{{\cos \alpha + \dfrac{1}{2}\left[ {\cos \left( {\alpha + 2\beta } \right) - \cos \alpha } \right]}}\)\( = \dfrac{{\sin \alpha + \dfrac{1}{2}\sin \left( {\alpha + 2\beta } \right) - \dfrac{1}{2}\sin \alpha }}{{\cos \alpha + \dfrac{1}{2}\cos \left( {\alpha + 2\beta } \right) - \dfrac{1}{2}\cos \alpha }}\)\( = \dfrac{{\dfrac{1}{2}\sin \alpha + \dfrac{1}{2}\sin \left( {\alpha + 2\beta } \right)}}{{\dfrac{1}{2}\cos \alpha + \dfrac{1}{2}\cos \left( {\alpha + 2\beta } \right)}}\) \( = \dfrac{{\sin \left( {\alpha + 2\beta } \right) + \sin \alpha }}{{\cos \left( {\alpha + 2\beta } \right) + \cos \alpha }}\) \( = \dfrac{{2\sin \dfrac{{\left( {\alpha + 2\beta + \alpha } \right)}}{2}\cos \dfrac{{\left( {\alpha + 2\beta - \alpha } \right)}}{2}}}{{2\cos \dfrac{{\left( {\alpha + 2\beta + \alpha } \right)}}{2}\cos \dfrac{{\left( {\alpha + 2\beta - \alpha } \right)}}{2}}}\)\( = \dfrac{{2\sin \left( {\alpha + \beta } \right)\cos \beta }}{{2\cos \left( {\alpha + \beta } \right)\cos \beta }} = \tan \left( {\alpha + \beta } \right)\)
Giá trị của biểu thức \(\cos \dfrac{{5x}}{2}\cos \dfrac{{3x}}{2} + \sin \dfrac{{7x}}{2}\sin \dfrac{x}{2} - \cos x\cos 2x\) bằng
Thực nghiệm \(\cos \dfrac{{5\pi }}{2}\cos \dfrac{{3\pi }}{2} + \sin \dfrac{{7\pi }}{2}\sin \dfrac{\pi }{2} - \cos \pi \cos 2\pi = 0\)
Biết rằng \({\sin ^4}x + {\cos ^4}x = m\cos 4x + n\left( {m,n \in \mathbb{Q}} \right)\). Tính tổng \(S = m + n\).
Ta có \({\sin ^4}x + {\cos ^4}x\) \( = {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^2} - 2{\sin ^2}x{\cos ^2}x\) $ = 1 - 2{\left( {\sin x\cos x} \right)^2}$ \( = 1 - 2{\left( {\dfrac{1}{2}\sin 2x} \right)^2}\)
$ = 1 - 2.\dfrac{1}{4}{\sin ^2}2x = 1 - \dfrac{1}{2}{\sin ^2}2x$
\( = 1 - \dfrac{1}{2}.\dfrac{{1 - \cos 4x}}{2} \)$ = 1 - \dfrac{1}{4}\left( {1 - \cos 4x} \right) = 1 - \dfrac{1}{4} + \dfrac{1}{4}\cos 4x$\(= \dfrac{1}{4}\cos 4x + \dfrac{3}{4}\)
\( \Rightarrow S = m + n = 1\)
Khi $\sin A = \dfrac{{\cos B + \cos C}}{{\sin B + \sin C}}$ thì tam giác $ABC$ là tam giác gì?
Ta có:
$\dfrac{{\cos B + \cos C}}{{\sin B + \sin C}}$ $= \dfrac{{2\cos \dfrac{{B + C}}{2}.\cos \dfrac{{B - C}}{2}}}{{2\sin \dfrac{{B + C}}{2}.\cos \dfrac{{B - C}}{2}}}$ $= \dfrac{{\cos \dfrac{{B + C}}{2}}}{{\sin \dfrac{{B + C}}{2}}}$ $= \dfrac{{\cos \left( {\dfrac{\pi }{2} - \dfrac{A}{2}} \right)}}{{\sin \left( {\dfrac{\pi }{2} - \dfrac{A}{2}} \right)}} $ $= \dfrac{{\sin \dfrac{A}{2}}}{{\cos \dfrac{A}{2}}}$ $\Rightarrow \sin A = \dfrac{{\sin \dfrac{A}{2}}}{{\cos \dfrac{A}{2}}}$
$\Rightarrow 2\sin \dfrac{A}{2}\cos \dfrac{A}{2} = \dfrac{{\sin \dfrac{A}{2}}}{{\cos \dfrac{A}{2}}} $ $\Rightarrow 2{\cos ^2}\dfrac{A}{2} = 1$ $\Rightarrow \cos A = 0 \Rightarrow A = {90^0}$
Nếu $\sin \left( {2\alpha + \beta } \right) = 3\sin \beta ;$ $\cos \alpha \ne 0;$ $\cos \left( {\alpha + \beta } \right) \ne 0$ thì $\tan \left( {\alpha + \beta } \right)$ bằng:
Ta có:
$\sin \left( {2\alpha + \beta } \right) = 3\sin \beta $ $ \Rightarrow \sin 2\alpha \cos \beta + \cos 2\alpha \sin \beta = 3\sin \beta $
$ \Rightarrow 2\sin \alpha \cos \alpha \cos \beta + \left( {2{{\cos }^2}\alpha - 1} \right)\sin \beta = 3\sin \beta $
