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$$ \begin{vmatrix} \cos 2x & \cos 2y & \cos 2u \\ \sin^2 x & \sin^2 y & \sin^2 u \\ 1 & 1 & 1 \end{vmatrix} = 0 $$

This is the conclusion where I got from another excercise and I want to prove me it is right. Please don't use the fact that $\cos(2x)=\cos^2x-\sin^2x$. We haven't learned it yet.

Alec Teal
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prishila
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  • What results do you know abt determinant?What've you tried? – Arpit Kansal Oct 14 '15 at 18:55
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    You're going to need to know some trig identities to prove this. If you haven't learned them, yet, then now's a good time. –  Oct 14 '15 at 18:56
  • I'm surprised that you're studying determinants before learning analytic trig identities... Anyway... Are you allowed to use the cosine sum formula? Because that gives: $\cos(2x) = \cos(x + x) = \cos(x)\cos(x) - \sin(x)\sin(x) = \cos^2(x)-\sin^2(x)$. – zahbaz Oct 14 '15 at 19:21
  • Any identity would qualify as a loophole, I think... – user170231 Oct 14 '15 at 19:22
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    I'm very interested to know where determinants are taught before trigonometry. In Croatia, trigonometry is taught at high school level, and you won't encounter (in standard program) any matrix before 1st year of university. – Ennar Oct 14 '15 at 19:52
  • @Ennar I tutor US high school students. Not many encounter matrices, and fewer study determinants. If they do, it's most likely in an upper level precalculus course, usually grade 11. A typical precalc course covers algebraic, exponential, and logarithmic functions with plotting, as well as analytic trig and trigonometric graphing. Conic sections are often taught as well in this year. If there is any additional time, the topics often are vectors, polar coords, complex numbers, and basic matrix operations with systems of equations. I think most see 3x3 determinants in 1st year university... – zahbaz Oct 14 '15 at 20:15
  • @zahbaz With exception of matrices, this sounds very similar to high school program in Croatia. Although linear systems (3x3) are taught, it's only at the level of Gauss elimination, not matrices. Trigonometry is taught at grade 10, first through definitions via ratios in right triangle, but later with addition and difference of angles, product to sum and sum to product identities, double-angle and half-angle formulae, transforms through quadrants, and trigonometric form of complex number. Thank you for the info, and not to drag this on too much, is any calculus taught in US high-school's? – Ennar Oct 14 '15 at 20:30
  • @Ennar. It sounds very similar. Trig is first seen in grade 10 but developed in grade 11. Students who take precalculus by grade 10 or 11 have the option of taking calculus. Some students opt to instead take a statistics course. Some take two courses in calculus if sufficiently advanced, but they do not reach multivariable studies. – zahbaz Oct 14 '15 at 20:36

5 Answers5

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Here's how easy it is if you do use $\cos(2x) = 1-2\sin^2(x)$:

$$\begin{align}\left|\begin{matrix} \cos(2x) & \cos(2y) & \cos(2u) \\ \sin^2(x) & \sin^2(y) & \sin^2(u) \\ 1 & 1 & 1\end{matrix} \right| &= \left|\begin{matrix} 1-2\sin^2(x) & 1-2\sin^2(y) & 1-2\sin^2(u) \\ \sin^2(x) & \sin^2(y) & \sin^2(u) \\ 1 & 1 & 1\end{matrix} \right| \\ &= \left|\begin{matrix} 1 & 1 & 1 \\ \sin^2(x) & \sin^2(y) & \sin^2(u) \\ 1 & 1 & 1\end{matrix} \right| - 2\left|\begin{matrix} \sin^2(x) & \sin^2(y) & \sin^2(u) \\ \sin^2(x) & \sin^2(y) & \sin^2(u) \\ 1 & 1 & 1\end{matrix} \right|\end{align}$$

This equals zero because the determinant of any matrix with a repeated row (or column) is $0$.

So maybe it's time to learn some trig.

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Via a cofactor expansion along the top row, you have $$\begin{align*}\begin{vmatrix}\cos2x&\cos2y&\cos2u\\\sin^2x&\sin^2y&\sin^2u\\1&1&1\end{vmatrix}&=\cos2x\begin{vmatrix}\sin^2y&\sin^2u\\1&1\end{vmatrix}-\cos2y\begin{vmatrix}\sin^2x&\sin^2u\\1&1\end{vmatrix}+\cos2u\begin{vmatrix}\sin^2x&\sin^2y\\1&1\end{vmatrix}\end{align*}$$ Since $\begin{vmatrix}a&b\\c&d\end{vmatrix}=ad-bc$, this is equivalent to $$\cos2x(\sin^2y-\sin^2u)-\cos2y(\sin^2x-\sin^2u)+\cos2u(\sin^2x-\sin^2y)$$ Assuming you know the half-angle identity for sine, i.e. $\dfrac{1-\cos2x}{2}=\sin^2x$, you can write this as $$\frac{\cos2x(\cos2u-\cos2y)-\cos2y(\cos2u-\cos2x)+\cos2u(\cos2y-\cos2x)}{2}$$ Notice the disappearing terms.

Zbigniew
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user170231
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If you replace $\cos$ by an arbitrary function $g$, the determinant (e.g. from expanding across the top row) is $$ g(2x) (\sin^2 y - \sin^2 u) + g(2y) (\sin^2 u - \sin^2 x) + g(2u) (\sin^2 x - \sin^2 y)$$ and this will not be $0$ in general. Take any $x,y,u$ such that no two of $\sin^2 x$, $\sin^2 y$ and $\sin^2 u$ are equal, and you have $a g(2x) + b g(2y) + c g(2z)$ for some nonzero $a,b,c$, and there's no reason for this to be $0$. So without some trig identity that can relate $g$ to $\sin$, it's impossible to do this problem.

Robert Israel
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$$\text{ Given determinant }=-\left|\begin{matrix}\sin^2x-\cos^2x&\sin^2y-\cos^2y&\sin^2u-\cos^2u\\\sin^2x&\sin^2y&\sin^2u\\1&1&1\end{matrix}\right|$$ $$=-\left|\begin{matrix}\sin^2x&\sin^2y&\sin^2u\\\sin^2x&\sin^2y&\sin^2u\\1&1&1\end{matrix}\right| \text{ , by using } R_1'=R_1-R_2+R_3 .$$ $$=0$$

Empty
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By using the equality $cos(2x) = 2cos^2(x)-1 =1-2sin^2(x)$, -2 second line + third line = first line

Tsemo Aristide
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