Giá trị lượng giác của các góc có liên quan đặc biệt

Ta có

Đáp án A:  $\left. \begin{array}{l}\cot (\pi  - \alpha ) =  - \cot \alpha \\\cot (\pi  + \alpha ) = \cot \alpha \end{array} \right\} \Rightarrow \cot (\pi  - \alpha ) \ne \cot (\pi  + \alpha )$. A sai

Đáp án B: $\left. \begin{array}{l}\cot (\dfrac{\pi }{2} - \alpha ) = \tan \alpha \\\tan (\pi  + \alpha ) = \tan \alpha \end{array} \right\} \Rightarrow \cot (\dfrac{\pi }{2} - \alpha ) = \tan (\pi  + \alpha )$. B đúng

Đáp án C: $\left. \begin{array}{l}\cot (\pi  + \alpha ) = \cot \alpha \\\cot ( - \alpha ) =  - \cot \alpha \end{array} \right\} \Rightarrow \cot (\pi  + \alpha ) \ne \cot ( - \alpha )$ . C sai

Đáp án D: $\left. \begin{array}{l}\cot (\pi  + \alpha ) = \cot \alpha \\\cot (\dfrac{\pi }{2} - \alpha ) = \tan \alpha \end{array} \right\} \Rightarrow \cot (\pi  + \alpha ) \ne \cot (\dfrac{\pi }{2} - \alpha )$. D sai

Đáp án A:  $\left. \begin{array}{l}\sin (\pi  - \alpha ) = \sin \alpha \\\sin (\pi  + \alpha ) =  - \sin \alpha \end{array} \right\} \Rightarrow \sin (\pi  - \alpha ) \ne \sin (\pi  + \alpha )$. A sai

Đáp án B: $\left. \begin{array}{l}\sin (\dfrac{\pi }{2} - \alpha ) = c{\rm{os}}\alpha \\\sin (\pi  + \alpha ) =  - \sin \alpha \end{array} \right\} \Rightarrow \sin (\dfrac{\pi }{2} - \alpha ) \ne \sin (\pi  + \alpha )$. B sai

Đáp án C: $\left. \begin{array}{l}\sin (\pi  + \alpha ) =  - \sin \alpha \\\sin ( - \alpha ) =  - \sin \alpha \end{array} \right\} \Rightarrow \sin (\pi  + \alpha ) = \sin ( - \alpha )$ . C đúng

Đáp án D: $\left. \begin{array}{l}\sin (\pi  + \alpha ) =  - \sin \alpha \\\cos (\dfrac{\pi }{2} - \alpha ) = \sin \alpha \end{array} \right\} \Rightarrow \sin (\pi  + \alpha ) \ne \cos (\dfrac{\pi }{2} - \alpha )$. D sai

Đáp án A: $\left. \begin{array}{l}\tan (\pi  - \alpha ) =  - \tan \alpha \\\tan (\dfrac{\pi }{2} - \alpha ) = \cot \alpha \end{array} \right\} \Rightarrow \tan (\pi  - \alpha ) \ne \tan (\dfrac{\pi }{2} - \alpha )$. A sai

Đáp án B: $\left. \begin{array}{l}\tan (\dfrac{\pi }{2} - \alpha ) = \cot \alpha \\\tan ( - \alpha ) =  - \tan \alpha \end{array} \right\} \Rightarrow \tan (\dfrac{\pi }{2} - \alpha ) \ne \tan ( - \alpha )$. B sai

Đáp án C: $\left. \begin{array}{l}\tan (\pi  - \alpha ) =  - \tan \alpha \\\tan (\dfrac{\pi }{2} + \alpha ) =  - \cot \alpha \end{array} \right\} \Rightarrow \tan (\pi  - \alpha ) \ne \tan (\dfrac{\pi }{2} + \alpha )$ . C sai

Đáp án D: $\left. \begin{array}{l}\tan (\dfrac{\pi }{2} - \alpha ) = \cot \alpha \\\cot (\pi  + \alpha ) = \cot \alpha \end{array} \right\} \Rightarrow \tan (\dfrac{\pi }{2} - \alpha ) = \cot (\pi  + \alpha )$. D đúng

Ta có $\tan (\pi  + \alpha ) = \tan (\alpha )$ nên A sai.

