Công thức biến đổi lượng giác

$\begin{array}{l}\cos \alpha =\dfrac{3}{4};\sin \alpha  > 0 \Rightarrow {\sin ^2}\alpha  = 1 - \dfrac{9}{{16}} = \dfrac{7}{{16}} \Rightarrow \sin \alpha  = \dfrac{{\sqrt 7 }}{4}\\{\cos 2\alpha} = 1 - 2{\sin ^2}\alpha  = 1 - 2.\dfrac{7}{{16}} = \dfrac{1}{8}\end{array}$

${\rm{cos}}\alpha {\rm{ = }}\dfrac{3}{4};\sin \alpha  > 0 \Rightarrow {\sin ^2}\alpha  = 1 - \dfrac{9}{{16}} = \dfrac{7}{{16}} \Rightarrow \sin \alpha  = \dfrac{{\sqrt 7 }}{4}$

${\rm{sin}}\beta {\rm{ = }}\dfrac{3}{4};cos\beta  < 0 \Rightarrow co{s^2}\beta  = 1 - \dfrac{9}{{16}} = \dfrac{7}{{16}} \Rightarrow cos\beta  =  - \dfrac{{\sqrt 7 }}{4}$

$\cos \left( {\alpha  + \beta } \right) = \cos \alpha \cos \beta  - \sin \alpha \sin \beta  = \dfrac{3}{4}.\left( { - \dfrac{{\sqrt 7 }}{4}} \right) - \dfrac{3}{4}.\left( {\dfrac{{\sqrt 7 }}{4}} \right) =  - \dfrac{{3\sqrt 7 }}{8}$

${\sin ^2}\dfrac{\alpha }{2} = \dfrac{{1 - \cos \alpha }}{2} = \dfrac{{1 - m}}{2}$

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

$A = {\sin ^2}\alpha  + {\sin ^2}\beta$ $+ 2\sin \alpha \sin \beta .\cos \left( {\alpha  + \beta } \right)$

$= {\sin ^2}\alpha  + {\sin ^2}\beta$ $+ 2\sin \alpha \sin \beta .\left( {\cos \alpha .\cos\beta  - \sin \alpha .\sin\beta } \right)$

$= {\sin ^2}\alpha  + {\sin ^2}\beta  - 2{\sin ^2}\alpha {\sin ^2}\beta$ $+ 2\sin \alpha \sin \beta \cos \alpha .\cos\beta$

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

$= {\sin ^2}\alpha {\cos ^2}\beta  + {\sin ^2}\beta {\cos ^2}\alpha$ $+ 2\sin \alpha \sin \beta \cos \alpha .\cos\beta$

$= {\left( {\sin \alpha \cos \beta  + \sin \beta \cos \alpha } \right)^2} = {\sin ^2}\left( {\alpha  + \beta } \right)$

Ta có:

$\sin\left( {\alpha  + 2\beta } \right) = \sin \alpha .\cos 2\beta  + \cos \alpha .\sin 2\beta$

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

$= \sin \alpha$ $+ 2\sin\beta \left( {\cos \alpha .\cos\beta  - \sin \alpha .\sin\beta } \right)$

$= \sin \alpha  + 2\sin\beta \cos \left( {\alpha  + \beta } \right)$

$= \sin \alpha$

Ta có:

$\dfrac{{2\sin \alpha  + 3\cos \alpha }}{{4\sin \alpha  - 5\cos \alpha }} = \dfrac{{2\tan \alpha  + 3}}{{4\tan \alpha  - 5}} = \dfrac{{2.3 + 3}}{{4.3 - 5}} = \dfrac{9}{7}$

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)$

Thực nghiệm $\cos \dfrac{{5\pi }}{2}\cos \dfrac{{3\pi }}{2} + \sin \dfrac{{7\pi }}{2}\sin \dfrac{\pi }{2} - \cos \pi \cos 2\pi  = 0$

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 ${\sin ^2}2\alpha  = 1$.

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.

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

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