Câu 6.32 trang 200 SBT Đại số 10 Nâng cao

LG a

\(\cos \left( {\alpha  - \dfrac{\pi }{2}} \right) + \sin \left( {\alpha  - \pi } \right);\)

Phương pháp giải:

\(\begin{array}{l}
\cos ( - x) = \cos x;\;\cos (\frac{\pi }{2} - x) = \sin x.\\
\sin ( - x) = - \sin x;\;\sin (\pi - x) = \sin x
\end{array}\)

Lời giải chi tiết:

\(\begin{array}{l}
\cos \left( {\alpha - \frac{\pi }{2}} \right) + \sin \left( {\alpha - \pi } \right) = \cos \left( {\frac{\pi }{2} - \alpha } \right) - \sin \left( {\pi - \alpha } \right)\\
= \sin \alpha - \sin \alpha = 0.
\end{array}\)

LG b

\(\cos \left( {\pi  - \alpha } \right) + \sin \left( {\alpha  + \dfrac{\pi }{2}} \right);\)

Lời giải chi tiết:

\(\cos \left( {\pi  - \alpha } \right) + \sin \left( {\alpha  + \frac{\pi }{2}} \right) =  - \cos \alpha  + \cos \left( { - \alpha } \right) =  - \cos \alpha  + \cos \alpha  = 0.\)

LG c

\(\cos \left( {\dfrac{\pi }{2} - \alpha } \right) + \sin \left( {\dfrac{\pi }{2} - \alpha } \right) - \cos \left( {\dfrac{\pi }{2} + \alpha } \right) - \sin \left( {\dfrac{\pi }{2} + \alpha } \right)\);

Lời giải chi tiết:

\(\begin{array}{l}
\cos \left( {\frac{\pi }{2} - \alpha } \right) + \sin \left( {\frac{\pi }{2} - \alpha } \right) - \cos \left( {\frac{\pi }{2} + \alpha } \right) - \sin \left( {\frac{\pi }{2} + \alpha } \right)\\
= \sin \alpha + \cos \alpha - \sin \left( { - \alpha } \right) - \cos \left( { - \alpha } \right)\\
= \sin \alpha + \cos \alpha - ( - \sin \alpha ) - \cos \alpha \\
= 2\sin \alpha
\end{array}\)

LG d

\(\cos \left( {\dfrac{{3\pi }}{2} - \alpha } \right) - \sin \left( {\dfrac{{3\pi }}{2} - \alpha } \right) + \cos \left( {\alpha  - \dfrac{{7\pi }}{2}} \right) - \sin \left( {\alpha  - \dfrac{{7\pi }}{2}} \right)\);

Lời giải chi tiết:

\(\begin{array}{l}
\cos \left( {\frac{{3\pi }}{2} - \alpha } \right) - \sin \left( {\frac{{3\pi }}{2} - \alpha } \right) + \cos \left( {\alpha - \frac{{7\pi }}{2}} \right) - \sin \left( {\alpha - \frac{{7\pi }}{2}} \right)\\
= \cos \left( {2\pi - \frac{\pi }{2} - \alpha } \right) - \sin \left( {2\pi - \frac{\pi }{2} - \alpha } \right) + \cos \left( {\alpha + \frac{\pi }{2} - 4\pi } \right) - \sin \left( {\alpha + \frac{\pi }{2} - 4\pi } \right)\\
= \cos \left( { - \frac{\pi }{2} - \alpha } \right) - \sin \left( { - \frac{\pi }{2} - \alpha } \right) + \cos \left( {\alpha + \frac{\pi }{2}} \right) - \sin \left( {\alpha + \frac{\pi }{2}} \right)\\
= \cos \left( {\frac{\pi }{2} + \alpha } \right) - \left( { - \sin \left( {\frac{\pi }{2} + \alpha } \right)} \right) + \cos \left( {\alpha + \frac{\pi }{2}} \right) - \sin \left( {\alpha + \frac{\pi }{2}} \right)\\
= 2\cos \left( {\alpha + \frac{\pi }{2}} \right) = 2\sin ( - \alpha ) = - 2\sin \alpha .
\end{array}\)

