Giá trị lượng giác của một góc bất kì từ 0 đến 180 độ

Ta có:

$E = \sin {36^0}\cos {6^0}$ $-\sin \left( {{{90}^0} + {{36}^0}} \right)\cos \left( {{{90}^0} - {6^0}} \right)$

$= \sin {36^0}\cos {6^0} - \cos {36^0}\sin {6^0}$

$= \sin \left( {{{36}^0} - {6^0}} \right) = \sin {30^0} = \dfrac{1}{2}$

$\begin{array}{l}A = \left( {{{\sin }^2}{{51}^0} + {{\sin }^2}{{39}^0}} \right) + \left( {{{\sin }^2}{{55}^0} + {{\sin }^2}{{35}^0}} \right)\\ = \left( {{{\sin }^2}{{51}^0} + {{\cos }^2}{{51}^0}} \right) + \left( {{{\sin }^2}{{55}^0} + {{\cos }^2}{{55}^0}} \right) = 2\end{array}$

Vì ${0^0} < \alpha  < \beta  < {90^0}$ nên:

$\begin{array}{l}0 < \sin \alpha  < \sin \beta ,\cos \alpha  > \cos \beta  > 0\\0 < \tan \alpha  < \tan \beta ,\cot \alpha  > \cot \beta  > 0\end{array}$

Ta có: $\cos B = \cos {30^0} = \dfrac{{\sqrt 3 }}{2}$.

Ta có:

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

Do ${90^0} < \alpha  < {180^0} \Rightarrow \tan \alpha  < 0$.

Ta có: $1 + {\tan ^2}\alpha  = \dfrac{1}{{{{\cos }^2}\alpha }}$$\Leftrightarrow {\tan ^2}\alpha  = \dfrac{5}{4}$$\Rightarrow \tan \alpha  =  - \dfrac{{\sqrt 5 }}{2}$.

$E = \dfrac{{\cot \alpha  + 3\tan \alpha }}{{2\cot \alpha  + \tan \alpha }}$$= \dfrac{{1 + 3{{\tan }^2}\alpha }}{{2 + {{\tan }^2}\alpha }}$$= \dfrac{{3\left( {{{\tan }^2}\alpha  + 1} \right) - 2}}{{1 + \left( {1 + {{\tan }^2}\alpha } \right)}}$

$= \dfrac{{\dfrac{3}{{{{\cos }^2}\alpha }} - 2}}{{\dfrac{1}{{{{\cos }^2}\alpha }} + 1}}$$= \dfrac{{3 - 2{{\cos }^2}\alpha }}{{1 + {{\cos }^2}\alpha }} = \dfrac{{19}}{{13}}$.

Ta có:

$\begin{array}{l} E = 2{\cos ^2}\alpha + 5\sin \alpha \cos \alpha + 1\\ = {\sin ^2}\alpha \left( {2.\frac{{{{\cos }^2}\alpha }}{{{{\sin }^2}\alpha }} + 5.\frac{{\cos \alpha }}{{\sin \alpha }} + \frac{1}{{{{\sin }^2}\alpha }}} \right)\\ = \left( {\frac{1}{{{{\cot }^2}\alpha + 1}}} \right)\left( {2{{\cot }^2}\alpha + 5\cot \alpha + \frac{1}{{\frac{1}{{{{\cot }^2}\alpha + 1}}}}} \right)\\ = \left( {\frac{1}{{{{\cot }^2}\alpha + 1}}} \right)\left( {2{{\cot }^2}\alpha + 5\cot \alpha + {{\cot }^2}\alpha + 1} \right)\\ = \left( {\frac{1}{{{{\cot }^2}\alpha + 1}}} \right)\left( {3{{\cot }^2}\alpha + 5\cot \alpha + 1} \right)\\ = \frac{1}{{{5^2} + 1}}\left( {{{3.5}^2} + 5.5 + 1} \right) = \frac{{101}}{{26}} \end{array}$

Ta có: ${\sin ^2}2\alpha  + {\cos ^2}2\alpha  = 1$ nên D đúng.

Ta có: $\tan \alpha .\cot \alpha  = 1$ nên C sai.

$A = \left( {\tan {1^0}.\tan {{89}^0}} \right).\left( {\tan {2^0}.\tan {{88}^0}} \right)...\left( {\tan {{44}^0}.\tan {{46}^0}} \right).\tan {45^0} = 1$.

$A = \dfrac{{{{\cot }^2}x - {{\cos }^2}x}}{{{{\cot }^2}x}} + \dfrac{{\sin x.\cos x}}{{\cot x}} = 1 - \dfrac{{{{\cos }^2}x}}{{{{\cot }^2}x}} + \dfrac{{\sin x.\cos x}}{{\cot x}} = 1 - {\sin ^2}x + {\sin ^2}x = 1$.

$7 = {\tan ^2}\alpha  + {\cot ^2}\alpha  = {\left( {\tan \alpha  + \cot \alpha } \right)^2} - 2$$\Rightarrow {m^2} = 9$$\Leftrightarrow m =  \pm 3$.

Ta có: $A = \dfrac{{3\sin \alpha  + 4\sin \alpha .\cot \alpha }}{{2\sin \alpha  - 5\sin \alpha .\cot \alpha }} = \dfrac{{3 + 4\cot \alpha }}{{2 - 5\cot \alpha }} = 13$.

Ta có: $\sin a + \cos a = \sqrt 2$$\Rightarrow 2 = {\left( {\sin a + \cos a} \right)^2}$ $\Leftrightarrow 2 = {\sin ^2}a + 2\sin a\cos a + {\cos ^2}a$ $\Leftrightarrow 2 = 1 + 2\sin a\cos a \Leftrightarrow 1 = 2\sin a\cos a$ $\Leftrightarrow \sin a\cos a = \dfrac{1}{2}$

Do đó ${\sin ^4}a + {\cos ^4}a = \left( {{{\sin }^2}a + {{\cos }^2}a} \right) - 2{\sin ^2}a{\cos ^2}a$ $= 1 - 2{\left( {\dfrac{1}{2}} \right)^2} = \dfrac{1}{2}$

${\sin ^4}x + {\cos ^4}x ={\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^2} - 2{\sin ^2}x{\cos ^2}x\\= 1 - 2{\sin ^2}x{\cos ^2}x$

$\begin{array}{l}{\sin ^6}x + {\cos ^6}x = {\left( {{{\sin }^2}x} \right)^3} + {\left( {{{\cos }^2}x} \right)^3}\\ = \left( {{{\sin }^2}x + {{\cos }^2}x} \right)\left[ {{{\sin }^4}x - {{\sin }^2}x{{\cos }^2}x + {{\cos }^4}x} \right]\\ = {\sin ^4}x - {\sin ^2}x{\cos ^2}x + {\cos ^4}x\\ = {\sin ^4}x + 2{\sin ^2}x{\cos ^2}x + {\cos ^4}x - 3{\sin ^2}x{\cos ^2}x\\ = {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^2} - 3{\sin ^2}x{\cos ^2}x\\ = 1 - 3{\sin ^2}x{\cos ^2}x\end{array}$

$f\left( x \right) = 3\left( {1 - 2{{\sin }^2}x{{\cos }^2}x} \right) - 2\left( {1 - 3{{\sin }^2}x{{\cos }^2}x} \right) = 1$