Câu 1:
a) Cho sina=2/3 với π/2 { "@context": "https://schema.org", "@type": "QAPage", "mainEntity": { "@type": "Question", "name": " Câu 1:
a) Cho sina=2/3 với π/2
0 bình luận về “Câu 1:
a) Cho sina=2/3 với π/2<a<π. Hãy tính cosa, tana, cota.
b) Rút gọn biểu thức P=(sinx + sin3x +sin5x) /(cosx + cos3x + cos5x)
Câu 2: Biểu thức”
Giải thích các bước giải:
\(\begin{array}{l} 1,\\ \dfrac{\pi }{2} < a < \pi \Rightarrow \left\{ \begin{array}{l} \sin a > 0\\ \cos a < 0 \end{array} \right.\\ \cos a < 0 \Rightarrow \cos a = – \sqrt {1 – {{\sin }^2}a} = – \dfrac{{\sqrt 5 }}{3}\\ \tan a = \dfrac{{\sin a}}{{\cos a}} = – \dfrac{2}{{\sqrt 5 }}\\ \cot a = \dfrac{{\cos a}}{{\sin a}} = – \dfrac{{\sqrt 5 }}{2}\\ b,\\ P = \dfrac{{\sin x + \sin 3x + \sin 5x}}{{\cos x + \cos 3x + \cos 5x}}\\ = \dfrac{{\left( {\sin x + \sin 5x} \right) + \sin 3x}}{{\left( {\cos x + \cos 5x} \right) + \cos 3x}}\\ = \dfrac{{2\sin 3x.\cos 2x + \sin 3x}}{{2.\cos 3x.\cos 2x + \cos 3x}}\\ = \dfrac{{\sin 3x\left( {2\cos 2x + 1} \right)}}{{\cos 3x\left( {2\cos 2x + 1} \right)}}\\ = \dfrac{{\sin 3x}}{{\cos 3x}} = \tan 3x\\ 2,\\ x \ge 5 \Rightarrow x > 4 \Rightarrow 4 – x < 0,\,\,\,\,\forall x \ge 5\\ 3,\\ \dfrac{{3x – 6}}{{5 + x}} \le 0 \Leftrightarrow \left\{ \begin{array}{l} x \ne – 5\\ \left( {3x – 6} \right)\left( {x + 5} \right) \le 0 \end{array} \right. \Leftrightarrow – 5 < x \le 2\\ x \in Z \Rightarrow x \in \left\{ { – 4; – 3; – 2; – 1;0;1;2} \right\} \end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
1,\\
\dfrac{\pi }{2} < a < \pi \Rightarrow \left\{ \begin{array}{l}
\sin a > 0\\
\cos a < 0
\end{array} \right.\\
\cos a < 0 \Rightarrow \cos a = – \sqrt {1 – {{\sin }^2}a} = – \dfrac{{\sqrt 5 }}{3}\\
\tan a = \dfrac{{\sin a}}{{\cos a}} = – \dfrac{2}{{\sqrt 5 }}\\
\cot a = \dfrac{{\cos a}}{{\sin a}} = – \dfrac{{\sqrt 5 }}{2}\\
b,\\
P = \dfrac{{\sin x + \sin 3x + \sin 5x}}{{\cos x + \cos 3x + \cos 5x}}\\
= \dfrac{{\left( {\sin x + \sin 5x} \right) + \sin 3x}}{{\left( {\cos x + \cos 5x} \right) + \cos 3x}}\\
= \dfrac{{2\sin 3x.\cos 2x + \sin 3x}}{{2.\cos 3x.\cos 2x + \cos 3x}}\\
= \dfrac{{\sin 3x\left( {2\cos 2x + 1} \right)}}{{\cos 3x\left( {2\cos 2x + 1} \right)}}\\
= \dfrac{{\sin 3x}}{{\cos 3x}} = \tan 3x\\
2,\\
x \ge 5 \Rightarrow x > 4 \Rightarrow 4 – x < 0,\,\,\,\,\forall x \ge 5\\
3,\\
\dfrac{{3x – 6}}{{5 + x}} \le 0 \Leftrightarrow \left\{ \begin{array}{l}
x \ne – 5\\
\left( {3x – 6} \right)\left( {x + 5} \right) \le 0
\end{array} \right. \Leftrightarrow – 5 < x \le 2\\
x \in Z \Rightarrow x \in \left\{ { – 4; – 3; – 2; – 1;0;1;2} \right\}
\end{array}\)