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[Solved] Symmetric trigonometry

 
Trigonometry 1 & 2
Last Post by Staff 4 years ago
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June 15, 2021 2:17 pm
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(@armor)
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June 15, 2021 2:18 pm
Topic starter
Staff 
(@armor)
Admin
  Honorable Member
317 Posts
156 153 0

Since the symmetric center of \small y=cosx    is   \small (k\Pi +\frac{\Pi }{2},0) (k\epsilon \mathbb{Z}),    and the symmetric axis equation is   \small k\Pi (k\epsilon \mathbb{Z}). Thus,

\small 2x-\frac{\Pi }{3}=k\Pi +\frac{\Pi }{2}\Rightarrow x=\frac{k\Pi }{2}+\frac{5\Pi }{12} (k\epsilon \mathbb{Z}).

Since \small 2x-\frac{\Pi }{3}=k\Pi ,   then  \small x=\frac{k\Pi }{2}+\frac{\Pi }{6} (k\epsilon \mathbb{Z}),  therefore the symmetric center of the function \small y=3-2cos(2x-\frac{\Pi }{3})    is \small (\frac{k\Pi }{2}+\frac{5\Pi }{12},3) (k\epsilon \mathbb{Z}),   and the symmetric axis equation is    \small x=\frac{k\Pi }{2}+\frac{\Pi }{6} (k\epsilon \mathbb{Z}).

When   \small 2x-\frac{\Pi }{3}=2kx\Rightarrow x=k\Pi +\frac{\Pi }{6} (k\epsilon \mathbb{Z}),   the minimum of \small y=3-2cos(2x-\frac{\Pi }{3})  is 1.

When \small 2x-\frac{\Pi }{3}=(2k+1)\Pi \Rightarrow x=k\Pi +\frac{2\Pi }{3} (k\epsilon \mathbb{Z}),  the maximum of \small y=3-2cos(2x-\frac{\Pi }{3})  is  5.

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