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

Ta có : \(\tan \alpha .\cot \alpha  = 1\)\( \Rightarrow \cot \alpha  = \dfrac{1}{{\tan \alpha }} = \dfrac{1}{{\dfrac{1}{2}}} = 2\).

Đặt \({\cos ^2}\alpha  = t \Rightarrow \dfrac{{{{\left( {1 - t} \right)}^2}}}{a} + \dfrac{{{t^2}}}{b} = \dfrac{1}{{a + b}}\)

\( \Leftrightarrow b{\left( {1 - t} \right)^2} + a{t^2} = \dfrac{{ab}}{{a + b}}\)\( \Leftrightarrow a{t^2} + b{t^2} - 2bt + b = \dfrac{{ab}}{{a + b}}\)\( \Leftrightarrow \left( {a + b} \right){t^2} - 2bt + b = \dfrac{{ab}}{{a + b}}\)\( \Leftrightarrow {\left( {a + b} \right)^2}{t^2} - 2b\left( {a + b} \right)t + {b^2} = 0\)\( \Leftrightarrow t = \dfrac{b}{{a + b}}\)

Suy ra ${\cos ^2}\alpha  = \dfrac{b}{{a + b}};\,{\sin ^2}\alpha  = \dfrac{a}{{a + b}}$

Vậy: \(\dfrac{{{{\sin }^8}\alpha }}{{{a^3}}} + \dfrac{{{{\cos }^8}\alpha }}{{{b^3}}} = \dfrac{a}{{{{\left( {a + b} \right)}^4}}} + \dfrac{b}{{{{\left( {a + b} \right)}^4}}} = \dfrac{1}{{{{\left( {a + b} \right)}^3}}}.\)

Ta có $C = 2{\left( {{{\sin }^4}x + {{\cos }^4}x + {{\sin }^2}x{{\cos }^2}x} \right)^2}-\left( {{{\sin }^8}x + {{\cos }^8}x} \right)$

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

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

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

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

$A = a{\cos ^2}x + 2b\sin x.cosx + c{\sin ^2}x$$ \Leftrightarrow \dfrac{A}{{{{\cos }^2}x}} = a + 2b\tan x + c{\tan ^2}x$

$ \Leftrightarrow A\left( {1 + {{\tan }^2}x} \right) = a + 2b\tan x + c{\tan ^2}x$$ \Leftrightarrow A\left( {1 + {{\left( {\dfrac{{2b}}{{a - c}}} \right)}^2}} \right) = a + 2b\dfrac{{2b}}{{a - c}} + c{\left( {\dfrac{{2b}}{{a - c}}} \right)^2}$

$ \Leftrightarrow A\dfrac{{{{\left( {a - c} \right)}^2} + {{\left( {2b} \right)}^2}}}{{{{\left( {a - c} \right)}^2}}} = \dfrac{{a{{\left( {a - c} \right)}^2} + 4{b^2}\left( {a - c} \right) + c4{b^2}}}{{{{\left( {a - c} \right)}^2}}}$

$ \Leftrightarrow A\dfrac{{{{\left( {a - c} \right)}^2} + {{\left( {2b} \right)}^2}}}{{{{\left( {a - c} \right)}^2}}} = \dfrac{{a{{\left( {a - c} \right)}^2} + 4{b^2}a}}{{{{\left( {a - c} \right)}^2}}} = \dfrac{{a.\left( {{{\left( {a - c} \right)}^2} + 4{b^2}} \right)}}{{{{\left( {a - c} \right)}^2}}}$$ \Leftrightarrow A = a$.

Vì $2\pi  < a < \dfrac{{5\pi }}{2}$ $ \Rightarrow \tan a > 0$, $\cot a > 0$.

$\sin x + \cos x = \dfrac{1}{2} \Rightarrow {\left( {\sin x + \cos x} \right)^2} = \dfrac{1}{4}$$ \Leftrightarrow 2\sin x.\cos x =  - \dfrac{3}{4}$$ \Rightarrow \sin x.\cos x =  - \dfrac{3}{8}$

Khi đó \(\sin x,\,\cos x\) là nghiệm của phương trình \({X^2} - \dfrac{1}{2}X - \dfrac{3}{8} = 0\)\( \Rightarrow \left[ \begin{array}{l}\sin x = \dfrac{{1 + \sqrt 7 }}{4}\\\sin x = \dfrac{{1 - \sqrt 7 }}{4}\end{array} \right.\)

Ta có $\sin x + \cos x = \dfrac{1}{2} \Rightarrow 2\left( {\sin x + \cos x} \right) = 1$

+) Với \(\sin x = \dfrac{{1 + \sqrt 7 }}{4}\)\( \Rightarrow 3\sin x + 2\cos x = \dfrac{{5 + \sqrt 7 }}{4}\)

+) Với \(\sin x = \dfrac{{1 - \sqrt 7 }}{4} \Rightarrow 3\sin x + 2\cos x = \dfrac{{5 - \sqrt 7 }}{4}\).

