Solve?
- Thread starter loser
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Is the "a - 7" in the numerator or the denominator? That is, do you mean this?
. . . . .\(\displaystyle \large{0.7\,= \,3.4 \sin{\left(\frac{2\pi}{12.3(a\,-\,7)}\right)}\,+\,4.1}\)
...or this?
. . . . .\(\displaystyle \large{0.7\,= \,3.4 \sin{\left(\frac{2\pi (a\,-\,7)}{12.3}\right)}\,+\,4.1}\)
...or something else?
Note: Either way, I would start by subtracting off the 4.1 and dividing through by the 3.4. Then take the inverse sine, and go from there.
Thank you.
Eliz.
. . . . .\(\displaystyle \large{0.7\,= \,3.4 \sin{\left(\frac{2\pi}{12.3(a\,-\,7)}\right)}\,+\,4.1}\)
...or this?
. . . . .\(\displaystyle \large{0.7\,= \,3.4 \sin{\left(\frac{2\pi (a\,-\,7)}{12.3}\right)}\,+\,4.1}\)
...or something else?
Note: Either way, I would start by subtracting off the 4.1 and dividing through by the 3.4. Then take the inverse sine, and go from there.
Thank you.
Eliz.
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It would help if you replied to the question, rather than editing your post, since edits are often overlooked and, in this case, the reformatting doesn't (I think) answer the question.in the edited post said:
Please reply with clarification of what the equation is. If you're not sure how to format the equation, please study the explanations found in the links in the "Forum Help" pull-down menu at the very top of the page. (Using the "Karls Notes" formatting would probably be quickest.)
When you reply, please include the steps you've tried thus far.
Thank you.
Eliz.
\(\displaystyle \mbox{ 0.7 = 3.4\sin{\left(\frac{2\pi}{12.3}(a - 7)\right)} + 4.1}\)
Solve for \(\displaystyle \mbox{\sin{\left(\frac{2\pi}{12.3}(a - 7)\right)}}\):
\(\displaystyle \mbox{ -3.4 = 3.4\sin{\left(\frac{2\pi}{12.3}(a - 7)\right)}}\)
\(\displaystyle \mbox{ \Rightarrow \sin{\left(\frac{2\pi}{12.3}(a - 7)\right)} = -1}\)
Can you proceed?
Solve for \(\displaystyle \mbox{\sin{\left(\frac{2\pi}{12.3}(a - 7)\right)}}\):
\(\displaystyle \mbox{ -3.4 = 3.4\sin{\left(\frac{2\pi}{12.3}(a - 7)\right)}}\)
\(\displaystyle \mbox{ \Rightarrow \sin{\left(\frac{2\pi}{12.3}(a - 7)\right)} = -1}\)
Can you proceed?
Thanks for all the help. I am unsure of when to use the inverse of sin because i had initially solved it and it became a negative number which cannnot be right. stapel... thanks for all your help and time but i dont really appreciate the directions in your reply and i take it as rudeness
\(\displaystyle \mbox{y = \sin{(x)}}\) reaches \(\displaystyle \mbox{y=-1}\) only once per cycle, so taking the arcsine alone is satisfactory (we don't need to worry about quadrants).
Even if there was a physical limitation preventing \(\displaystyle \mbox{a}\) from being negative, what do you get \(\displaystyle \mbox{a}\) to equal when
\(\displaystyle \L \mbox{\frac{2\pi}{12.3}\left(a - 7\right) = \frac{-\pi}{2}}\) ?
Even if there was a physical limitation preventing \(\displaystyle \mbox{a}\) from being negative, what do you get \(\displaystyle \mbox{a}\) to equal when
\(\displaystyle \L \mbox{\frac{2\pi}{12.3}\left(a - 7\right) = \frac{-\pi}{2}}\) ?
pka
Elite Member
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Of course the equation \(\displaystyle \L
\frac{{2\pi }}{{12.3}}\left( {a - 7} \right) = \frac{{ - \pi }}{2}\) has a positive solution a~3.925.
One can get a general solution to \(\displaystyle \L
\frac{{2\pi }}{{12.3}}\left( {a - 7} \right) = \frac{{(4k - 1)\pi }}{2}\) where k in an integer, because \(\displaystyle \L
\sin \left[ {\frac{{(4k - 1)\pi }}{2}} \right] = - 1\).
\frac{{2\pi }}{{12.3}}\left( {a - 7} \right) = \frac{{ - \pi }}{2}\) has a positive solution a~3.925.
One can get a general solution to \(\displaystyle \L
\frac{{2\pi }}{{12.3}}\left( {a - 7} \right) = \frac{{(4k - 1)\pi }}{2}\) where k in an integer, because \(\displaystyle \L
\sin \left[ {\frac{{(4k - 1)\pi }}{2}} \right] = - 1\).