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

Ta có: $\sin \left( {{{180}^0}-a} \right) = \sin a$

Ta có: $\sin \left( {\dfrac{\pi }{2} - x} \right) = \cos x$ nên A đúng.

Đáp án B: $\sin \left( {\dfrac{\pi }{2} + x} \right) = \sin \left( {\pi  - \dfrac{\pi }{2} - x} \right)$ $ = \sin \left( {\dfrac{\pi }{2} - x} \right) = \cos x$ nên B đúng.

Đáp án C: $\tan \left( {\dfrac{\pi }{2} - x} \right) = \cot x$ nên C đúng.

Đáp án D: $\tan \left( {\dfrac{\pi }{2} + x} \right) = \tan \left( {\dfrac{\pi }{2} + x - \pi } \right)$$ = \tan \left( {x - \dfrac{\pi }{2}} \right) =  - \tan \left( {\dfrac{\pi }{2} - x} \right)$ $ =  - \cot x \ne \cot x$ nên D sai.

$A = {\cos ^2}\dfrac{\pi }{8} + {\cos ^2}\dfrac{{3\pi }}{8} + {\cos ^2}\dfrac{{3\pi }}{8} + {\cos ^2}\dfrac{\pi }{8}$ $ \Leftrightarrow A = 2\left( {{{\cos }^2}\dfrac{\pi }{8} + {{\cos }^2}\dfrac{{3\pi }}{8}} \right)$

$ \Leftrightarrow A = 2\left( {{{\cos }^2}\dfrac{\pi }{8} + {{\sin }^2}\dfrac{\pi }{8}} \right) = 2$.

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

$A = -\dfrac{{\cos {{750}^0} + \sin {{420}^0}}}{{\sin \left( { - {{330}^0}} \right) - \cos \left( { - {{390}^0}} \right)}}$ \( = -\dfrac{{\cos \left( {{{30}^0} + {{2.360}^0}} \right) + \sin \left( {{{60}^0} + {{360}^0}} \right)}}{{\sin \left( {{{30}^0} - {{360}^0}} \right) - \cos \left( { - 30 - {{360}^0}} \right)}}\) \( = -\dfrac{{\cos {{30}^0} + \sin {{60}^0}}}{{\sin {{30}^0} - \cos \left( { - {{30}^0}} \right)}}\)

$ = -\dfrac{{\cos {{30}^0} + \sin {{60}^0}}}{{\sin {{30}^0} - \cos {{30}^0}}} = -\dfrac{{2\sqrt 3 }}{{1 - \sqrt 3 }} $ $=  \dfrac{{2\sqrt 3 }}{{ \sqrt 3 }-1} $.

$B = \dfrac{{\left( {\cot {{44}^0} + \tan {{46}^0}} \right).\cos {{46}^0}}}{{\cos {{44}^0}}} $ $- \cot {72^0}.\tan{72^0}$ $ = \dfrac{{\left( {\tan {{46}^0} + \tan {{46}^0}} \right).\cos {{46}^0}}}{{\sin {{46}^0}}} - 1$\( = \dfrac{{2\tan {{46}^0}.\cos {{46}^0}}}{{\sin {{46}^0}}} - 1\) $ = \dfrac{{2\sin {{46}^0}}}{{\sin {{46}^0}}} - 1 = 2 - 1 = 1$.

\(\begin{array}{l}A = \cos \left( {\alpha  - \dfrac{\pi }{2}} \right) + \sin (\dfrac{\pi }{2} - \alpha ) - \cos \left( {\dfrac{\pi }{2} + \alpha } \right) - \sin \left( {\dfrac{\pi }{2} + \alpha } \right)\\{\rm{    }} = \sin \alpha  + \cos \alpha  + \sin \alpha  - \cos \alpha  = 2\sin \alpha \end{array}\)

$A = \dfrac{{\sin {{155}^0}.\cos {{115}^0} + \cot {{42}^0}.\cot {{48}^0}}}{{\cot {{55}^0}.\cot \left( { - {{145}^0}} \right) + \tan {{17}^0}.cot{{17}^0}}}$$ \Leftrightarrow A = \dfrac{{\sin {{25}^0}.\left( { - \sin {{25}^0}} \right) + \cot {{42}^0}.tan{{42}^0}}}{{\cot {{55}^0}.tan{{55}^0} + 1}}$

                 $ \Leftrightarrow A = \dfrac{{ - {{\sin }^2}{{25}^0} + 1}}{2}$$ \Leftrightarrow A = \dfrac{{{\rm{co}}{{\rm{s}}^2}{{25}^0}}}{2}$.

