Tính \(B = \cos \dfrac{\pi }{{11}} + \cos \dfrac{{3\pi }}{{11}} + \cos \dfrac{{5\pi }}{{11}} + \cos \dfrac{{7\pi }}{{11}} + \cos \dfrac{{9\pi }}{{11}}\)
Với $k = 1,2,3,4,5$ ta có
\(\cos \dfrac{{\left( {2k - 1} \right)\pi }}{{11}}\sin \dfrac{\pi }{{11}} = \dfrac{1}{2}\left[ {\sin \dfrac{{2k\pi }}{{11}} - \sin \dfrac{{\left( {2k - 2} \right)\pi }}{{11}}} \right]\)
\(\begin{array}{l} \Rightarrow B.\sin \dfrac{\pi }{{11}} = \dfrac{1}{2}\left[ {\left( {\sin \dfrac{{2\pi }}{{11}} - \sin 0} \right) + \left( {\sin \dfrac{{4\pi }}{{11}} - \sin \dfrac{{2\pi }}{{11}}} \right) + ... + \left( {\sin \dfrac{{10\pi }}{{11}} - \sin \dfrac{{8\pi }}{{11}}} \right)} \right]\\ = \dfrac{1}{2}\sin \dfrac{{10\pi }}{{11}} = \dfrac{1}{2}\sin \dfrac{\pi }{{11}}\\ \Rightarrow B = \dfrac{1}{2}\end{array}\)
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 ${\sin ^2}2\alpha = 1$.
Cho \(\tan \alpha + \cot \alpha = m\left( {\left| m \right| \ge 2} \right)\). Tính theo $m$ giá trị của \(A = \left| {\tan \alpha - \cot \alpha } \right|\)
Ta có:
\({\left( {\tan \alpha + \cot \alpha } \right)^2}\) \( = {\tan ^2}\alpha + {\cot ^2}\alpha + 2\tan \alpha .\cot \alpha \) \( \Rightarrow {\tan ^2}\alpha + {\cot ^2}\alpha \) \(= {\left( {\tan \alpha + \cot \alpha } \right)^2} - 2\tan \alpha \cot \alpha \) \( = {m^2} - 2\) (do \(\tan \alpha .\cot \alpha = 1\))
Do đó:
\({\left( {\tan \alpha - \cot \alpha } \right)^2}\)\( = {\tan ^2}\alpha + {\cot ^2}\alpha - 2\tan \alpha \cot \alpha \)\( = {m^2} - 2 - 2 = {m^2} - 4\)
Vậy \(\left| {\tan \alpha - \cot \alpha } \right| = \sqrt {{m^2} - 4} \)
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: \(\tan \left( {a + b} \right) = \dfrac{{\tan a + \tan b}}{{1 - \tan a\tan b}} = \dfrac{{\dfrac{1}{7} + \dfrac{3}{4}}}{{1 - \dfrac{1}{7}.\dfrac{3}{4}}} = 1\)
\( \Rightarrow \left[ \begin{array}{l}a + b = \dfrac{\pi }{4}\\a + b = \dfrac{{5\pi }}{4}\end{array} \right. \Rightarrow a + b = \dfrac{\pi }{4}\,\,\,\left( {do\,\,\,0 < a,\,\,b < \dfrac{\pi }{2} \Rightarrow 0 < a + b < \pi } \right)\)
\(\sin A + \sin B = \cos A + \cos B \Leftrightarrow \sin \dfrac{{A + B}}{2}\cos \dfrac{{A - B}}{2} = \cos \dfrac{{A + B}}{2}\cos \dfrac{{A - B}}{2}\,\,\,\,\left( 1 \right)\)
TH1: \(\cos \dfrac{{A - B}}{2} = 0 \Rightarrow \dfrac{{A - B}}{2} = {90^o} \Rightarrow A - B = {180^o} = A + B + C \Leftrightarrow 2B + C = 0\) vô lý
TH2: \(\cos \dfrac{{A - B}}{2} \ne 0\) khi đó \(\left( 1 \right) \Leftrightarrow \sin \dfrac{{A + B}}{2} = \cos \dfrac{{A + B}}{2} \Leftrightarrow \sin \dfrac{{A + B}}{2} = \sin \dfrac{C}{2}\,\,\,\,\left( {do\,\,\,\dfrac{{A + B}}{2} + \dfrac{C}{2} = {{90}^o}} \right)\)
\( \Rightarrow \dfrac{{A + B}}{2} = \dfrac{C}{2} \Leftrightarrow A + B = C \Leftrightarrow {180^o} - C = C \Leftrightarrow 2C = {180^o} \Leftrightarrow C = {90^o}\)
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.
