Một số công thức biến đổi lượng giác

Ta có $\sin a + \cos a = \dfrac{5}{4}$ $\Leftrightarrow {\left( {\sin a + \cos a} \right)^2} = \dfrac{{25}}{{16}}$ $\Leftrightarrow 1 + 2\sin a\cos a = \dfrac{{25}}{{16}}$ $\Leftrightarrow \sin a\cos a = \dfrac{9}{{32}}$

Dễ thấy với $\left\{ \begin{array}{l}\dfrac{\pi }{2} < \alpha  < \pi \\\sin \alpha  = \dfrac{{\sqrt 3 }}{2}\end{array} \right.$ $\Rightarrow \alpha  = \dfrac{{2\pi }}{3} \Rightarrow 2\alpha  = \dfrac{{4\pi }}{3}$ $\Rightarrow \cos \left( {2\alpha  - \dfrac{\pi }{3}} \right) = \cos \pi  =  - 1$

Ta có $P = \dfrac{{\sin 3\alpha  - \sin \alpha }}{{\sin 2\alpha }} = \dfrac{{2.\cos 2\alpha .\sin \alpha }}{{2.\sin \alpha .\cos \alpha }} = \dfrac{{\cos 2\alpha }}{{\cos \alpha }} = \dfrac{{2.{{\cos }^2}\alpha  - 1}}{{\cos \alpha }} =  - \dfrac{7}{3}$.

Ta có $P = \dfrac{{\sin 2a.\sin a}}{{1 + \cos 2a}} = \dfrac{{2\sin a\cos a.\sin a}}{{2{{\cos }^2}a}}$ $= \dfrac{{2{{\sin }^2}a\cos a}}{{2{{\cos }^2}a}} = \dfrac{{2\cos a\left( {1 - {{\cos }^2}a} \right)}}{{2{{\cos }^2}a}} =  - \dfrac{5}{6}$

Thực nghiệm $T = \dfrac{{\cos \left( {\pi  + 0} \right)\cos \left( {\pi  - 0} \right) + 1}}{{{{\cos }^2}\pi  + {{\cos }^2}0}} = 1$

Tự luận:

$\begin{array}{l}T = \dfrac{{\cos \left( {a + b} \right)\cos \left( {a - b} \right) + 1}}{{{{\cos }^2}a + {{\cos }^2}b}}\\ = \dfrac{{\dfrac{1}{2}\left\{ {\cos \left[ {\left( {a + b} \right) + \left( {a - b} \right)} \right] + \cos \left[ {\left( {a + b} \right) - \left( {a - b} \right)} \right]} \right\} + 1}}{{{{\cos }^2}a + {{\cos }^2}b}}\\ = \dfrac{{\dfrac{1}{2}\left( {\cos 2a + \cos 2b} \right) + 1}}{{{{\cos }^2}a + {{\cos }^2}b}}\\ = \dfrac{1}{2}.\dfrac{{\left( {\cos 2a + \cos 2b} \right) + 2}}{{{{\cos }^2}a + {{\cos }^2}b}}\\ = \dfrac{1}{2}.\dfrac{{(\cos 2a + 1) + (\cos 2b + 1)}}{{{{\cos }^2}a + {{\cos }^2}b}}\\ = \dfrac{1}{2}.\dfrac{{2{{\cos }^2}a + 2{{\cos }^2}b}}{{{{\cos }^2}a + {{\cos }^2}b}} = 1\end{array}$

Ta có: $A = {\left( {\sin a\cos b + \cos a\sin b} \right)^2} - {\sin ^2}a - {\sin ^2}b$

$\begin{array}{l} = {\sin ^2}a{\cos ^2}b + 2\sin a\cos a\sin b\cos b + {\cos ^2}a{\sin ^2}b - {\sin ^2}a - {\sin ^2}b\\ = {\sin ^2}a\left( {{{\cos }^2}b - 1} \right) + {\sin ^2}b\left( {{{\cos }^2}a - 1} \right) + 2\sin a\cos a\sin b\cos b\\ = 2\sin a\cos a\sin b\cos b - 2{\sin ^2}a{\sin ^2}b\\ = 2\sin a\sin b\left( {\cos a\cos b - \sin a\sin b} \right) = 2\sin a\sin b\cos \left( {a + b} \right)\end{array}$

Ta có $P = {\left( {\sin a + \sin b} \right)^2} + {\left( {\cos a + \cos b} \right)^2}$

$\begin{array}{l} = {\sin ^2}a + 2\sin a\sin b + {\sin ^2}b + {\cos ^2}a + 2\cos a\cos b + {\cos ^2}b\\ = 2 + 2\left( {\sin a.\sin b + \cos a.\cos b} \right)\\ = 2 + 2.\cos \left( {a - b} \right) = 2 + 2\cos \dfrac{\pi }{4} = 2 + \sqrt 2 \end{array}$

Ta có $\sin \left( {a + b} \right)\sin \left( {a - b} \right) = \dfrac{1}{2}\left( {\cos 2b - \cos 2a} \right)$

$= \dfrac{1}{2}\left[ {\left( {2{{\cos }^2}b - 1} \right) - \left( {2{{\cos }^2}a - 1} \right)} \right] = {\cos ^2}b - {\cos ^2}a$.

