a,cho cosx=2sinx(pi 01/10/2021 Bởi Claire a,cho cosx=2sinx(pi { "@context": "https://schema.org", "@type": "QAPage", "mainEntity": { "@type": "Question", "name": " a,cho cosx=2sinx(pi
Giải thích các bước giải: Ta có: \(\begin{array}{l}a,\\\pi < x < \dfrac{{3\pi }}{2} \Rightarrow \left\{ \begin{array}{l}\sin x < 0\\\cos x < 0\end{array} \right.\\{\sin ^2}x + {\cos ^2}x = 1\\ \Leftrightarrow {\sin ^2}x + {\left( {2\sin x} \right)^2} = 1\\ \Leftrightarrow 5{\sin ^2}x = 1\\ \Leftrightarrow {\sin ^2}x = \dfrac{1}{5}\\\sin x < 0 \Rightarrow \sin x = – \dfrac{1}{{\sqrt 5 }} \Rightarrow \cos x = – \dfrac{2}{{\sqrt 5 }}\\\cos 2x = 2{\cos ^2}x – 1 = 2.{\left( { – \dfrac{2}{{\sqrt 5 }}} \right)^2} – 1 = \dfrac{3}{5}\\b,\\\tan x = – 3 \Leftrightarrow \dfrac{{\sin x}}{{\cos x}} = – 3 \Leftrightarrow \sin x = – 3\cos x\\{\sin ^2}x + {\cos ^2}x = 1\\ \Leftrightarrow {\left( { – 3\cos x} \right)^2} + {\cos ^2}x = 1\\ \Leftrightarrow {\cos ^2}x = \dfrac{1}{{10}}\\P = 3{\sin ^2}x + 2\sin x.\cos x – {\cos ^2}x\\ = 3.{\left( { – 3\cos x} \right)^2} + 2.\left( { – 3\cos x} \right).\cos x – {\cos ^2}x\\ = 20{\cos ^2}x = 20.\dfrac{1}{{10}} = 2\\c,\\A{B^2} = \sqrt {B{C^2} – A{C^2}} = \sqrt {{{10}^2} – {6^2}} = 8\left( {cm} \right)\\{S_{ABC}} = \dfrac{1}{2}AB.AC = \dfrac{1}{2}.6.8 = 24\left( {c{m^2}} \right)\\{S_{ABC}} = \dfrac{{AB + BC + CA}}{2}.r\\ \Leftrightarrow 24 = \dfrac{{6 + 8 + 10}}{2}.r\\ \Leftrightarrow r = 2\left( {cm} \right)\end{array}\) Bình luận
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\pi < x < \dfrac{{3\pi }}{2} \Rightarrow \left\{ \begin{array}{l}
\sin x < 0\\
\cos x < 0
\end{array} \right.\\
{\sin ^2}x + {\cos ^2}x = 1\\
\Leftrightarrow {\sin ^2}x + {\left( {2\sin x} \right)^2} = 1\\
\Leftrightarrow 5{\sin ^2}x = 1\\
\Leftrightarrow {\sin ^2}x = \dfrac{1}{5}\\
\sin x < 0 \Rightarrow \sin x = – \dfrac{1}{{\sqrt 5 }} \Rightarrow \cos x = – \dfrac{2}{{\sqrt 5 }}\\
\cos 2x = 2{\cos ^2}x – 1 = 2.{\left( { – \dfrac{2}{{\sqrt 5 }}} \right)^2} – 1 = \dfrac{3}{5}\\
b,\\
\tan x = – 3 \Leftrightarrow \dfrac{{\sin x}}{{\cos x}} = – 3 \Leftrightarrow \sin x = – 3\cos x\\
{\sin ^2}x + {\cos ^2}x = 1\\
\Leftrightarrow {\left( { – 3\cos x} \right)^2} + {\cos ^2}x = 1\\
\Leftrightarrow {\cos ^2}x = \dfrac{1}{{10}}\\
P = 3{\sin ^2}x + 2\sin x.\cos x – {\cos ^2}x\\
= 3.{\left( { – 3\cos x} \right)^2} + 2.\left( { – 3\cos x} \right).\cos x – {\cos ^2}x\\
= 20{\cos ^2}x = 20.\dfrac{1}{{10}} = 2\\
c,\\
A{B^2} = \sqrt {B{C^2} – A{C^2}} = \sqrt {{{10}^2} – {6^2}} = 8\left( {cm} \right)\\
{S_{ABC}} = \dfrac{1}{2}AB.AC = \dfrac{1}{2}.6.8 = 24\left( {c{m^2}} \right)\\
{S_{ABC}} = \dfrac{{AB + BC + CA}}{2}.r\\
\Leftrightarrow 24 = \dfrac{{6 + 8 + 10}}{2}.r\\
\Leftrightarrow r = 2\left( {cm} \right)
\end{array}\)
Đáp án:
Giải thích các bước giải:
Here you are