Ta có: \(\cos 2\alpha .\sin 5\alpha = \dfrac{1}{2}\left[ {\sin 7\alpha - \sin \left( { - 3\alpha } \right)} \right] = \dfrac{1}{2}\left( {\sin 7\alpha + \sin 3\alpha } \right)\)
Vậy D sai.
Biết A, B, C là các góc trong của tam giác ABC
\(\begin{array}{l} \Rightarrow A + B + C = {180^o} \Rightarrow \dfrac{{A + B}}{2} = {90^o} - \dfrac{C}{2}\\ \Rightarrow \cos \left( {\dfrac{{A + B}}{2}} \right) = \cos \left( {{{90}^o} - \dfrac{C}{2}} \right) = \sin \dfrac{C}{2}.\end{array}\)
\(B = \tan \alpha \left( {\dfrac{{1 + {{\cos }^2}\alpha }}{{\sin \alpha }} - \sin \alpha } \right) \\= \dfrac{{\sin \alpha }}{{\cos \alpha }}.\dfrac{{{{\cos }^2}\alpha + {{\sin }^2}\alpha + {{\cos }^2}\alpha - {{\sin }^2}\alpha }}{{\sin \alpha }}\\ = \dfrac{{2{{\cos }^2}\alpha }}{{\cos \alpha }} = 2\cos \alpha \)
\(\sin 4x\cos 5x - \cos 4x\sin 5x = \sin \left( {4x - 5x} \right) = \sin \left( { - x} \right) = - \sin x\)
Với giá trị nào của n thì đẳng thức sau luôn đúng?
\(\sqrt {\dfrac{1}{2} + \dfrac{1}{2}\sqrt {\dfrac{1}{2} + \dfrac{1}{2}\sqrt {\dfrac{1}{2} + \dfrac{1}{2}\cos 12x} } } = \cos \dfrac{x}{{2n}}\)\(\,\,0 < x < \dfrac{\pi }{{12}}\).
Ta có: \(0 < x < \dfrac{\pi }{{12}} \Rightarrow 0 < \dfrac{{3x}}{2} < 3x < 6x < \dfrac{\pi }{2} \Rightarrow 0 < \cos 6x < \cos 3x < \cos \dfrac{{3x}}{2} < 1\) (do hàm số \(y = \cos x\) là hàm số nghịch biến).
\(\begin{array}{l}\sqrt {\dfrac{1}{2} + \dfrac{1}{2}\sqrt {\dfrac{1}{2} + \dfrac{1}{2}\sqrt {\dfrac{1}{2} + \dfrac{1}{2}\cos 12x} } } \\ = \sqrt {\dfrac{1}{2} + \dfrac{1}{2}\sqrt {\dfrac{1}{2} + \dfrac{1}{2}\sqrt {\dfrac{1}{2} + \dfrac{1}{2}\left( {2{{\cos }^2}6x - 1} \right)} } } \\ = \sqrt {\dfrac{1}{2} + \dfrac{1}{2}\sqrt {\dfrac{1}{2} + \dfrac{1}{2}\sqrt {\dfrac{1}{2} + {{\cos }^2}6x - \dfrac{1}{2}} } } \\ = \sqrt {\dfrac{1}{2} + \dfrac{1}{2}\sqrt {\dfrac{1}{2} + \dfrac{1}{2}\sqrt {{{\cos }^2}6x} } } \\ = \sqrt {\dfrac{1}{2} + \dfrac{1}{2}\sqrt {\dfrac{1}{2} + \dfrac{1}{2}\cos 6x} } \left( {do\cos 6x > 0} \right)\\ = \sqrt {\dfrac{1}{2} + \dfrac{1}{2}\sqrt {\dfrac{1}{2} + \dfrac{1}{2}\left( {2{{\cos }^2}3x - 1} \right)} } \\ = \sqrt {\dfrac{1}{2} + \dfrac{1}{2}\sqrt {{{\cos }^2}3x} } \end{array}\)
\(\begin{array}{l} = \sqrt {\dfrac{1}{2} + \dfrac{1}{2}\cos 3x} \left( {do\cos 3x > 0} \right)\\ = \sqrt {\dfrac{1}{2} + \dfrac{1}{2}\left( {2{{\cos }^2}\dfrac{{3x}}{2} - 1} \right)} \\ = \sqrt {{{\cos }^2}\dfrac{{3x}}{2}} = \cos \dfrac{{3x}}{2}\left( {do\cos \dfrac{{3x}}{2} > 0} \right)\\ \Rightarrow \cos \dfrac{{3x}}{2} = \cos \dfrac{x}{{2n}}\left( 1 \right)\end{array}\)
Để (1) luôn đúng \( \Rightarrow \dfrac{{3x}}{2} = \dfrac{x}{{2n}} \Leftrightarrow n = \dfrac{1}{3}\)
