Đáp án:
$\begin{array}{l}
b)B = 2\sin \alpha \\
c)C = \cot x\\
d)D = \cos x
\end{array}$
Giải thích các bước giải:
$\begin{array}{l}
b)B = \cos \left( {\dfrac{{3\pi }}{2} - \alpha } \right) - \sin \left( {\dfrac{{3\pi }}{2} - \alpha } \right) + \cos \left( {\alpha - \dfrac{{7\pi }}{2}} \right) - \sin \left( {\alpha - \dfrac{{7\pi }}{2}} \right)\\
= -\cos \left( {\dfrac{\pi }{2} - \alpha } \right) + \sin \left( {\dfrac{\pi }{2} - \alpha } \right) + \cos \left( {\alpha - \dfrac{\pi }{2}} \right) - \sin \left( {\alpha - \dfrac{\pi }{2}} \right)\\
= -\sin \alpha + \cos \alpha + \sin \alpha + \cos \alpha \\
= 2\cos \alpha \\
c)2\cos x + 3\cos \left( {\pi - x} \right) - \sin \left( {\dfrac{{7\pi }}{2} - x} \right) + \tan \left( {\dfrac{{3\pi }}{2} - x} \right)\\
= 2\cos x - 3\cos x + \sin \left( {\dfrac{\pi }{2} - x} \right) + \tan \left( {\dfrac{\pi }{2} - x} \right)\\
= - \cos x + \cos x + \cot x\\
= \cot x\\
d)D = 2\sin \left( {\dfrac{\pi }{2} + x} \right) + \sin \left( {5\pi - x} \right) + \sin \left( {\dfrac{{3\pi }}{2} + x} \right) + \cos \left( {\dfrac{\pi }{2} + x} \right)\\
= 2\cos \left( { - x} \right) + \sin \left( {\pi - x} \right) - \sin \left( {\dfrac{\pi }{2} + x} \right) + \sin \left( { - x} \right)\\
= 2\cos x + \sin x - \cos \left( { - x} \right) - \sin x\\
= 2\cos x + \sin x - \cos x - \sin x\\
= \cos x
\end{array}$
