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<math class="inline">S_1 \cup S_2 = \left\{ \sin (t),\sin (\frac{t}{2}), \cos (t)\right\}</math>  
 
<math class="inline">S_1 \cup S_2 = \left\{ \sin (t),\sin (\frac{t}{2}), \cos (t)\right\}</math>  
  
<math class="inline">S_1 \cup S_2</math> is a set because the union of two sets is the set of all distinct elements from those two sets.  In this case because <math class="inline"> \sin (t+\frac{\pi}{2}) </math> and <math>\cos (t)</math> are equivalent, we only need to include one of these elements in our union set.
+
<math class="inline">S_1 \cup S_2</math> is a set because the union of two sets is the set of all distinct elements from those two sets.  In this case because <math class="inline"> \sin (t+\frac{\pi}{2}) </math> and <math class="inline"> \cos (t)</math> are equivalent, we only need to include one of these elements in our union set.
  
 
===Answer 3===
 
===Answer 3===

Revision as of 10:31, 10 January 2013

Practice Problemon set operations


Consider the following sets:

$ \begin{align} S_1 &= \left\{ \sin (t), \cos (t)\right\}, \\ S_2 & = \left\{ \sin (\frac{t}{2}), \sin (t+\frac{\pi}{2})\right\}. \\ \end{align} $

Write $ S_1 \cup S_2 $ explicitely. Is $ S_1 \cup S_2 $ a set?


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Answer 1

No, because a set must have unique elements; sin(t+pi/2) is basically cos(t). The union of both sets is a set with elements from both S1 and S2. S1 U S2 = {sin(t),cos(t),sin(t/2),sin(t+pi/2)}

Answer 2

$ S_1 \cup S_2 = \left\{ \sin (t),\sin (\frac{t}{2}), \cos (t)\right\} $

$ S_1 \cup S_2 $ is a set because the union of two sets is the set of all distinct elements from those two sets. In this case because $ \sin (t+\frac{\pi}{2}) $ and $ \cos (t) $ are equivalent, we only need to include one of these elements in our union set.

Answer 3

Write it here.


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