Một số công thức biến đổi lượng giác

Ta có \(A = {\sin ^4}x + {\cos ^4}x - \dfrac{1}{4}\cos 4x\) \( = {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^2} - 2{\sin ^2}x{\cos ^2}x - \dfrac{1}{4}\cos 4x\)

\( = 1 - \dfrac{1}{2}{\sin ^2}2x - \dfrac{1}{4}\cos 4x\) \( = 1 - \dfrac{1}{4}(1 - \cos 4x) - \dfrac{1}{4}\cos 4x = \dfrac{3}{4}\)

Ta có: \(A = \dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{\dfrac{{{{\cos }^2}x}}{{{{\sin }^2}x}} - \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}}}}\) \( = \dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{{{\cos }^4}x - {{\sin }^4}x}}.{\sin ^2}x{\cos ^2}x\) \( = \dfrac{{{{\sin }^2}x{{\cos }^2}x}}{{{{\sin }^2}x + {{\cos }^2}x}} = \dfrac{1}{4}{\left( {2\sin x\cos x} \right)^2}\)

\( = \dfrac{1}{4}{\sin ^2}2x\).

\(C = \sin \left( {a + b} \right) + \sin \left( {\dfrac{\pi }{2} - a} \right)\sin \left( { - b} \right)\)\( = \sin a\cos b + \cos a\sin b - \cos a\sin b\)\( = \sin a\cos b\)

Ta có: \(A = \dfrac{{1 + 2\sin x\cos x}}{{{{\cos }^2}x - {{\sin }^2}x}} = \dfrac{{{{\sin }^2}x + 2\sin x\cos x + {{\cos }^2}x}}{{\cos 2x}} = \dfrac{{{{\left( {\sin x + \cos x} \right)}^2}}}{{\left( {\cos x - \sin x} \right)\left( {\cos x + \sin x} \right)}}\)

\( = \dfrac{{\sin x + \cos x}}{{\cos x - \sin x}} = \dfrac{{\sqrt 2 \sin \left( {x + \dfrac{\pi }{4}} \right)}}{{\sqrt 2 \cos \left( {x + \dfrac{\pi }{4}} \right)}} = \tan \left( {x + \dfrac{\pi }{4}} \right)\).

Ta có \({\sin ^6}x + {\cos ^6}x\) \( = {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^3}\) \( - 3{\sin ^2}x{\cos ^2}x\left( {{{\sin }^2}x + {{\cos }^2}x} \right)\)

\( = 1 - 3{\left( {\dfrac{1}{2}\sin 2x} \right)^2}\) \( = 1 - \dfrac{3}{4}\dfrac{{1 - \cos 4x}}{2} = \dfrac{3}{8}\cos 4x + \dfrac{5}{8}\) \( \Rightarrow S = m + n = 1\)

Ta có:

\(\begin{array}{l}3{\sin ^4}x + 2{\cos ^4}x = \dfrac{{98}}{{81}}\\2{\sin ^4}x + 3{\cos ^4}x = A\end{array}\)

Trừ vế cho vế hai đẳng thức trên ta được:

\(\begin{array}{l}{\sin ^4}x - {\cos ^4}x = \dfrac{{98}}{{81}} - A\\ \Leftrightarrow \left( {{{\sin }^2}x - {{\cos }^2}x} \right)\left( {{{\sin }^2}x + {{\cos }^2}x} \right) = \dfrac{{98}}{{81}} - A\\ \Leftrightarrow {\sin ^2}x - {\cos ^2}x = \dfrac{{98}}{{81}} - A\\ \Leftrightarrow {\cos ^2}x - {\sin ^2}x = A - \dfrac{{98}}{{81}}\\ \Leftrightarrow \cos 2x = A - \dfrac{{98}}{{81}}\end{array}\)

Cộng vế với vế hai đẳng thức đầu bài ta được:

