Exact values of a Trigonometric function
- Thread starter anna_sims
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This one is fun:
\(\displaystyle (\frac{4}{3}*\pi+4*asin(\frac{3}{5}),0)\)
With what values of cosine are you familiar? Find values for 'x' that produce those known reference angles.
240º = \(\displaystyle \frac{4}{3}\pi\)
600º = \(\displaystyle \frac{10}{3}\pi\)
\(\displaystyle \frac{2\pi}{\frac{1}{4}} = 8\pi\) <== The Period of your function.
\(\displaystyle \frac{3}{2}\pi\), for example. cos(\(\displaystyle \frac{3}{2}\pi\)) = 0
\(\displaystyle \frac{1}{4}*(x-\frac{4}{3}\pi) = \frac{3}{2}\pi\)
\(\displaystyle x-\frac{4}{3}\pi = 6\pi\)
\(\displaystyle x = \frac{22}{3}\pi\)
By Symmetry, this works out to \(\displaystyle \frac{10}{3}\pi\) which you state we don't get to use.
Pick another one.
\(\displaystyle cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}\)
\(\displaystyle \frac{\pi}{6} = \frac{1}{4}*(x-\frac{4}{3}\pi)\)
This leads to x = \(\displaystyle 2\pi\) which is in your Domain. \(\displaystyle (2\pi,3-\frac{5}{2}\sqrt{3})\)
\(\displaystyle (\frac{4}{3}*\pi+4*asin(\frac{3}{5}),0)\)
With what values of cosine are you familiar? Find values for 'x' that produce those known reference angles.
240º = \(\displaystyle \frac{4}{3}\pi\)
600º = \(\displaystyle \frac{10}{3}\pi\)
\(\displaystyle \frac{2\pi}{\frac{1}{4}} = 8\pi\) <== The Period of your function.
\(\displaystyle \frac{3}{2}\pi\), for example. cos(\(\displaystyle \frac{3}{2}\pi\)) = 0
\(\displaystyle \frac{1}{4}*(x-\frac{4}{3}\pi) = \frac{3}{2}\pi\)
\(\displaystyle x-\frac{4}{3}\pi = 6\pi\)
\(\displaystyle x = \frac{22}{3}\pi\)
By Symmetry, this works out to \(\displaystyle \frac{10}{3}\pi\) which you state we don't get to use.
Pick another one.
\(\displaystyle cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}\)
\(\displaystyle \frac{\pi}{6} = \frac{1}{4}*(x-\frac{4}{3}\pi)\)
This leads to x = \(\displaystyle 2\pi\) which is in your Domain. \(\displaystyle (2\pi,3-\frac{5}{2}\sqrt{3})\)