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Chapter 4: Problem 269

The value of determinant \(\left|\begin{array}{ccc}\cos ^{2}[(\pi / 2)+x] & \cos ^{2}[(3 \pi / 2)+x] & \cos ^{2}[(5 \pi / 2)+x] \\ \cos [(\pi / 2)+x] & \cos [(3 \pi / 2)+x] & \cos [(5 \pi / 2)+x] \\ \cos [(\pi / 2)-x] & \cos [(3 \pi / 2)-x] & \cos [(5 \pi / 2)-x]\end{array}\right|\) is \(\ldots \ldots\) (a) 0 (b) \(\cos ^{2}[3 \mathrm{x}-(9 \pi / 2)]\) (c) \(\sin ^{2}[(3 \pi / 2)+\mathrm{x}]\) (d) \(\cos ^{2}[(15 \pi / 2)-\mathrm{x}]\)

Short Answer

Expert verified
The value of the determinant is \(\boxed{\textbf{(a) } 0}\).

Step by step solution

01

Write down the determinant in LaTeX form

To make the working more comfortable, we write down the determinant in LaTeX form, then proceed as follows: \[ D = \left|\begin{array}{ccc} \cos^{2}\left(\frac{\pi}{2}+x\right) & \cos^{2}\left(\frac{3\pi}{2}+x\right) & \cos^{2}\left(\frac{5\pi}{2}+x\right) \\ \cos\left(\frac{\pi}{2}+x\right) & \cos\left(\frac{3\pi}{2}+x\right) & \cos\left(\frac{5\pi}{2}+x\right) \\ \cos\left(\frac{\pi}{2}-x\right) & \cos\left(\frac{3\pi}{2}-x\right) & \cos\left(\frac{5\pi}{2}-x\right) \end{array}\right| \]
02

Simplify the trigonometric expressions

Now we will use some trigonometric identities to simplify the given expressions. \[\cos\left(\frac{\pi}{2} + x\right) = -\sin x\] \[\cos\left(\frac{3\pi}{2} + x\right) = \sin x\] \[\cos\left(\frac{5\pi}{2} + x\right) = -\sin x\] \[\cos^{2}\left(\frac{\pi}{2} + x\right) = \sin^{2} x\] \[\cos^{2}\left(\frac{3\pi}{2} + x\right) = \sin^{2} x\] \[\cos^{2}\left(\frac{5\pi}{2} + x\right) = \sin^{2} x\]
03

Substitute the simplified expressions

Substitute these simplified expressions back into the determinant. \[D = \left|\begin{array}{ccc} \sin^2 x & \sin^2 x & \sin^2 x \\ -\sin x & \sin x & -\sin x \\ \cos x & -\cos x & \cos x \end{array}\right|\]
04

Evaluate the determinant

To evaluate the determinant, we will use the first row expansion. \[ D = \sin^2 x \left|\begin{array}{cc} \sin x & -\sin x \\ -\cos x & \cos x \end{array}\right| - \sin^2 x \left|\begin{array}{cc} -\sin x & -\sin x \\ \cos x & \cos x \end{array}\right| + \sin^2 x \left|\begin{array}{cc} -\sin x & \sin x \\ \cos x & -\cos x \end{array}\right| \] Now expand the 2x2 determinants. \[ D = \sin^2 x ((\sin x)(\cos x) - (-\sin x)(-\cos x)) - \sin^2 x ((-\sin x)(\cos x) - (-\sin x)(\cos x)) + \sin^2 x ((-\sin x)(-cos x) - (\cos x)(\sin x)) \] \[ D = \sin^2 x (0) - \sin^2 x (0) + \sin^2 x (0) \]
05

Final Answer

Now we get: \[D = 0\] Thus, the value of the determinant is 0, \[\boxed{\textbf{(a) } 0}\].

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Most popular questions from this chapter

If \(\mathrm{A}=\left|\begin{array}{ll}\alpha & 0 \\ 2 & 3\end{array}\right|\) and \(\mathrm{I}=\left|\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right|\) then \(\mathrm{A}^{2}=91\) for (a) \(\alpha=4\) (b) \(\alpha=3\) (c) \(\alpha=-3\) (d) no \(\alpha\)

\(\underline{A}=\left|\begin{array}{ccc}-1 & 2-3 i & 3+4 i \\ 2+3 i & 5 & 1+i \\\ 3-4 i & 1-i & 4\end{array}\right|\) then det \(\mathrm{A}\) is \(\ldots .\) (a) purely real (b) purely imaginary (c) complex number (d) 0

If \(\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}, \ldots\) are in GP, then $$ \left|\begin{array}{ccc} \log a_{n} & \log a_{n+1} & \log a_{n+2} \\ \log a_{n+3} & \log a_{n+4} & \log a_{n+5} \\ \log a_{n+6} & \log a_{n+7} & \log a_{n+8} \end{array}\right|=\ldots \ldots $$ (a) 0 (b) 1 (c) 2 (d) 4

If $$ \mathrm{A}=\left|\begin{array}{cc} 1 & 0 \\ -1 & 7 \end{array}\right| $$ and \(A^{2}=8 \mathrm{~A}+\mathrm{kl}_{2}\), then \(\mathrm{k}=\ldots\) (a) 1 (b) \(-1\) (c) 7 (d) \(-7\)

The set of natural numbers \(\mathrm{N}\) is partitioned into arrays of rows and columns in the form of matrices as \(\mathrm{M}_{1}=|1|, \mathrm{M}_{2}=\left|\begin{array}{ll}2 & 3 \\ 4 & 5\end{array}\right|, \mathrm{M}_{3}=\left|\begin{array}{ccc}6 & 7 & 8 \\ 9 & 10 & 11 \\ 12 & 13 & 14\end{array}\right|, \ldots, \mathrm{M}_{\mathrm{n}}=[\ldots]\) and so on. Find the sum of elements of the diagonal in \(\mathrm{M}_{6}\). (a) 144 (b) 441 (c) 321 (d) 461
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