\(\begin{array}{l}5{\sin ^4}x + 5{\cos ^4}x = A + \dfrac{{98}}{{81}}\\ \Leftrightarrow 5\left( {{{\sin }^4}x + {{\cos }^4}x} \right) = A + \dfrac{{98}}{{81}}\\ \Leftrightarrow 5\left[ {{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}^2} - 2{{\sin }^2}x{{\cos }^2}x} \right] = A + \dfrac{{98}}{{81}}\\ \Leftrightarrow 5\left( {1 - \dfrac{1}{2}.4{{\sin }^2}x{{\cos }^2}x} \right) = A + \dfrac{{98}}{{81}}\\ \Leftrightarrow 5\left( {1 - \dfrac{1}{2}{{\sin }^2}2x} \right) = A + \dfrac{{98}}{{81}}\\ \Leftrightarrow 1 - \dfrac{1}{2}{\sin ^2}2x = \dfrac{1}{5}\left( {A + \dfrac{{98}}{{81}}} \right)\\ \Leftrightarrow 1 - \dfrac{1}{2}\left( {1 - {{\cos }^2}2x} \right) = \dfrac{1}{5}\left( {A + \dfrac{{98}}{{81}}} \right)\\ \Leftrightarrow \dfrac{1}{2} + \dfrac{1}{2}{\cos ^2}2x = \dfrac{1}{5}\left( {A + \dfrac{{98}}{{81}}} \right)\\ \Leftrightarrow 1 + {\cos ^2}2x = \dfrac{2}{5}\left( {A + \dfrac{{98}}{{81}}} \right)\end{array}\)

Thay \(\cos 2x = A - \dfrac{{98}}{{81}}\) ta được:

\( 1 + {\left( {A - \dfrac{{98}}{{81}}} \right)^2} = \dfrac{2}{5}\left( {A + \dfrac{{98}}{{81}}} \right) \) \(= \dfrac{2}{5}\left( {A - \dfrac{{98}}{{81}}} \right) + \dfrac{{392}}{{405}}\)

Đặt \(A - \dfrac{{98}}{{81}} = t\)\( \Rightarrow {t^2} - \dfrac{2}{5}t + \dfrac{{13}}{{405}} = 0\)\( \Leftrightarrow \left[ \begin{array}{l}t = \dfrac{{13}}{{45}}\\t = \dfrac{1}{9}\end{array} \right.\)

+) \(t = \dfrac{{13}}{{45}} \Rightarrow A = \dfrac{{607}}{{405}}\)

+) \(t = \dfrac{1}{9} \Rightarrow A = \dfrac{{107}}{{81}}.\)

Ta có \(\tan a + \tan b = \dfrac{{\sin a}}{{\cos a}} + \dfrac{{\sin b}}{{\cos b}}\) \( = \dfrac{{\sin a\cos b + \sin b\cos a}}{{\cos a\cos b}} = \dfrac{{\sin \left( {a + b} \right)}}{{\cos a\cos b}}\) suy ra A đúng

Tương tự ta có B đúng.

\(\tan a + \cot a = \dfrac{{\sin a}}{{\cos a}} + \dfrac{{\cos a}}{{\sin a}}\) \( = \dfrac{{{{\sin }^2}a + {{\cos }^2}a}}{{\sin a\cos a}} = \dfrac{2}{{\sin 2a}}\) nên D đúng.

\(\cot a + \cot b = \dfrac{{\cos a}}{{\sin a}} + \dfrac{{\cos b}}{{\sin b}} = \dfrac{{\sin \left( {a + b} \right)}}{{\sin a\sin b}}\) nên C sai.

Ta có tam giác ABC vuông cân tại A và AB = 2, M  là trung điểm AB

\(\begin{array}{l} \Rightarrow MA = \dfrac{1}{2}AB = 1;AC = AB = 2\\ \Rightarrow \tan \angle ACB = \dfrac{{AB}}{{AC}} = 1\\\tan \angle MCA = \dfrac{{AM}}{{AC}} = \dfrac{1}{2}\end{array}\) 

Mặt khác \(\tan \angle ACB = \dfrac{{\tan \angle MCA + \tan \angle MCB}}{{1 - \tan \angle MCA.\tan \angle MCB}}\)

Hay \(1 = \dfrac{{\dfrac{1}{2} + \tan \angle MCB}}{{1 - \dfrac{1}{2}.\tan \angle MCB}}\)\( \Leftrightarrow 1 - \dfrac{1}{2}\tan \angle MCB = \dfrac{1}{2} + \tan \angle MCB\)

\( \Leftrightarrow \dfrac{3}{2}\tan \angle MCB = \dfrac{1}{2} \Leftrightarrow \tan \angle MCB = \dfrac{1}{3}\)

