Một số công thức biến đổi lượng giác

Với $k = 1,2,3,4,5$ ta có

$\cos \dfrac{{\left( {2k - 1} \right)\pi }}{{11}}\sin \dfrac{\pi }{{11}} = \dfrac{1}{2}\left[ {\sin \dfrac{{2k\pi }}{{11}} - \sin \dfrac{{\left( {2k - 2} \right)\pi }}{{11}}} \right]$

$\begin{array}{l} \Rightarrow B.\sin \dfrac{\pi }{{11}} = \dfrac{1}{2}\left[ {\left( {\sin \dfrac{{2\pi }}{{11}} - \sin 0} \right) + \left( {\sin \dfrac{{4\pi }}{{11}} - \sin \dfrac{{2\pi }}{{11}}} \right) + ... + \left( {\sin \dfrac{{10\pi }}{{11}} - \sin \dfrac{{8\pi }}{{11}}} \right)} \right]\\ = \dfrac{1}{2}\sin \dfrac{{10\pi }}{{11}} = \dfrac{1}{2}\sin \dfrac{\pi }{{11}}\\ \Rightarrow B = \dfrac{1}{2}\end{array}$

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}$

$\Rightarrow S = m + n = 1$

Ta có:

$\dfrac{{\cos B + \cos C}}{{\sin B + \sin C}}$ $= \dfrac{{2\cos \dfrac{{B + C}}{2}.\cos \dfrac{{B - C}}{2}}}{{2\sin \dfrac{{B + C}}{2}.\cos \dfrac{{B - C}}{2}}}$ $= \dfrac{{\cos \dfrac{{B + C}}{2}}}{{\sin \dfrac{{B + C}}{2}}}$ $= \dfrac{{\cos \left( {\dfrac{\pi }{2} - \dfrac{A}{2}} \right)}}{{\sin \left( {\dfrac{\pi }{2} - \dfrac{A}{2}} \right)}}$ $= \dfrac{{\sin \dfrac{A}{2}}}{{\cos \dfrac{A}{2}}}$ $\Rightarrow \sin A = \dfrac{{\sin \dfrac{A}{2}}}{{\cos \dfrac{A}{2}}}$

$\Rightarrow 2\sin \dfrac{A}{2}\cos \dfrac{A}{2} = \dfrac{{\sin \dfrac{A}{2}}}{{\cos \dfrac{A}{2}}}$ $\Rightarrow 2{\cos ^2}\dfrac{A}{2} = 1$ $\Rightarrow \cos A = 0 \Rightarrow A = {90^0}$

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$

$\begin{array}{l}A = {\sin ^6}\alpha  + {\cos ^6}\alpha  = {\left( {{{\sin }^2}\alpha  + {{\cos }^2}\alpha } \right)^3} - 3{\sin ^2}\alpha {\cos ^2}\alpha \left( {{{\sin }^2}\alpha  + {{\cos }^2}\alpha } \right)\\ = 1 - 3{\sin ^2}\alpha {\cos ^2}\alpha  = 1 - \dfrac{3}{4}{\sin ^2}2\alpha \end{array}$

Vì $0 \le {\sin ^2}2\alpha  \le 1 \Rightarrow A \ge \dfrac{1}{4}$ nên $\min A = \dfrac{1}{4}$ khi ${\sin ^2}2\alpha  = 1$.

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

$\begin{array}{l} \Rightarrow A + B + C = {180^o} \Rightarrow \dfrac{{A + B}}{2} = {90^o} - \dfrac{C}{2}\\ \Rightarrow \cos \left( {\dfrac{{A + B}}{2}} \right) = \cos \left( {{{90}^o} - \dfrac{C}{2}} \right) = \sin \dfrac{C}{2}.\end{array}$

$B = \tan \alpha \left( {\dfrac{{1 + {{\cos }^2}\alpha }}{{\sin \alpha }} - \sin \alpha } \right) \\= \dfrac{{\sin \alpha }}{{\cos \alpha }}.\dfrac{{{{\cos }^2}\alpha  + {{\sin }^2}\alpha  + {{\cos }^2}\alpha  - {{\sin }^2}\alpha }}{{\sin \alpha }}\\ = \dfrac{{2{{\cos }^2}\alpha }}{{\cos \alpha }} = 2\cos \alpha$

$\sin 4x\cos 5x - \cos 4x\sin 5x = \sin \left( {4x - 5x} \right) = \sin \left( { - x} \right) =  - \sin x$

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 $\Rightarrow \dfrac{{3x}}{2} = \dfrac{x}{{2n}} \Leftrightarrow n = \dfrac{1}{3}$