$ \Rightarrow 2\sin \alpha \cos \alpha \cos \beta + 2{\cos ^2}\alpha \sin \beta = 4\sin \beta $
$ \Rightarrow 2\cos \alpha \left( {\sin \alpha \cos \beta + \sin \beta \cos \alpha } \right) = 4\sin \beta $
$ \Rightarrow \cos \alpha \sin \left( {\alpha + \beta } \right) = 2\sin \beta $
Lại có:
$\sin \left( {2\alpha + \beta } \right) = 3\sin \beta $ $ \Rightarrow \sin 2\alpha \cos \beta + \cos 2\alpha \sin \beta = 3\sin \beta $
$ \Rightarrow 2\sin \alpha \cos \alpha \cos \beta + \left( {1 - 2{{\sin }^2}\alpha } \right)\sin \beta = 3\sin \beta $
$ \Rightarrow 2\sin \alpha \cos \alpha \cos \beta - 2{\sin ^2}\alpha \sin \beta = 2\sin \beta $
$ \Rightarrow 2\sin \alpha \left( {\cos \alpha \cos \beta - \sin \beta \sin \alpha } \right) = 2\sin \beta $
$ \Rightarrow \sin \alpha \cos \left( {\alpha + \beta } \right) = \sin \beta $
Từ đó suy ra \(\dfrac{{\cos \alpha \sin \left( {\alpha + \beta } \right)}}{{\sin \alpha \cos \left( {\alpha + \beta } \right)}} = \dfrac{{2\sin \beta }}{{\sin \beta }}\) hay $\cot \alpha \tan \left( {\alpha + \beta } \right) = 2$$ \Rightarrow \tan \left( {\alpha + \beta } \right) = 2\tan \alpha $
Tìm giá trị nhỏ nhất của biểu thức \({\sin ^6}\alpha + {\cos ^6}\alpha \)
$\begin{array}{l}A = {\sin ^6}\alpha + {\cos ^6}\alpha = {\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right)^3} - 3{\sin ^2}\alpha {\cos ^2}\alpha \left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right)\\ = 1 - 3{\sin ^2}\alpha {\cos ^2}\alpha = 1 - \dfrac{3}{4}{\sin ^2}2\alpha \end{array}$
Vì $0 \le {\sin ^2}2\alpha \le 1 \Rightarrow A \ge \dfrac{1}{4}$ nên \(\min A = \dfrac{1}{4}\) khi ${\sin ^2}2\alpha = 1$.
Trong các đáp án chỉ có đáp án C sai, công thức đúng: \(\tan \left( {\pi + \alpha } \right) = \tan \alpha .\)
\(\sin \alpha + \cos \alpha = \dfrac{3}{4} \Rightarrow \cos \alpha = \dfrac{3}{4} - \sin \alpha .\)
Lại có: \({\sin ^2}\alpha + {\cos ^2}\alpha = 1\)
\( \Rightarrow {\sin ^2}\alpha + {\left( {\dfrac{3}{4} - \sin \alpha } \right)^2} = 1 \Rightarrow 2{\sin ^2}\alpha - \dfrac{3}{2}\sin \alpha - \dfrac{7}{{16}} = 0\)
\( \Rightarrow \sin \alpha = \dfrac{{3 + \sqrt {23} }}{8}\) (vì với \(\dfrac{\pi }{2} < \alpha < \pi \) thì \(\sin \alpha > 0)\).
\( \Rightarrow \cos \alpha = \dfrac{3}{4} - \sin \alpha = \dfrac{3}{4} - \dfrac{{3 + \sqrt {23} }}{8} = \dfrac{{3 - \sqrt {23} }}{8}\) \( \Rightarrow \cos \alpha - \sin \alpha = - \dfrac{{\sqrt {23} }}{4}.\)
Ta có: \(\cos a - \cos b = - 2\sin \dfrac{{a + b}}{2}\sin \dfrac{{a - b}}{2}\)
Vậy C sai.
Ta có: \(\cos 2a = {\cos ^2}a - {\sin ^2}a\) \( = \left( {{{\cos }^2}a - {{\sin }^2}a} \right)\left( {{{\cos }^2}a + {{\sin }^2}a} \right)\)\( = {\cos ^4}a - {\sin ^4}a\)
Vậy B đúng.
\(B = \tan \alpha \left( {\dfrac{{1 + {{\cos }^2}\alpha }}{{\sin \alpha }} - \sin \alpha } \right) \\= \dfrac{{\sin \alpha }}{{\cos \alpha }}.\dfrac{{{{\cos }^2}\alpha + {{\sin }^2}\alpha + {{\cos }^2}\alpha - {{\sin }^2}\alpha }}{{\sin \alpha }}\\ = \dfrac{{2{{\cos }^2}\alpha }}{{\cos \alpha }} = 2\cos \alpha \)
\(\sin 4x\cos 5x - \cos 4x\sin 5x = \sin \left( {4x - 5x} \right) = \sin \left( { - x} \right) = - \sin x\)