Ta có $P = \dfrac{{1 + \cos \alpha  + \cos 2\alpha }}{{\sin \alpha  + \sin 2\alpha }}$ $= \dfrac{{2{{\cos }^2}\alpha  + \cos \alpha }}{{\sin \alpha  + 2\sin \alpha .\cos \alpha }}$ $= \dfrac{{\cos \alpha \left( {1 + 2\cos \alpha } \right)}}{{\sin \alpha \left( {1 + 2\cos \alpha } \right)}}$ $= \cot \alpha  = \dfrac{1}{2}$

Ta có $\cos \alpha  =  - \dfrac{2}{{\sqrt 5 }} \Rightarrow {\cos ^2}\alpha  = \dfrac{4}{5}$$\Rightarrow {\sin ^2}\alpha  = 1 - \dfrac{4}{5} = \dfrac{1}{5}$

Do $\pi  < \alpha  < \dfrac{{3\pi }}{2} \Rightarrow \sin \alpha  < 0$$\Rightarrow \sin \alpha  =  - \dfrac{1}{{\sqrt 5 }} \Rightarrow \tan \alpha  = \dfrac{{\sin \alpha }}{{\cos \alpha }} = \dfrac{1}{2}$

Ta có:

$\begin{array}{l}\sin \dfrac{\pi }{7}\left( {\sin \dfrac{{2\pi }}{7} + \sin \dfrac{{4\pi }}{7} + \sin \dfrac{{6\pi }}{7}} \right) = \sin \dfrac{\pi }{7}\sin \dfrac{{2\pi }}{7} + \sin \dfrac{\pi }{7}\sin \dfrac{{4\pi }}{7} + \sin \dfrac{\pi }{7}\sin \dfrac{{6\pi }}{7}\\ = \dfrac{1}{2}\left( {\cos \dfrac{\pi }{7} - \cos \dfrac{{3\pi }}{7}} \right) + \dfrac{1}{2}\left( {\cos \dfrac{{3\pi }}{7} - \cos \dfrac{{5\pi }}{7}} \right) + \dfrac{1}{2}\left( {\cos \dfrac{{5\pi }}{7} - \cos \dfrac{{7\pi }}{7}} \right)\\ = \dfrac{1}{2}\cos \dfrac{\pi }{7} + \dfrac{1}{2} = {\cos ^2}\dfrac{\pi }{{14}}\\\sin \dfrac{\pi }{7} = 2\sin \dfrac{\pi }{{14}}\cos \dfrac{\pi }{{14}} \Rightarrow \left( {\sin \dfrac{{2\pi }}{7} + \sin \dfrac{{4\pi }}{7} + \sin \dfrac{{6\pi }}{7}} \right) = \dfrac{1}{2}\cot \dfrac{\pi }{{14}}\end{array}$

$M = \dfrac{{\sin a}}{{{{\sin }^3}a + 2{{\cos }^3}a}} \Rightarrow \dfrac{1}{M} = \dfrac{{{{\sin }^3}a + 2{{\cos }^3}a}}{{\sin a}} = {\sin ^2}a + 2\dfrac{{{{\cos }^3}a}}{{\sin a}}$

Do $\tan a = \dfrac{{\sin a}}{{\cos a}} = 2 \Rightarrow \dfrac{{\cos a}}{{\sin a}} = \dfrac{1}{2}$ $\Rightarrow \dfrac{1}{M} = {\sin ^2}a + 2.\dfrac{1}{2}.{\cos ^2}a = {\sin ^2}a + {\cos ^2}a = 1$

$\Rightarrow M = 1$

$A = \cos \alpha {\rm{ + }}\cos \left( {\alpha  + \dfrac{\pi }{5}} \right) + ... + \cos \left( {\alpha  + \dfrac{{9\pi }}{5}} \right)$$A = \left[ {\cos \alpha  + \cos \left( {\alpha  + \dfrac{{9\pi }}{5}} \right)} \right] + ... + \left[ {\cos \left( {\alpha  + \dfrac{{4\pi }}{5}} \right) + \cos \left( {\alpha  + \dfrac{{5\pi }}{5}} \right)} \right]$$A = 2\cos \left( {\alpha  + \dfrac{{9\pi }}{{10}}} \right)\cos \dfrac{{9\pi }}{{10}} + 2\cos \left( {\alpha  + \dfrac{{9\pi }}{{10}}} \right)\cos \dfrac{{7\pi }}{{10}} + ... + 2\cos \left( {\alpha  + \dfrac{{9\pi }}{{10}}} \right)\cos \dfrac{\pi }{{10}}$$A = 2\cos \left( {\alpha  + \dfrac{{9\pi }}{{10}}} \right)\left( {\cos \dfrac{{9\pi }}{{10}} + \cos \dfrac{{7\pi }}{{10}} + \cos \dfrac{{5\pi }}{{10}} + \cos \dfrac{{3\pi }}{{10}} + \cos \dfrac{\pi }{{10}}} \right)$$A = 2\cos \left( {\alpha  + \dfrac{{9\pi }}{{10}}} \right)\left( {2\cos \dfrac{\pi }{2}\cos \dfrac{{2\pi }}{5} + 2\cos \dfrac{\pi }{2}\cos \dfrac{\pi }{5} + \cos \dfrac{\pi }{2}} \right)$$\Leftrightarrow A = 2\cos \left( {\alpha  + \dfrac{{9\pi }}{{10}}} \right).0 = 0.$

