MIDTERMS I NEED HEL0 UNDERSTANDING! THE UNIT CIRCLE.Help Let P(t) be the point on the unit circle U that corresponds to t.If P (t) has the given retangular coordnates find a)P(T+PIE) B)P(t-pie) c)P(-t) d)P(-t-pie) for (3/5,4/5)
help with exams is prohibited
please read the anti-cheating policy under the terms and conditions
unless you have something else to say @Mertsj
I AM PRACTICING FOR EXAMS I DONT HAVE A MIDTERM IN FRONT OF ME
next time rephrase your question as it is very misleading
i need help understanding how is that cheating exactly?
because the point of an exam/test is to test your own understanding of the subject
i didnt know people have take home midterm test for maths ...n e ways can someone please help me
The wording of this question is confusing... So at the end there, is that an ordered pair like this? \[\large \left(t,\;p(t)\right) \qquad=\qquad \left(\frac{3}{5},\;\frac{4}{5}\right)\]
Oh p(t) is the given `rectangular` coordinates.. so that's (x,y) i guess? This is worded so strange :( I can't make sense of it.
yes i think so
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Does this drawing make sense to you? I've drawn the unit circle, and labeled our x and y coordinates. This particular location we're calling \(\large P(t)\).
with this can you just plot the graph to see uwhat u r doing also right?
Ya it helps to visualize it I think :)
yes it makes sense
So if this is our point \(\large P(t)\), and \(\large t\) represents some distance along the unit circle, what would adding \(\large \pi\) to our \(\large t\) do? So if you remember back to the unit circle, one full rotation is \(\large 2\pi\). So a half-rotation is just \(\large \pi\). So if we add \(\large \pi\) to our \(\large t\), it's another way of saying "rotate your point halfway around the circle"
ok
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