Câu 6.33 trang 201 SBT Đại số 10 Nâng cao

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) $\cos \left( {\alpha  - \dfrac{{2\pi }}{3}} \right) = \cos \left( {\dfrac{{4\pi }}{3} + \alpha } \right).$

Giải:

a)

$\begin{array}{l}\sin \left( {\dfrac{{5\pi }}{4} + \alpha } \right) = \sin \left( {2\pi  - \dfrac{{3\pi }}{4} + \alpha } \right)\\ = \sin \left( { - \dfrac{{3\pi }}{4} + \alpha } \right) =  - \sin \left( {\dfrac{{3\pi }}{4} - \alpha } \right)\end{array}$

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)

$\begin{array}{l}\cos \left( {\alpha  - \dfrac{{2\pi }}{3}} \right) = \cos \left( {\alpha  + \dfrac{{4\pi }}{3} - 2\pi } \right)\\ = \cos \left( {\alpha  + \dfrac{{4\pi }}{3}} \right)\end{array}$

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