a) Cho \(\sin x = \dfrac{1}{3}\) với \(\dfrac{\pi }{2} < x < \pi .\) Tính \(\cos x,\cos \left( {2x +
a) Cho \(\sin x = \dfrac{1}{3}\) với \(\dfrac{\pi }{2} < x < \pi .\) Tính \(\cos x,\cos \left( {2x + \dfrac{\pi }{3}} \right)\)
b) Cho \(\sin x = \dfrac{7}{{25}}\) và \(0 < x < \dfrac{\pi }{2}.\) Tính \(\cos \left( {x + \dfrac{\pi }{6}} \right)\)
c) Cho \(\tan \alpha = - 3\) với \(\dfrac{{3\pi }}{2} < \alpha < 2\pi \). Tính \(\sin \left( {\alpha + \dfrac{\pi }{4}} \right)\)
a) + Ta có: \({\sin ^2}x + {\cos ^2}x = 1 \Leftrightarrow {\cos ^2}x = 1 - {\sin ^2}x = \dfrac{8}{9}\)
+ Do \(\dfrac{\pi }{2} < x < \pi \Rightarrow \cos x < 0\)\( \Rightarrow \cos x = - \sqrt {\dfrac{8}{9}} = - \dfrac{{2\sqrt 2 }}{3}\)
+ Khi đó: \(\cos \left( {2x + \dfrac{\pi }{3}} \right) = \cos 2x\cos \dfrac{\pi }{3} - \sin 2x\sin \dfrac{\pi }{3} = \left( {2{{\cos }^2}x - 1} \right).\cos \dfrac{\pi }{3} - 2\sin x\cos x\sin \dfrac{\pi }{3}\)
\( = \left[ {2.{{\left( { - \dfrac{{2\sqrt 2 }}{3}} \right)}^2} - 1} \right].\dfrac{1}{2} - 2.\dfrac{1}{3}.\left( { - \dfrac{{2\sqrt 2 }}{3}} \right).\dfrac{{\sqrt 3 }}{2} = \dfrac{{7 + 4\sqrt 6 }}{{18}}\)
b) + Ta có: \({\sin ^2}x + {\cos ^2}x = 1 \Leftrightarrow {\cos ^2}x = 1 - {\sin ^2}x = \dfrac{{576}}{{625}}\)
+ Do \(0 < x < \dfrac{\pi }{2} \Rightarrow \cos x > 0\)\( \Rightarrow \cos x = \sqrt {\dfrac{{576}}{{625}}} = \dfrac{{24}}{{25}}\)
+ Khi đó: \(\cos \left( {x + \dfrac{\pi }{6}} \right) = \cos x\cos \dfrac{\pi }{6} - \sin x\sin \dfrac{\pi }{6} = \dfrac{{24}}{{25}}.\dfrac{{\sqrt 3 }}{2} - \dfrac{7}{{25}}.\dfrac{1}{2} = \dfrac{{ - 7 + 24\sqrt 3 }}{{50}}\)
c) + Ta có: \(1 + {\tan ^2}\alpha = \dfrac{1}{{{{\cos }^2}\alpha }} \Rightarrow {\cos ^2}\alpha = \dfrac{1}{{1 + {{\tan }^2}\alpha }} = \dfrac{1}{{10}}\)
+ Do \(\dfrac{{3\pi }}{2} < \alpha < 2\pi \Rightarrow \cos \alpha > 0 \Rightarrow \cos \alpha = \sqrt {\dfrac{1}{{10}}} = \dfrac{{\sqrt {10} }}{{10}}\)
+ Lại có: \(\tan \alpha = \dfrac{{\sin \alpha }}{{\cos \alpha }} \Rightarrow \sin \alpha = \tan \alpha \cos \alpha = - \dfrac{{3\sqrt {10} }}{{10}}\)
+ Khi đó: \(\sin \left( {\alpha + \dfrac{\pi }{4}} \right) = \sin \alpha \cos \dfrac{\pi }{4} + \cos \alpha \sin \dfrac{\pi }{4} = - \dfrac{{3\sqrt {10} }}{{10}}.\dfrac{{\sqrt 2 }}{2} + \dfrac{{\sqrt {10} }}{{10}}.\dfrac{{\sqrt 2 }}{2} = - \dfrac{{\sqrt 5 }}{5}\)
Hỗ trợ - Hướng dẫn
-
024.7300.7989
-
1800.6947
(Thời gian hỗ trợ từ 7h đến 22h)
Email: lienhe@tuyensinh247.com