Question

$${ 9 }^{ 2 } = { 12 }^{ 2 } + { 6 }^{ 2 } -2(12)(6) \cos y$$

Answer

y=sequence(2*PI*n+arccos(11/16),in(n,Z)),sequence(2*PI*n+2*PI-arccos(11/16),in(n,Z))

Solution


Simplify  \({9}^{2}\)  to  \(81\).
\[81={12}^{2}+{6}^{2}-2\times 12\times 6\cos{y}\]
Simplify  \({12}^{2}\)  to  \(144\).
\[81=144+{6}^{2}-2\times 12\times 6\cos{y}\]
Simplify  \({6}^{2}\)  to  \(36\).
\[81=144+36-2\times 12\times 6\cos{y}\]
Simplify  \(2\times 12\times 6\cos{y}\)  to  \(144\cos{y}\).
\[81=144+36-144\cos{y}\]
Simplify  \(144+36-144\cos{y}\)  to  \(-144\cos{y}+180\).
\[81=-144\cos{y}+180\]
Subtract \(180\) from both sides.
\[81-180=-144\cos{y}\]
Simplify  \(81-180\)  to  \(-99\).
\[-99=-144\cos{y}\]
Divide both sides by \(-144\).
\[\frac{-99}{-144}=\cos{y}\]
Two negatives make a positive.
\[\frac{99}{144}=\cos{y}\]
Simplify  \(\frac{99}{144}\)  to  \(\frac{11}{16}\).
\[\frac{11}{16}=\cos{y}\]
Switch sides.
\[\cos{y}=\frac{11}{16}\]
Ask: What values of \(y\) will make \(\cos{y}\) equal \(\frac{11}{16}\)?
\[y=\cos^{-1}{(\frac{11}{16})},2\pi -\cos^{-1}{(\frac{11}{16})}\]
Since cos is a periodic function, add the periodicity.
\[\begin{aligned}&y=2\pi n+\cos^{-1}{(\frac{11}{16})},n \in Z\\&y=2\pi n+2\pi -\cos^{-1}{(\frac{11}{16})},n \in Z\end{aligned}\]