Một số công thức biến đổi lượng giác

Ta có: \(\sin \alpha  + \cos \alpha  = 1 \Leftrightarrow \sin \alpha .\dfrac{{\sqrt 2 }}{2} + \cos \alpha .\dfrac{{\sqrt 2 }}{2} = \dfrac{{\sqrt 2 }}{2} \Leftrightarrow \sin \left( {\alpha  + \dfrac{\pi }{4}} \right) = \dfrac{{\sqrt 2 }}{2}\)

Ta có: \(0 < \alpha  < \dfrac{\pi }{2} \Rightarrow \left\{ \begin{array}{l}\cos \alpha  > 0\\\sin \alpha  > 0\end{array} \right.\)

\(\sqrt {\dfrac{{1 + \sin \alpha }}{{1 - \sin \alpha }}}  + \sqrt {\dfrac{{1 - \sin \alpha }}{{1 + \sin \alpha }}} \)\( = \dfrac{{1 + \sin \alpha  + 1 - \sin \alpha }}{{\sqrt {1 - {{\sin }^2}\alpha } }} = \dfrac{2}{{\sqrt {{{\cos }^2}\alpha } }}\)\( = \dfrac{2}{{\cos \alpha }}\)

Xét \(\Delta OAB\) và \(\Delta COD\) có:

\(\begin{array}{l}\angle OBA = \angle CDO = {90^o}\left( {gt} \right)\\B = CD\left( {gt} \right)\\AB = OD\left( {gt} \right)\\ \Rightarrow \Delta OAB = \Delta COD\left( {c - g - c} \right)\end{array}\)

\( \Rightarrow OA = OC\) (2 cạnh tương ứng)

\( \Rightarrow OA.OC = O{A^2}\)\( = O{B^2} + A{B^2} = {a^2} + {b^2}\) (Pytago)

\(\cos \angle AOC = \cos \left( {\angle AOB - \angle COD} \right)\)\( = \cos \angle AOB\cos \angle COD\)\( + \sin \angle AOB\sin \angle COD\)

\( = \dfrac{{OB}}{{OA}}.\dfrac{{OD}}{{OC}} + \dfrac{{AB}}{{OA}}.\dfrac{{CD}}{{OC}}\)\( = \dfrac{{OB.OD + AB.CD}}{{OA.OC}} = \dfrac{{ab + ab}}{{{a^2} + {b^2}}} = \dfrac{{2ab}}{{{a^2} + {b^2}}}\) 

Ta có: \(\pi  < \alpha  < \dfrac{{3\pi }}{2} \Rightarrow \sin \alpha  < 0\)

\( \Rightarrow \sin \alpha  =  - \sqrt {1 - {{\cos }^2}\alpha }  =  - \sqrt {1 - \dfrac{9}{{25}}}  =  - \dfrac{4}{5}\)

\(\cos a - \cos b =  - 2\sin \dfrac{{a + b}}{2}.\sin \dfrac{{a - b}}{2}\)

Vậy công thức D sai.

\(\begin{array}{l}A = \dfrac{{\sin 6x + \sin 7x + \sin 8x}}{{\cos 6x + \cos 7x + \cos 8x}} = \dfrac{{\left( {\sin 8x + \sin 6x} \right) + \sin 7x}}{{\left( {\cos 8x + \cos 6x} \right) + \cos 7x}}\\ = \dfrac{{2\sin 7x.\cos x + \sin 7x}}{{2\cos 7x.\cos x + \cos 7x}} = \dfrac{{\sin 7x\left( {2\cos x + 1} \right)}}{{\cos 7x\left( {2\cos x + 1} \right)}} = \dfrac{{\sin 7x}}{{\cos 7x}} = \tan 7x.\end{array}\)

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

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

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

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

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

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

$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:

\(\cos \alpha  + \cos \beta  = m;\sin \alpha  + \sin \beta  = n\)