$A = \dfrac{{\sin {{15}^o} + \sin {{45}^o} + \sin {{75}^o}}}{{\cos {{15}^o} + \cos {{45}^o} + \cos {{75}^o}}}$$= \dfrac{{\left( {\sin {{15}^o} + \sin {{75}^o}} \right) + \sin {{45}^o}}}{{\left( {\cos {{15}^o} + \cos {{75}^o}} \right) + \cos {{45}^o}}}$$= \dfrac{{2\sin {{45}^o}.\cos {{30}^o} + \sin {{45}^o}}}{{2\cos {{45}^o}.\cos {{30}^o} + \cos {{45}^o}}}$

$\,\,\,\, = \dfrac{{\sin {{45}^o}\left( {2\cos {{30}^o} + 1} \right)}}{{\cos {{45}^o}\left( {2\cos {{30}^o} + 1} \right)}} = \dfrac{{\sin {{45}^o}}}{{\cos {{45}^o}}}$$= \tan {45^o} = 1$

Thực nghiệm ${\cos ^3}\pi \cos 3\pi  - {\sin ^3}\pi \sin 3\pi  - \dfrac{3}{4}\cos 4\pi  = \dfrac{1}{4}$

Ta có: ${\left( {\sin \alpha  + \cos \alpha \;} \right)^2} = {\sin ^2}\alpha  + {\cos ^2}\alpha  + 2\sin \alpha \cos \alpha  = 1 + \sin 2\alpha  = 1 + a$

Vì  ${0^o} < \alpha  < {90^o} \Rightarrow {0^o} < 2\alpha  < {180^o} \Rightarrow a > 0 \Rightarrow 1 + a > 0$

Mặt khác ${0^o} < \alpha  < {90^o} \Rightarrow \sin \alpha  + \cos \alpha  > 0 \Rightarrow \sin \alpha  + \cos \alpha  = \sqrt {a + 1}$

$A = \dfrac{{1 - \cos \dfrac{\pi }{4}}}{2} + \dfrac{{1 - \cos \dfrac{{3\pi }}{4}}}{2} + \dfrac{{1 - \cos \dfrac{{5\pi }}{4}}}{2} + \dfrac{{1 - \cos \dfrac{{7\pi }}{4}}}{2}$$= 2 - \dfrac{1}{2}\left( {\cos \dfrac{\pi }{4} + \cos \dfrac{{3\pi }}{4} + \cos \dfrac{{5\pi }}{4} + \cos \dfrac{{7\pi }}{4}} \right)$$= 2 - \dfrac{1}{2}\left( {\cos \dfrac{\pi }{4} + \cos \dfrac{{3\pi }}{4} - \cos \dfrac{{3\pi }}{4} - \cos \dfrac{\pi }{4}} \right) = 2.$

Ta có: $A = \cos \left( {x - a} \right)\left[ {\cos \left( {x - a} \right) - 2\cos a\cos x} \right] + {\cos ^2}x$

$\begin{array}{l} = \cos \left( {x - a} \right)\left( { - \cos x\cos a + \sin x\sin a} \right) + {\cos ^2}x =  - \cos \left( {x - a} \right).\cos \left( {x + a} \right) + {\cos ^2}x\\ =  - \dfrac{1}{2}\left( {\cos 2x + \cos 2a} \right) + \dfrac{{1 + \cos 2x}}{2} = \dfrac{{1 - \cos 2a}}{2} = {\sin ^2}a\end{array}$

Ta có: $\cos 2\alpha  = 1 - 2{\sin ^2}\alpha  = 1 - {2.0,6^2} = 0,28$

Với $k = 1,2,3,4,5$  ta có

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

$\begin{array}{l} \Rightarrow C.\sin \dfrac{\pi }{{11}} = \dfrac{1}{2}\left[ {\left( {\sin \dfrac{{3\pi }}{{11}} - \sin \dfrac{\pi }{{11}}} \right) + \left( {\sin \dfrac{{5\pi }}{{11}} - \sin \dfrac{{3\pi }}{{11}}} \right) + ... + \left( {\sin \dfrac{{11\pi }}{{11}} - \sin \dfrac{{9\pi }}{{11}}} \right)} \right]\\ =  - \dfrac{1}{2}\sin \dfrac{\pi }{{11}}\\ \Rightarrow C =  - \dfrac{1}{2}\end{array}$