Giá trị lượng giác của một góc (cung) lượng giác

Ta có $\sin {743^0} = \sin ({23^0} + {2.360^0}) = \sin 23{}^0$.

\(\begin{array}{l}A = \dfrac{{\cos {{750}^0} + \sin {{420}^0}}}{{\sin ( - {{330}^0}) - \cos ( - {{390}^0})}}\\{\rm{   }} = \dfrac{{{\rm{cos(3}}{{\rm{0}}^{\rm{0}}} + {{2.360}^0}) + \sin ({{60}^0} + {{360}^0})}}{{ - \sin ( - {{30}^0} + {{360}^0}) - \cos ({{30}^0} + {{360}^0})}}\\{\rm{   }} = \dfrac{{\cos {{30}^0} + \sin {{60}^0}}}{{\sin {{30}^0} - \cos {{30}^0}}}\\{\rm{   }} = \dfrac{{\dfrac{{\sqrt {\rm{3}} }}{{\rm{2}}} + \dfrac{{\sqrt 3 }}{2}}}{{\dfrac{1}{2} - \dfrac{{\sqrt 3 }}{2}}} = \dfrac{{2\sqrt 3 }}{{1 - \sqrt 3 }} =  - 3 - \sqrt 3 \end{array}\)

Vì ${0^0} < x < {90^0}$ nên $\sin x > 0,\cos x > 0,\tan {\rm{ }}x > 0,\cot x > 0$

Suy ra $\cos x < 0$ sai.

    Đáp án B: $1 + {\tan ^2}\alpha  = \dfrac{1}{{{{\cos }^2}\alpha }},{\rm{ }}\alpha  \ne k\pi ,{\rm{ }}k \in Z$ sai vì $\cos x \ne 0{\rm{ }} \Leftrightarrow {\rm{ }}x \ne \dfrac{\pi }{2} + k\pi ,{\rm{ }}k \in Z$

    Đáp án C: ${\sin ^2}\alpha  + {\cos ^2}\beta  = 1$ sai vì \(\alpha  \ne \beta \).

    Đáp án D: $1 + {\cot ^2}\alpha  = \dfrac{1}{{{{\sin }^2}\alpha }},{\rm{ }}\alpha  \ne \dfrac{\pi }{2}{\rm{ + k}}\pi {\rm{, k}} \in Z$ sai vì $\sin x \ne 0{\rm{ }} \Leftrightarrow {\rm{ }}x \ne k\pi ,{\rm{ }}k \in Z$

Ta có ${\sin ^2}\alpha  + {\cos ^2}\alpha  = 1 \Rightarrow {\sin ^2}\alpha  = 1 - {\cos ^2}\alpha  = 1 - {\left( {\dfrac{{ - 12}}{{13}}} \right)^2} = \dfrac{{25}}{{169}} \Rightarrow {\rm{ }}\sin \alpha  =  \pm \dfrac{5}{{13}}$

Vì $\dfrac{\pi }{2} < \alpha  < \pi $ nên $\sin \alpha  > 0 \Rightarrow {\rm{sin}}\alpha  = \dfrac{5}{{13}} \Rightarrow \tan \alpha  = \dfrac{{\sin \alpha }}{{{\rm{cos}}\alpha }} = \dfrac{{ - 5}}{{12}}$.

$P = m\sin {0^0} + {\rm{ }}n\cos {0^0}{\rm{ }} + {\rm{ }}p\sin {90^0} = {\rm{ }}m.0{\rm{ }} + {\rm{ }}n.1{\rm{ }} + {\rm{ }}p.1{\rm{ }} = n{\rm{ }} + {\rm{ }}p$.

$S = 3 - {\sin ^2}{90^0} + 2{\cos ^2}{60^0} - 3{\tan ^2}{45^0} = 3 - {1^2} + 2.{\left( {\dfrac{1}{2}} \right)^2} - {3.1^2} = \dfrac{{ - 1}}{2}$.

$P = 3{\sin ^2}x + 4{\cos ^2}x = 3({\sin ^2}x + {\cos ^2}x) + {\cos ^2}x = 3 + {\left( {\dfrac{1}{2}} \right)^2} = \dfrac{{13}}{4}$.