Ta có \(\dfrac{{3\sin a - 2\cos a}}{{12{{\sin }^3}a + 4{{\cos }^3}a}}\) \( = \dfrac{{\dfrac{3}{{{{\sin }^2}a}} - 2\dfrac{{\cos a}}{{\sin a}}.\dfrac{1}{{{{\sin }^2}a}}}}{{12 + 4\dfrac{{{{\cos }^3}a}}{{{{\sin }^3}a}}}}\) \( = \dfrac{{3\left( {1 + {{\cot }^2}a} \right) - 2\cot a\left( {1 + {{\cot }^2}a} \right)}}{{12 + 4{{\cot }^3}a}}\) \( =  - \dfrac{1}{4}\)

$A = \dfrac{{{{\tan }^2}a - {{\sin }^2}a}}{{{{\cot }^2}a - {{\cos }^2}a}}$$ \Leftrightarrow A = \dfrac{{{{\sin }^2}a\left( {\dfrac{1}{{{{\cos }^2}a}} - 1} \right)}}{{{{\cos }^2}\left( {\dfrac{1}{{{{\sin }^2}a}} - 1} \right)}} = \dfrac{{{{\tan }^2}a.{{\tan }^2}a}}{{{{\cot }^2}a}} = {\tan ^6}a$.

Ta có \({\sin ^2}a.{\tan ^2}a + 4{\sin ^2}a - {\tan ^2}a + 3{\cos ^2}a = {\sin ^2}a\left( {\dfrac{1}{{{{\cos }^2}a}} - 1} \right) + 4{\sin ^2}a - {\tan ^2}a + 3{\cos ^2}a\)

\( = \dfrac{{{{\sin }^2}a}}{{{{\cos }^2}a}} - {\sin ^2}a + 4{\sin ^2}a - {\tan ^2}a + 3{\cos ^2}a = 3{\sin ^2}a + 3{\cos ^2}a = 3\)

Cho \(\dfrac{\pi }{2} < \alpha  < \pi  \Rightarrow \sin \alpha  > 0\)

\(\cot \alpha  = \dfrac{1}{{\tan \alpha }} = 2\)

Ta có \(\dfrac{\pi }{2} < \alpha  < \pi  \Rightarrow \sin \alpha  > 0\)

Vậy D đúng.

\(\alpha \) là góc tù \( \Rightarrow \dfrac{\pi }{2} < \alpha  < \pi \) \( \Rightarrow \left\{ \begin{array}{l}\sin \alpha  > 0\\\cos \alpha  < 0\end{array} \right.\)\( \Rightarrow \tan \alpha  = \dfrac{{\sin \alpha }}{{\cos \alpha }} < 0\)

Ta có: \(\cos x = \dfrac{2}{{\sqrt 5 }} \Rightarrow {\cos ^2}x = \dfrac{4}{5} \Rightarrow {\sin ^2}x = 1 - {\cos ^2}x = 1 - \dfrac{4}{5} = \dfrac{1}{5}\)

Vì \( - \dfrac{\pi }{2} < x < 0 \Rightarrow \sin x < 0\) \( \Rightarrow \sin x =  - \dfrac{1}{{\sqrt 5 }}\)

Ta có: \(\cos x =  - \dfrac{2}{5} \Rightarrow {\tan ^2}x = \dfrac{1}{{{{\cos }^2}x}} - 1 = \dfrac{1}{{{{\left( { - \dfrac{2}{5}} \right)}^2}}} - 1 = \dfrac{{21}}{4}.\)

Khi \(\pi  < x < \dfrac{{3\pi }}{2} \Rightarrow \left\{ \begin{array}{l}\cos x < 0\\\sin x < 0\end{array} \right. \Rightarrow \tan x > 0\) \( \Rightarrow \tan x = \sqrt {\dfrac{{21}}{4}}  = \dfrac{{\sqrt {21} }}{2}.\)

Ta có \( - \dfrac{\pi }{2} < \alpha  < 0 \Rightarrow \)điểm cuối của cung có số đo \(\alpha \) thuộc vào góc phần tư thứ IV

\( \Rightarrow \sin \alpha  < 0,\cos \alpha  > 0\) 

Biến đổi  $\cot \dfrac{{89\pi }}{6} = \cot \left( { - \dfrac{\pi }{6} + 15\pi } \right)$ $ = \cot \left( { - \dfrac{\pi }{6}} \right) =  - \cot \dfrac{\pi }{6} =  - \sqrt 3 $

Vì $\dfrac{\pi }{2} < a < \pi $$ \Rightarrow \sin a > 0$, $\cos a < 0$.

D sai vì : \(\tan \alpha .\cot \alpha  = 1\,\left( {\alpha  \ne \dfrac{{k\pi }}{2},\,k \in \mathbb{Z}} \right)\).

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