# Proving that two determinants are equal without expanding them

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.

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

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

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

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

HINT: Relate the third column to the first two using the trigonometric identity $$\sin(\theta+\delta)=\sin\theta\cos\delta+\cos\theta\sin\delta.$$

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