Trả lời bởi giáo viên
Đáp án đúng: b
$\sin \dfrac{{7\pi }}{{12}} = \sin \left( {\dfrac{\pi }{3} + \dfrac{\pi }{4}} \right)$$ = \sin \dfrac{\pi }{3}\cos \dfrac{\pi }{4} + \sin \dfrac{\pi }{4}\cos \dfrac{\pi }{3}$$ = \dfrac{{\sqrt 3 }}{2}.\dfrac{{\sqrt 2 }}{2} + \dfrac{1}{2}.\dfrac{{\sqrt 2 }}{2} = \dfrac{{\sqrt 6 + \sqrt 2 }}{4}$
$\cos {285^0} = \cos \left( {{{360}^0} - {{75}^0}} \right)$$ = \cos {75^0} = \cos \left( {{{30}^0} + {{45}^0}} \right)$$ = \cos {30^0}\cos {45^0} - \sin {30^0}\sin {45^0}$$ = \dfrac{{\sqrt 3 }}{2}.\dfrac{{\sqrt 2 }}{2} - \dfrac{1}{2}.\dfrac{{\sqrt 2 }}{2}$$ = \dfrac{{\sqrt 6 - \sqrt 2 }}{4}$
$\sin \dfrac{\pi }{{12}} = \sin \left( {\dfrac{\pi }{3} - \dfrac{\pi }{4}} \right)$$ = \sin \dfrac{\pi }{3}\cos \dfrac{\pi }{4} - \cos \dfrac{\pi }{3}\sin \dfrac{\pi }{4}$$ = \dfrac{{\sqrt 3 }}{2}.\dfrac{{\sqrt 2 }}{2} - \dfrac{1}{2}.\dfrac{{\sqrt 2 }}{2}$$ = \dfrac{{\sqrt 6 - \sqrt 2 }}{4}$
$\sin \dfrac{{103\pi }}{{12}} = \sin \left( {8\pi + \dfrac{{7\pi }}{{12}}} \right)$$ = \sin \dfrac{{7\pi }}{{12}} = \dfrac{{\sqrt 6 + \sqrt 2 }}{4}$
Hướng dẫn giải:
Sử dụng các công thức biến đổi:
$\begin{array}{l}\sin (a + b) = \sin a\cos b + \sin b\cos a\\\sin (a - b) = \sin a\cos b - \sin b\cos a\end{array}$
$\begin{array}{l}\cos (a + b) = \cos a\cos b - \sin a\sin b\\\cos (a - b) = \cos a\cos b + \sin a\sin b\end{array}$