Do \(ABC\) là tam giác nên \(A + B + C = {180^0}\) và \(\dfrac{{A + C}}{2} = {90^0} - \dfrac{B}{2}\)

Khi đó: $\sin \dfrac{{A + C}}{2} = \sin \left( {{{90}^0} - \dfrac{B}{2}} \right) = \cos \dfrac{B}{2}$ nên A đúng.

$\cos \dfrac{{A + C}}{2} = \cos \left( {{{90}^0} - \dfrac{B}{2}} \right) = \sin \dfrac{B}{2}$ nên B đúng.

$\sin \left( {A + B} \right) = \sin \left( {{{180}^0} - C} \right) = \sin C$ nên C đúng.

${\rm{cos}}\left( {A + B} \right) = \cos \left( {{{180}^0} - C} \right) =  - \cos C$ nên D sai.

\(\sin \dfrac{{47\pi }}{6} = \sin \left( {8\pi  - \dfrac{\pi }{6}} \right)\)\( = \sin \left( { - \dfrac{\pi }{6}} \right) =  - \sin \dfrac{\pi }{6} =  - \dfrac{1}{2}\)

Ta có: \(\sin \alpha  = 1\, \Leftrightarrow \sin \alpha  = \sin \dfrac{\pi }{2} \Leftrightarrow \alpha  = \dfrac{\pi }{2} + k2\pi \,\,\,\,do\,\,\,\sin \left( {\alpha  + k2\pi } \right) = \sin \alpha .\)

Ta có: \(\cos \left( { - \alpha } \right) = \cos \alpha \) 

Nhìn đường tròn lượng giác ta thấy các góc cách nhau một số lẻ \(\pi \) sẽ có giá trị sin, cos đối nhau và tan, cot bằng nhau nên hệ thức \(\cos \left( {\alpha  + k\pi } \right) = \cos \alpha \) sai

\(\cos \left( {\alpha  + k\pi } \right) = \left\{ \begin{array}{l}\cos \alpha \,\,\,khi\,\,\,k = 2n\\ - \cos \alpha \,\,\,\,khi\,\,\,k = 2n + 1\end{array} \right..\)

\(\sin \left( {14\pi  - \alpha } \right) + 3\cos \left( {\dfrac{{21\pi }}{2} + \alpha } \right)\)\( - 2\sin \left( {\alpha  + 5\pi } \right) - \cos \left( {\dfrac{\pi }{2} + \alpha } \right)\)

\( = \sin \left( { - \alpha } \right) + 3\cos \left( {\dfrac{\pi }{2} + \alpha  + 10\pi } \right)\)\( - 2\sin \left( {\alpha  + \pi  + 4\pi } \right) - \sin \left( { - \alpha } \right)\)

\( = \sin \left( { - \alpha } \right) + 3\cos \left( {\dfrac{\pi }{2} + \alpha } \right)\)\( - 2\sin \left( {\alpha  + \pi } \right) - \sin \left( { - \alpha } \right)\)

\( =  - \sin \alpha  - 3\sin \alpha  + 2\sin \alpha  + \sin \alpha \)\( =  - \sin \alpha .\)

\(\begin{array}{l}N = {\left[ {\sin \left( {\dfrac{\pi }{2} - x} \right) + \cos \left( {9\pi  - x} \right)} \right]^2} + {\left[ {\cos \left( {\dfrac{\pi }{2} - x} \right)} \right]^2} \\= {\left[ {\cos x + \cos \left( {\pi  - x} \right)} \right]^2} + {\sin ^2}x\\ = {\left[ {\cos x - \cos x} \right]^2} + {\sin ^2}x = {\sin ^2}x.\end{array}\)

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.

Ta có $\cot (\pi  + \alpha ) = \cot (\alpha )$ nên C sai.