$$29.7=21.9 \sin(\tfrac{\pi}{6}.1-\phi)+51.6$$
Here $\phi=\tfrac{2\pi}{3}$
Graph this equation and the regression equation on the same graph.
Can someone help me find theta algebraically. I know what the answer is but I don't know how to get to it. IF you could show every step. I'm confused how to get rid of the sin and parentheses to get theta by itself.
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$\begingroup$$$29.7=21.9\sin(\frac{\pi}{6}*1-\phi)+51.6$$Find $\phi$
Okay so we can do the arithemtic, once we're done with that we're left with $$-1=\sin(\frac{\pi}{6}*1-\phi)$$
Due to the nature of the $\sin$ function we have multiple values for $\sin(x)=-1$ **
Since $sin$ equals $-1$ once every cycle, and every cycle is of length $2\pi$, we have $x=\frac{3\pi}{2}+2\pi k$ for ** where k is any integer
From here we can see that in your problem we have $$\frac{\pi}{6}*1-\phi=\frac{3\pi}{2}+2\pi k$$
So $$\phi=\frac{\pi}{6}-\frac{3\pi}{2}-2\pi k$$
So doing the arithmetic on the left side we have:$$\phi=-\frac{4\pi}{3}-2\pi k$$
So we have $\phi=\frac{2\pi}{3}$ when $k=-1$
$\endgroup$ $\begingroup$I would do it this way:\begin{align} 29.7= &21.9 \sin \Bigl(\frac{\pi}{6} \cdot 1 - \phi \Bigr) + 51.6\iff 21.9 \sin \Bigl(\frac{\pi}{6} - \phi \Bigr)=29.7-51.6=-21.9 \\ \iff &\sin \Bigl(\frac{\pi}{6} - \phi \Bigr)=-1\iff \frac{\pi}{6} - \phi\equiv -\frac{\pi}{2}\mod 2\pi \\ \iff &\phi\equiv \frac{\pi}{6}+\frac{\pi}{2}=\frac{2\pi}{3}\mod 2\pi \end{align}
$\endgroup$ $\begingroup$$$29.7 = 21.9 \sin \left(\frac{\pi}{6} \cdot 1 - \phi \right) + 51.6$$
$$29.7 - 51.6 = 21.9 \sin \left(\frac{\pi}{6} \cdot 1 - \phi \right)$$
$$\frac{29.7 - 51.6}{21.9} = \sin \left(\frac{\pi}{6} \cdot 1 - \phi \right)$$
$$\sin^{-1} \left( \frac{29.7 - 51.6}{21.9} \right) = \frac{\pi}{6} \cdot 1 - \phi$$
$$\sin^{-1} \left( \frac{29.7 - 51.6}{21.9} \right) - \frac{\pi}{6} = - \phi$$
$$ -\sin^{-1} \left( \frac{29.7 - 51.6}{21.9} \right) + \frac{\pi}{6}= \phi$$
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