\(\dfrac{1}{{{{\sin }^2}\alpha }} = 1 + {\cot ^2}\alpha  = 1 + 18 = 19\)\( \to {\sin ^2}\alpha  = \dfrac{1}{{19}}\)\( \to \sin \alpha  =  \pm \dfrac{1}{{\sqrt {19} }}\)

Vì  $\dfrac{\pi }{2} < \alpha  < \pi $$ \Rightarrow \sin \alpha  > 0$$ \Rightarrow \sin \alpha  = \dfrac{1}{{\sqrt {19} }}$

Suy ra $\tan \dfrac{\alpha }{2} + \cot \dfrac{\alpha }{2} = \dfrac{{{{\sin }^2}\dfrac{\alpha }{2} + {{\cos }^2}\dfrac{\alpha }{2}}}{{\sin \dfrac{\alpha }{2}\cos \dfrac{\alpha }{2}}} = \dfrac{2}{{\sin \alpha }} = 2\sqrt {19} $.

A đúng vì \(VT = \dfrac{{\tan x + \tan y}}{{\dfrac{1}{{\tan x}} + \dfrac{1}{{{\mathop{\rm tany}\nolimits} }}}} = \tan x.\tan y = VP\)

B đúng vì

\(VT = \dfrac{{1 + \sin a}}{{1 - \sin a}} + \dfrac{{1 - \sin a}}{{1 + \sin a}} - 2\) \( = \dfrac{{{{\left( {1 + \sin a} \right)}^2} + {{\left( {1 - \sin a} \right)}^2}}}{{1 - {{\sin }^2}a}} - 2\) \( = \dfrac{{2 + 2{{\sin }^2}a}}{{{{\cos }^2}a}} - 2 = 4{\tan ^2}a = VP\)

C đúng vì \(VT = \dfrac{{ - {{\sin }^2}\alpha  - {{\cos }^2}\alpha }}{{{{\cos }^2}\alpha  - {{\sin }^2}\alpha }} = \dfrac{{{{\sin }^2}\alpha  + {{\cos }^2}\alpha }}{{{{\sin }^2}\alpha  - {{\cos }^2}\alpha }} = \dfrac{{1 + {{\cot }^2}\alpha }}{{1 - {{\cot }^2}\alpha }} = VP\).

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

\(\begin{array}{l} \Rightarrow \cos \alpha  =  - \sqrt {1 - {{\sin }^2}\alpha }  =  - \sqrt {1 - \dfrac{1}{9}}  \\=  - \sqrt {\dfrac{8}{9}}  =  - \dfrac{{2\sqrt 2 }}{3}\end{array}\)

$\Rightarrow \sin 2\alpha  = 2\sin \alpha \cos \alpha $$=2.\left( -\dfrac{1}{3}\right) .\left( -\dfrac{{2\sqrt 2 }}{3}\right) $$= 2.\dfrac{1}{3}.\dfrac{{2\sqrt 2 }}{3} $$= \dfrac{{4\sqrt 2 }}{9}$

Ta có: \(\sin 2a = 2\sin a\cos a;\,\,\,\cos 2a = 2{\cos ^2}a - 1 = 1 - 2{\sin ^2}a = {\cos ^2}a - {\sin ^2}a.\)

Vậy B sai

\(\cos 2\alpha  = 2{\cos ^2}\alpha  - 1 = 2.\dfrac{1}{9} - 1 =  - \dfrac{7}{9}\)            

\(\sin 2a = 2\sin a\cos a = 2.\dfrac{1}{{\sqrt 2 }}.\dfrac{{\sqrt 2 }}{2} = 1\)

Ta có: \(T = \cos \left( {a + b} \right)\cos \left( {a - b} \right) - \sin \left( {a + b} \right)\sin \left( {a - b} \right) = \cos \left[ {\left( {a + b} \right) + \left( {a - b} \right)} \right] = \cos 2a\)

Ta có: \(\sin \alpha \cos \beta  + \sin \beta \cos \alpha  = \sin \left( {\alpha  + \beta } \right) = \sin {90^o} = 1\)

Ta có: \(\dfrac{1}{2}\left[ {\cos \left( {\dfrac{\pi }{3} - 2x} \right) - \cos \left( {\dfrac{\pi }{2} + 2x} \right)} \right]\)\( - \sin \dfrac{\pi }{{12}}.\cos \left( {\dfrac{\pi }{{12}} + 2x} \right) = \sin \left( {ax + b\pi } \right)\)

