$\lceil$ Chứng!minh $\rfloor$ ($\sqrt{2}\angle\pi/12-\angle\pi/3=-\angle\pi/6$)
$$\sqrt{2}\,\cos(\,4\pi w+ \pi/12\,)- \cos(\,4\pi w+ \pi/3\,)= \cos(\,4\pi w- \pi/6\,)$$
$\lceil$ Chứng!minh $\rfloor$ ($\sqrt{2}\angle\pi/12-\angle\pi/3=-\angle\pi/6$)
$$\sqrt{2}\,\cos(\,4\pi w+ \pi/12\,)- \cos(\,4\pi w+ \pi/3\,)= \cos(\,4\pi w- \pi/6\,)$$
$\lceil$ Chứng!minh $\rfloor$ ($\sqrt{2}\angle\pi/12-\angle\pi/3=-\angle\pi/6$)
$$\sqrt{2}\,\cos(\,4\pi w+ \pi/12\,)- \cos(\,4\pi w+ \pi/3\,)= \cos(\,4\pi w- \pi/6\,)$$
\[\begin{array}{l}
\sqrt 2 \cos \left( {4\pi \omega + \frac{\pi }{{12}}} \right) - \cos \left( {4\pi \omega + \frac{\pi }{3}} \right)\\
= \sqrt 2 \cos \left( {4\pi \omega + \frac{\pi }{{12}}} \right) + \cos \left( {4\pi \omega - \frac{{2\pi }}{3}} \right)\\
= \sqrt 2 \left[ {\cos \left( {4\pi \omega } \right)\cos \left( {\frac{\pi }{{12}}} \right) - \sin \left( {4\pi \omega } \right)\sin \left( {\frac{\pi }{{12}}} \right)} \right] + \left[ {\cos \left( {4\pi \omega } \right)\cos \left( {\frac{{ - 2\pi }}{3}} \right) - \sin \left( {4\pi \omega } \right)\sin \left( {\frac{{ - 2\pi }}{3}} \right)} \right]\\
= \cos \left( {4\pi \omega } \right)\left[ {\sqrt 2 \cos \left( {\frac{\pi }{{12}}} \right) + \cos \left( {\frac{{ - 2\pi }}{3}} \right)} \right] - \sin \left( {4\pi \omega } \right)\left[ {\sqrt 2 \sin \left( {\frac{\pi }{{12}}} \right) + \sin \left( {\frac{{ - 2\pi }}{3}} \right)} \right]
\end{array}\]
Để ý:
$$\cos {\left( {4\pi \omega } \right)^2} + \sin {\left( {4\pi \omega } \right)^2} = 1$$
$${\left[ {\sqrt 2 \cos \left( {\frac{\pi }{{12}}} \right) + \cos \left( {\frac{{ - 2\pi }}{3}} \right)} \right]^2} + {\left[ {\sqrt 2 \sin \left( {\frac{\pi }{{12}}} \right) + \sin \left( {\frac{{ - 2\pi }}{3}} \right)} \right]^2} = 1$$
Do đó, đặt:
$$\sqrt 2 \cos \left( {\frac{\pi }{{12}}} \right) + \cos \left( {\frac{{ - 2\pi }}{3}} \right) = \cos \left( \alpha \right)$$
$$\sqrt 2 \sin \left( {\frac{\pi }{{12}}} \right) + \sin \left( {\frac{{ - 2\pi }}{3}} \right) = \sin \left( \alpha \right)$$
$$\Rightarrow \tan \left( \alpha \right) = \frac{{\sqrt 2 \sin \left( {\frac{\pi }{{12}}} \right) + \sin \left( {\frac{{ - 2\pi }}{3}} \right)}}{{\sqrt 2 \cos \left( {\frac{\pi }{{12}}} \right) + \cos \left( {\frac{{ - 2\pi }}{3}} \right)}} = \frac{{ - 1}}{{\sqrt 3 }} \Rightarrow \alpha = \frac{{ - \pi }}{6}$$Do đó:
\[\sqrt 2 \cos \left( {4\pi \omega + \frac{\pi }{{12}}} \right) - \cos \left( {4\pi \omega + \frac{\pi }{3}} \right) = \cos \left( {4\pi \omega } \right)\cos \left( \alpha \right) - \sin \left( {4\pi \omega } \right)\sin \left( \alpha \right) = \cos \left( {4\pi \omega + \alpha } \right) = \cos \left( {4\pi \omega - \frac{\pi }{6}} \right)\]
Bài viết đã được chỉnh sửa nội dung bởi tr2512: 02-06-2019 - 19:11
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