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So I need to prove that

$$ \begin{vmatrix} \sin^2(\alpha) & \cos(2\alpha) & \cos^2(\alpha) \\ \sin^2(\beta) & \cos(2\beta) & \cos^2(\beta) \\ \sin^2(\gamma) & \cos(2\gamma) & \cos^2(\gamma) \\ \end{vmatrix} $$ $$ = \begin{vmatrix} \sin(\alpha) & \cos(\alpha) & \sin(\alpha + \delta) \\ \sin(\beta) & \cos(\beta) & \sin(\beta + \delta) \\ \sin(\gamma) & \cos(\gamma) & \sin(\gamma + \delta) \\ \end{vmatrix} $$

Now, $$ \begin{vmatrix} \sin^2(\alpha) & \cos(2\alpha) & \cos^2(\alpha) \\ \sin^2(\beta) & \cos(2\beta) & \cos^2(\beta) \\ \sin^2(\gamma) & \cos(2\gamma) & \cos^2(\gamma) \\ \end{vmatrix} = \begin{vmatrix} \sin^2(\alpha) & \cos^2(\alpha) - \sin^2(\alpha) & \cos^2(\alpha) \\ \sin^2(\beta) & \cos^2(\beta) - \sin^2(\beta) & \cos^2(\beta) \\ \sin^2(\gamma) & \cos^2(\gamma) - \sin^2(\gamma) & \cos^2(\gamma) \\ \end{vmatrix} $$

Adding column $1$ to column $2$ then makes column $2$ and column $3$ equal and hence the first determinant $ = 0$.

I'm stuck in trying to prove that the second determinant is also zero, so I need help in that.

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    $\begingroup$ Expand $\sin(\alpha+\delta)$, note relation to first two columns. $\endgroup$ – Gerry Myerson Feb 16 '19 at 11:42
  • $\begingroup$ @Gerry I tried doing that, however it didn't lead me anywhere. What do I do after that? $\endgroup$ – Helix Feb 16 '19 at 11:43
  • $\begingroup$ What did you get when you expanded $\sin(\alpha+\delta)$? $\endgroup$ – Gerry Myerson Feb 16 '19 at 11:46
  • $\begingroup$ $\sin(\alpha)\cos(\delta) + \sin(\delta)cos(\alpha)$. Still kind of confused as to how to use the relation $\endgroup$ – Helix Feb 16 '19 at 11:47
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$$\begin{vmatrix} \sin(\alpha) & \cos(\alpha) & \sin(\alpha + \delta) \\ \sin(\beta) & \cos(\beta) & \sin(\beta + \delta) \\ \sin(\gamma) & \cos(\gamma) & \sin(\gamma + \delta) \\ \end{vmatrix}=\begin{vmatrix} \sin(\alpha) & \cos(\alpha) & \sin\alpha\color{red}{\cos\delta}+\color{red}{\sin\delta}\cos\alpha \\ \sin(\beta) & \cos(\beta) & \sin\beta\color{red}{\cos\delta}+\color{red}{\sin\delta}\cos\beta \\ \sin(\gamma) & \cos(\gamma) & \sin\gamma\color{red}{\cos\delta}+\color{red}{\sin\delta}\cos\gamma \\ \end{vmatrix}$$$${}$$

$$=\begin{vmatrix} \sin(\alpha) & \cos(\alpha) & \sin\alpha \\ \sin(\beta) & \cos(\beta) & \sin\beta\\ \sin(\gamma) & \cos(\gamma) & \sin\gamma \\ \end{vmatrix}\color{red}{\cos\delta}+\begin{vmatrix} \sin(\alpha) & \cos(\alpha) & \cos\alpha \\ \sin(\beta) & \cos(\beta) & \cos\beta\\ \sin(\gamma) & \cos(\gamma) & \cos\gamma \\ \end{vmatrix}\color{red}{\sin\delta}=0+0=0$$

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  • $\begingroup$ I completely forgot about that property, thank you. $\endgroup$ – Helix Feb 16 '19 at 11:52
  • $\begingroup$ ...but that's precisely the property you used to prove that the first determinant equals $0$... $\endgroup$ – Servaes Feb 16 '19 at 11:53
  • $\begingroup$ Servaes is right. More than forget that property I think all those sines and cosines together confused you. $\endgroup$ – DonAntonio Feb 16 '19 at 11:54
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    $\begingroup$ @Servaes I meant the one where you can make two determinants from one by splitting the elements of a column or row $\endgroup$ – Helix Feb 16 '19 at 11:54
  • $\begingroup$ @DonAntonio That's probably it, thanks anyways. $\endgroup$ – Helix Feb 16 '19 at 11:56
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HINT: Relate the third column to the first two using the trigonometric identity $$\sin(\theta+\delta)=\sin\theta\cos\delta+\cos\theta\sin\delta.$$

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  • $\begingroup$ That's what I said, innit? $\endgroup$ – Gerry Myerson Feb 16 '19 at 11:46
  • $\begingroup$ @GerryMyerson I guess it pretty much is. Would you like the give the answer? I can delete this one. $\endgroup$ – Servaes Feb 16 '19 at 11:48

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