LG e

\(\cos \left( {\dfrac{\pi }{2} - \alpha } \right) + \cos \left( {\pi  - \alpha } \right) + \cos \left( {\dfrac{{3\pi }}{2} - \alpha } \right) + \cos \left( {2\pi  - \alpha } \right)\);

Lời giải chi tiết:

\(\begin{array}{l}
\cos \left( {\frac{\pi }{2} - \alpha } \right) + \cos \left( {\pi - \alpha } \right) + \cos \left( {\frac{{3\pi }}{2} - \alpha } \right) + \cos \left( {2\pi - \alpha } \right)\\
= \sin \alpha + \left( { - \cos \alpha } \right) + \cos \left( {2\pi - \frac{\pi }{2} - \alpha } \right) + \cos \left( { - \alpha } \right)\\
= \sin \alpha - \cos \alpha + \cos \left( { - \frac{\pi }{2} - \alpha } \right) + \cos \alpha \\
= \sin \alpha + \cos \left( {\frac{\pi }{2} + \alpha } \right) = \sin \alpha + \sin \left( { - \alpha } \right)\\
= \sin \alpha - \sin \alpha = 0
\end{array}\)

LG f

\(\sin \left( {\dfrac{{5\pi }}{2} - \alpha } \right) - \cos \left( {\dfrac{{13\pi }}{2} - \alpha } \right) - 3\sin \left( {\alpha  - 5\pi } \right) - 2\sin \alpha  - \cos \alpha ;\)

Lời giải chi tiết:

\(\begin{array}{l}
\sin \left( {\frac{{5\pi }}{2} - \alpha } \right) - \cos \left( {\frac{{13\pi }}{2} - \alpha } \right) - 3\sin \left( {\alpha - 5\pi } \right) - 2\sin \alpha - \cos \alpha \\
= \sin \left( {2\pi + \frac{\pi }{2} - \alpha } \right) - \cos \left( {6\pi + \frac{\pi }{2} - \alpha } \right) - 3\sin \left( {\alpha - \pi - 4\pi } \right) - 2\sin \alpha - \cos \alpha \\
= \sin \left( {\frac{\pi }{2} - \alpha } \right) - \cos \left( {\frac{\pi }{2} - \alpha } \right) - 3\sin \left( {\alpha - \pi } \right) - 2\sin \alpha - \cos \alpha \\
= \cos \alpha - \sin \alpha - 3\left( { - \sin \left( {\pi - \alpha } \right)} \right) - 2\sin \alpha - \cos \alpha \\
= - 3\sin \alpha - 3\left( { - \sin \alpha } \right) = 0.
\end{array}\)

LG g

\(\cos \left( {5\pi  + \alpha } \right) - 2\sin \left( {\dfrac{{11\pi }}{2} - \alpha } \right) - \sin \left( {\dfrac{{11\pi }}{2} + \alpha } \right)\).

Lời giải chi tiết:

\(\begin{array}{l}
\cos \left( {5\pi + \alpha } \right) - 2\sin \left( {\frac{{11\pi }}{2} - \alpha } \right) - \sin \left( {\frac{{11\pi }}{2} + \alpha } \right)\\
= \cos \left( {4\pi + \pi + \alpha } \right) - 2\sin \left( {6\pi - \frac{\pi }{2} - \alpha } \right) - \sin \left( {6\pi - \frac{\pi }{2} + \alpha } \right)\\
= \cos \left( {\pi + \alpha } \right) - 2\sin \left( { - \frac{\pi }{2} - \alpha } \right) - \sin \left( { - \frac{\pi }{2} + \alpha } \right)\\
= - \cos \left( { - \alpha } \right) - 2\left( { - \sin \left( {\frac{\pi }{2} + \alpha } \right)} \right) - \left( { - \sin \left( {\frac{\pi }{2} - \alpha } \right)} \right)\\
= - \cos \alpha - 2\left( { - \cos \left( { - \alpha } \right)} \right) - \left( { - \cos \alpha } \right)\\
= - \cos \alpha - 2\left( { - \cos \alpha } \right) - \left( { - \cos \alpha } \right)\\
= 2\cos \alpha
\end{array}\)