${\rm{S}} = {\cos ^2}{12^0} + {\cos ^2}{78^0} + {\cos ^2}{1^0} + {\cos ^2}{89^0}$

${\rm{    = (si}}{{\rm{n}}^2}{78^0} + {\cos ^2}{78^0}) + ({\sin ^2}{89^0} + {\cos ^2}{89^0})$

${\rm{    = 1 + 1 = 2}}$

${\rm{S}} = \cos {\rm{(9}}{{\rm{0}}^0} - x)\sin \left( {{{180}^0} - x} \right)$ $ - {\rm{\sin(9}}{{\rm{0}}^0} - x)\cos \left( {{{180}^0} - x} \right)$

$\begin{array}{l}{\rm{    = sinx}}.\sin {\rm{x}} - \cos {\rm{x}}{\rm{.}}\left( { - \cos x} \right)\\{\rm{    = si}}{{\rm{n}}^2}x + {\cos ^2}x\\{\rm{  }} = 1\end{array}$

$\sin {225^0} = \sin ({180^0} + {45^0}) $ $=  - \sin {45^0} = \dfrac{{ - \sqrt 2 }}{2}$

 A đúng

${\rm{\cos 22}}{{\rm{5}}^0} = \cos {\rm{(18}}{{\rm{0}}^0} + {45^0}) $ $=  - \cos {\rm{4}}{{\rm{5}}^0} =  - \dfrac{{\sqrt 2 }}{2}$

B đúng

$\tan {225^0} = \tan ({180^0} + {45^0}) = \tan {45^0} = 1$

C sai

$\cot {225^0} = \cot ({180^0} + {45^0}) = \cot {45^0} = 1$

D đúng

\(A = \cos \left( {\alpha  - \dfrac{\pi }{2}} \right) + \sin (\alpha  - \pi )\) \(=\cos \left( { \dfrac{\pi }{2}}-\alpha  \right) - \sin (\pi - \alpha  ) \) \(= \sin \alpha  - \sin \alpha  = 0\).

\(\widehat A + \widehat B + \widehat C = {180^0} \Rightarrow \cos \dfrac{{B + C}}{2} = \cos \left( {{{90}^0} - \dfrac{A}{2}} \right) = \sin \dfrac{A}{2}\)

A đúng

\(\widehat A + \widehat B + \widehat C = {180^0} \Rightarrow \sin \left( {A + C} \right) = \sin \left( {{{180}^0} - B} \right) = \sin B\)

B sai

\(\widehat A + \widehat B + \widehat C = {180^0} \Rightarrow \cos \left( {A + B + 2C} \right) = \cos \left( {{{180}^0} + C} \right) =  - \cos C\)

C đúng

\(\widehat A + \widehat B + \widehat C = {180^0} \Rightarrow \cos \left( {A + B} \right) = \cos \left( {{{180}^0} - C} \right) =  - \cos C\)

D đúng

\(\hat A + \hat B + \hat C \) \(= {180^0} \Rightarrow \cos \dfrac{{B + C}}{2} \) \(= \cos \left( {{{90}^0} - \dfrac{A}{2}} \right)\) \( = \sin \dfrac{A}{2}\)

(I) đúng

\((II){\rm{  }}\tan \dfrac{{A + B}}{2}.\tan \dfrac{C}{2}\) \( = \tan ({90^0} - \dfrac{C}{2}).\tan \dfrac{C}{2}\) \( = \cot \dfrac{C}{2}.\tan \dfrac{C}{2} = 1\)

(II) đúng

\((III){\rm{  }}\cos (A + B - C) = \cos \left( {{{180}^0} - 2C} \right) =  - \cos 2C\)

(III) sai

\(\begin{array}{l}A = \sin \left( {\pi  + x} \right) + \cos \left( {\dfrac{\pi }{2} - x} \right) + \cot \left( {2\pi  - x} \right) + \tan \left( {\dfrac{{3\pi }}{2} + x} \right)\\A =  - \sin x + \sin x - \cot x + \tan \left( { - \dfrac{\pi }{2} + x} \right)\\A =  - \cot x - \tan \left( {\dfrac{\pi }{2} - x} \right)\\A =  - \cot x - \cot x\\A =  - 2\cot x.\end{array}\)

Ta có: \(\cos \left( {3\pi  + \alpha } \right) = \cos \left( {\pi  + \alpha } \right) =  - \cos \alpha  =  - \dfrac{1}{3}\)

Ta có \(\Delta ABC \Rightarrow A + B + C = {180^o}\)  (định lý tổng 3 góc trong tam giác)

\( \Rightarrow \sin \left( {A + B} \right) = \sin \left( {{{180}^0} - A - B} \right) = \sin C\)

Vậy C đúng.