${\sin ^6}x + {\cos ^6}x = {({\sin ^2}x + {\cos ^2}x)^3} - 3{\sin ^2}x{\cos ^2}x({\sin ^2}x + {\cos ^2}x) = 1 - 3{\sin ^2}x{\cos ^2}x$

$\tan \alpha  + \cot \alpha  = 2 \Rightarrow {(\tan \alpha  + \cot \alpha )^2} = 4 \Rightarrow {\tan ^2}\alpha  + 2\tan \alpha \cot \alpha  + {\cot ^2}\alpha  = 4 \Rightarrow {\tan ^2}\alpha  + {\cot ^2}\alpha  = 2$

Vì $\pi  < \alpha  < \dfrac{{3\pi }}{2}{\rm{ }} \Rightarrow \dfrac{\pi }{2}{\rm{ + }}\pi {\rm{  < }}\dfrac{\pi }{2} + \alpha  < \dfrac{\pi }{2} + \dfrac{{3\pi }}{2} \Rightarrow \dfrac{{3\pi }}{2} < \dfrac{\pi }{2} + \alpha  < 2\pi  \Rightarrow {\rm{sin}}\left( {\dfrac{\pi }{2} + \alpha } \right) < 0$

Ta có ${\cos ^2}\alpha  = 1 - {\sin ^2}\alpha  \Rightarrow {\cos ^2}\alpha  = \dfrac{8}{9} \Rightarrow \cos \alpha  =  \pm \dfrac{{2\sqrt 2 }}{3}$

Vì $\dfrac{\pi }{2} < \alpha  < \pi  \Rightarrow \cos \alpha  = \dfrac{{ - 2\sqrt 2 }}{3} \Rightarrow \tan \alpha  = \dfrac{{\sin \alpha }}{{{\rm{cos}}\alpha }} = \dfrac{{ - \sqrt 2 }}{4}$.

Ta có ${\sin ^2}\alpha  = 1 - {\cos ^2}\alpha  \Rightarrow {\sin ^2}\alpha  = \dfrac{5}{9} \Rightarrow \sin \alpha  =  \pm \dfrac{{\sqrt 5 }}{3}$

Vì ${180^0} < \alpha  < {270^0} \Rightarrow \sin \alpha  = \dfrac{{ - \sqrt 5 }}{3} \Rightarrow \cot \alpha  = \dfrac{{{\rm{cos}}\alpha }}{{\sin \alpha }} = \dfrac{{2\sqrt 5 }}{5}$.

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

$= {\left[ {\left( {\sin \alpha + \dfrac{{\sin \alpha }}{{\cos \alpha }}} \right):\left( {\cos \alpha + 1} \right)} \right]^2} + 1$

$= {\left[ {\sin \alpha \left( {1 + \dfrac{1}{{\cos \alpha }}} \right).\dfrac{1}{{\cos \alpha + 1}}} \right]^2} $$+ 1$

$= {\left[ {\sin \alpha .\dfrac{{\cos \alpha + 1}}{{\cos \alpha }}.\dfrac{1}{{\cos \alpha + 1}}} \right]^2} $$+ 1$

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

$= \dfrac{{{{\sin }^2}\alpha + {{\cos }^2}\alpha }}{{{{\cos }^2}\alpha }}\\
= \dfrac{1}{{{{\cos }^2}\alpha }}$

Vì $\cos {90^0} = 0$ nên $A = \cos {235^0}.\sin {60^0}.\tan {125^0}.\cos {90^0} = 0$.

 $\begin{array}{l}P = {\cos ^2}x.{\cot ^2}x{\rm{ }} + 3{\cos ^2}x - {\cot ^2}x + 2{\sin ^2}x\\{\rm{    }} = {\cot ^2}x\left( {{{\cos }^2}x - 1} \right) + {\cos ^2}x + 2\left( {{{\cos }^2}x + {{\sin }^2}x} \right)\\{\rm{    }} = \dfrac{{{{\cos }^2}x}}{{{{\sin }^2}x}}.( - {\sin ^2}x) + {\cos ^2}x + 2\\{\rm{    }} =  - {\cos ^2}x + {\cos ^2}x + 2 = 2\end{array}$

Ta có:

$6{\cos ^2}x + 6{\sin }x - 2$ $= 6(1 - {\sin ^2}x) + 6\sin x - 2$ $=  - 6{\sin ^2}x + 6\sin x + 4$ $=  - 6({\sin ^2}x - \sin x) + 4$ $=  - 6{\left( {\sin x - \dfrac{1}{2}} \right)^2} + \dfrac{{11}}{2} \le \dfrac{{11}}{2}$

Dấu $“=”$ xảy ra khi \(\sin x = \dfrac{1}{2}\).