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Chứng minh rằng với mọi \(\alpha \) ta có:
a) \(\sin \left( {\dfrac{{5\pi }}{4} + \alpha } \right) = – \sin \left( {\dfrac{{3\pi }}{4} – \alpha } \right)\);
b) \(\cos \left( {\alpha – \dfrac{{2\pi }}{3}} \right) = – \cos \left( {\dfrac{\pi }{3} + \alpha } \right)\);
c) \(\cos \left( {\alpha – \dfrac{{2\pi }}{3}} \right) = \cos \left( {\dfrac{{4\pi }}{3} + \alpha } \right).\)
a)
\(\begin{array}{l}\sin \left( {\dfrac{{5\pi }}{4} + \alpha } \right) = \sin \left( {2\pi – \dfrac{{3\pi }}{4} + \alpha } \right)\\ = \sin \left( { – \dfrac{{3\pi }}{4} + \alpha } \right) = – \sin \left( {\dfrac{{3\pi }}{4} – \alpha } \right)\end{array}\)
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b)
\(\begin{array}{l}\cos \left( {\alpha – \dfrac{{2\pi }}{3}} \right) = – \cos \left( {\alpha – \dfrac{{2\pi }}{3} + \pi } \right)\\ = – \cos \left( {\alpha + \dfrac{\pi }{3}} \right)\end{array}\)
c)
\(\begin{array}{l}\cos \left( {\alpha – \dfrac{{2\pi }}{3}} \right) = \cos \left( {\alpha + \dfrac{{4\pi }}{3} – 2\pi } \right)\\ = \cos \left( {\alpha + \dfrac{{4\pi }}{3}} \right)\end{array}\)