\(\begin{array}{l} \Leftrightarrow \dfrac{1}{2}.\left( { - 2} \right).\sin \left( {\dfrac{{\dfrac{\pi }{3} - 2x + \dfrac{\pi }{2} + 2x}}{2}} \right).\\\sin \left( {\dfrac{{\dfrac{\pi }{3} - 2x - \dfrac{\pi }{2} - 2x}}{2}} \right) \\- \sin \dfrac{\pi }{{12}}.\cos \left( {\dfrac{\pi }{{12}} + 2x} \right) = \sin \left( {ax + b\pi } \right)\\ \Leftrightarrow  - \sin \dfrac{{5\pi }}{{12}}.\sin \left( { - \dfrac{\pi }{{12}} - 2x} \right) \\- \sin \dfrac{\pi }{{12}}.\cos \left( {\dfrac{\pi }{{12}} + 2x} \right) = \sin \left( {ax + b\pi } \right)\\ \Leftrightarrow \sin \left( {\dfrac{\pi }{2} - \dfrac{\pi }{{12}}} \right).\sin \left( {\dfrac{\pi }{{12}} + 2x} \right)\\ - \sin \dfrac{\pi }{{12}}.\cos \left( {\dfrac{\pi }{{12}} + 2x} \right) = \sin \left( {ax + b\pi } \right)\\ \Leftrightarrow \cos \dfrac{\pi }{{12}}.\sin \left( {\dfrac{\pi }{{12}} + 2x} \right)\\ - \sin \dfrac{\pi }{{12}}.\cos \left( {\dfrac{\pi }{{12}} + 2x} \right) = \sin \left( {ax + b\pi } \right)\\ \Leftrightarrow \sin \left( {\dfrac{\pi }{{12}} + 2x - \dfrac{\pi }{{12}}} \right) = \sin \left( {ax + b\pi } \right) \\\Leftrightarrow \sin 2x = \sin \left( {ax + b\pi } \right)\\ \Rightarrow \left\{ \begin{array}{l}a = 2\\b = 2k\,\,\,\,\left( {k \in Z} \right)\end{array} \right.\,\,\,\,\,\,\\Do\,\,\,\,b \in \left[ {0;\dfrac{1}{2}} \right] \Rightarrow b = 0 \Rightarrow a + b = 2.\end{array}\)

Ta có: \(t = \dfrac{{\sin x}}{{\cos x}} = \tan x \Rightarrow \dfrac{1}{{{{\cos }^2}x}} = 1 + {\tan ^2}x = 1 + {t^2} \Rightarrow {\cos ^2}x = \dfrac{1}{{1 + {t^2}}}\)

Với \(x \ne \dfrac{\pi }{2} + k\pi  \Rightarrow \cos x \ne 0 \Rightarrow {\cos ^2}x \ne 0\). Chia cả 2 vế của biểu thức cho \({\cos ^2}x \ne 0\) ta được:

\(\dfrac{P}{{{{\cos }^2}x}} = 3.\dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} + 2.\dfrac{{\sin x}}{{\cos x}} - 1 \Leftrightarrow \left( {1 + {t^2}} \right)P = 3{t^2} + 2t - 1 \Leftrightarrow P = \dfrac{{3{t^2} + 2t - 1}}{{1 + {t^2}}}\) 

\(\begin{array}{l}S = {\sin ^2}{5^o} + {\sin ^2}{10^o} + {\sin ^2}{15^o} + ... + {\sin ^2}{85^o}\\ = \left( {{{\sin }^2}{5^o} + {{\sin }^2}{{85}^o}} \right) + \left( {{{\sin }^2}{{10}^o} + {{\sin }^2}{{80}^o}} \right) + ... + \left( {{{\sin }^2}{{40}^o} + {{\sin }^2}{{50}^o}} \right) + {\sin ^2}{45^o}\\ = \left( {{{\sin }^2}{5^o} + {{\cos }^2}{5^o}} \right) + \left( {{{\sin }^2}{{10}^o} + {{\cos }^2}{{10}^o}} \right) + ... + \left( {{{\sin }^2}{{40}^o} + {{\cos }^2}{{40}^o}} \right) + {\sin ^2}{45^o}\\ = 8 + {\left( {\dfrac{{\sqrt 2 }}{2}} \right)^2} = 8 + \dfrac{1}{2} = \dfrac{{17}}{2}.\end{array}\)