\( \Rightarrow {m^2} + {n^2}\) \( = {\left( {\cos \alpha  + \cos \beta } \right)^2} + {\left( {\sin \alpha  + \sin \beta } \right)^2}\)

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

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

\( = 1 + 1 + 2\cos \left( {\alpha  - \beta } \right)\) \( = 2 + 2\cos \left( {\alpha  - \beta } \right)\)

Do đó \(\cos \left( {\alpha  - \beta } \right) = \dfrac{{{m^2} + {n^2} - 2}}{2}\)

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:

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

\(\begin{array}{l}A\sin \dfrac{{2\pi }}{9} \\= \sin \dfrac{{2\pi }}{9}\cos \dfrac{{2\pi }}{9}\cos \dfrac{{4\pi }}{9}\cos \dfrac{{8\pi }}{9} \\ = \frac{1}{2}.2\sin \frac{{2\pi }}{9}\cos \frac{{2\pi }}{9}\cos \frac{{4\pi }}{9}\cos \frac{{8\pi }}{9}\\= \dfrac{1}{2}\sin \dfrac{{4\pi }}{9}\cos \dfrac{{4\pi }}{9}\cos \dfrac{{8\pi }}{9}\\ = \frac{1}{2}.\frac{1}{2}.2\sin \frac{{4\pi }}{9}\cos \frac{{4\pi }}{9}\cos \frac{{8\pi }}{9}\\= \dfrac{1}{4}\sin \dfrac{{8\pi }}{9}\cos \dfrac{{8\pi }}{9} \\ = \frac{1}{4}.\frac{1}{2}.2\sin \frac{{8\pi }}{9}\cos \frac{{8\pi }}{9}\\= \dfrac{1}{8}\sin \dfrac{{16\pi }}{9} \\= \dfrac{1}{8}\sin \left( {2\pi  - \dfrac{{2\pi }}{9}} \right) \\=  - \dfrac{1}{8}\sin \dfrac{{2\pi }}{9}\\ \Rightarrow A = \dfrac{{ - 1}}{8}\end{array}\)

Ta có \(A = \dfrac{{\sin \dfrac{\pi }{9} + \sin \dfrac{{5\pi }}{9}}}{{\cos \dfrac{\pi }{9} + \cos \dfrac{{5\pi }}{9}}} = \dfrac{{2\sin \left( {\dfrac{\pi }{3}} \right)\cos \left( {\dfrac{{2\pi }}{9}} \right)}}{{2\cos \left( {\dfrac{\pi }{3}} \right)\cos \left( {\dfrac{{2\pi }}{9}} \right)}} = \tan \left( {\dfrac{\pi }{3}} \right) = \sqrt 3 \)

Ta có

$\begin{array}{l}4\cos {15^0}\cos {24^0}\cos {21^0} - \cos {\rm{1}}{{\rm{2}}^0} - \cos {18^0}\\ = 2\cos {15^0}\left( {\cos \left( {{{24}^0} + {{21}^0}} \right) + \cos \left( {{{24}^0} - {{21}^0}} \right)} \right) - 2\cos \left( {\dfrac{{{\rm{1}}{{\rm{2}}^0} + 18{}^0}}{2}} \right)\cos \left( {\dfrac{{{{12}^0} - {{18}^0}}}{2}} \right)\\ = 2\cos {15^0}\left( {\cos {{45}^0} + \cos {3^0}} \right) - 2\cos {15^0}\cos {3^0}\\ = 2\cos {15^0}.\cos {45^0} = \cos {60^0} + \cos {30^0} = \dfrac{1}{2} + \dfrac{{\sqrt 3 }}{2}\end{array}$

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 \(\cos \dfrac{{5\pi }}{2}\cos \dfrac{{3\pi }}{2} + \sin \dfrac{{7\pi }}{2}\sin \dfrac{\pi }{2} - \cos \pi \cos 2\pi  = 0\)