Q5
1 markMCQSection A

One of the values of x for which ∣cos⁡xsin⁡x−cos⁡xsin⁡x∣=1\begin{vmatrix}\cos x & \sin x\\-\cos x & \sin x\end{vmatrix} = 1 is

Determinants
Determinant of a Trigonometric Matrix

Options

(A)0
(B)π4\frac{\pi}{4}
(C)π3\frac{\pi}{3}
(D)π2\frac{\pi}{2}
Official Answer

Correct option: (B) x=π4x = \frac{\pi}{4}


Expanding the determinant: cos⁡x⋅sin⁡x−sin⁡x⋅(−cos⁡x)=2sin⁡xcos⁡x=sin⁡2x\cos x\cdot\sin x - \sin x\cdot(-\cos x) = 2\sin x \cos x = \sin 2x. Setting sin⁡2x=1 gives 2x=π2\sin 2x = 1 \text{ gives } 2x = \frac{\pi}{2}, i.e. x=π4x = \frac{\pi}{4}.

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Marking Scheme

  • 11 mark: correct option (B) selected — requires correct determinant expansion (noting the −cos⁡x-\cos x entry) and identity recognition.

Hint

Expand the determinant carefully with the sign of the (2,1) entry (−cos⁡x)(-\cos x); it simplifies to 2sin⁡xcos⁡x=sin⁡2x2\sin x \cos x = \sin 2x, so solve sin⁡2x=1\sin 2x = 1.

Quick Oral Answer

The determinant expands to 2sin⁡xcos⁡x=sin⁡2x2\sin x \cos x = \sin 2x; setting sin⁡2x=1 gives 2x=π2, so x=π4\sin 2x = 1 \text{ gives } 2x = \frac{\pi}{2}, \text{ so } x = \frac{\pi}{4}, option (B).

Analysis & Explanation

This tests determinant expansion combined with the double-angle identity and solving a simple trigonometric equation. Note the second row is (−cos⁡x,sin⁡x)(-\cos x, \sin x), so the sign must be handled carefully.


Expanding the determinant

  • ∣cos⁡xsin⁡x−cos⁡xsin⁡x∣=(cos⁡x)(sin⁡x)−(sin⁡x)(−cos⁡x)=sin⁡xcos⁡x+sin⁡xcos⁡x=2sin⁡xcos⁡x\begin{vmatrix}\cos x & \sin x\\-\cos x & \sin x\end{vmatrix} = (\cos x)(\sin x) - (\sin x)(-\cos x) = \sin x \cos x + \sin x \cos x = 2\sin x \cos x.
  • By the double-angle identity, 2sin⁡xcos⁡x=sin⁡2x2\sin x \cos x = \sin 2x.

Solving sin 2x = 1

  • sin 2x = 1 when 2x=π2+2nπ2x = \frac{\pi}{2} + 2n\pi (n an integer), i.e. x = π/4 + nπ. Taking n=0 gives x=π4\text{Taking } n = 0 \text{ gives } x = \frac{\pi}{4}.

Why (B) is correct

  • Substituting x=π4:cos⁡(π4)=sin⁡(π4)=12, so the determinant=(12)(12)−(12)(−12)=12+12=1\text{Substituting } x = \frac{\pi}{4}: \cos\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}, \text{ so the determinant} = \left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right) - \left(\frac{1}{\sqrt{2}}\right)\left(-\frac{1}{\sqrt{2}}\right) = \frac{1}{2} + \frac{1}{2} = 1 This matches the required value exactly.

Why the other options are wrong

  • (A) x=0:sin⁡0=0≠1 (determinant=1⋅0−0⋅(−1)=0)x = 0: \sin 0 = 0 \ne 1\ (\text{determinant} = 1\cdot 0 - 0\cdot(-1) = 0).
  • (C) x=π3:sin⁡(2π3)=32≠1x = \frac{\pi}{3}: \sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2} \ne 1.
  • (D) x=π2:sin⁡(π)=0≠1x = \frac{\pi}{2}: \sin(\pi) = 0 \ne 1.

Common Mistakes

  1. 1Missing the negative sign of the (2,1) entry (−cos⁡x)(-\cos x) and expanding as cos⁡2x−sin⁡2x\cos^2 x - \sin^2 x instead of 2sin⁡xcos⁡x2\sin x \cos x.
  2. 2Not recognizing 2sin⁡xcos⁡x as sin⁡2x2\sin x \cos x \text{ as } \sin 2x and instead trying to solve the equation term by term.
  3. 3Selecting x=0x = 0 by reflex; here the sign in the second row makes the value sin⁡2x\sin 2x, which equals 0 (not 1) at x=0x = 0.

Interesting Facts

The double-angle identity sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x \cos x follows directly from the sine addition formula sin⁡(x+x)=sin⁡xcos⁡x+cos⁡xsin⁡x\sin(x+x) = \sin x \cos x + \cos x \sin x.

A determinant of the form ∣cos⁡xsin⁡x−cos⁡xsin⁡x∣ equals sin⁡2x\begin{vmatrix}\cos x & \sin x\\-\cos x & \sin x\end{vmatrix} \text{ equals } \sin 2x, whereas the rotation matrix ∣cos⁡x−sin⁡xsin⁡xcos⁡x∣ always equals 1 for every x\begin{vmatrix}\cos x & -\sin x\\\sin x & \cos x\end{vmatrix} \text{ always equals } 1 \text{ for every } x — small sign changes drastically alter the result.

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Frequently Asked Questions

How do you expand this 2×22\times 2 determinant?

For ∣abcd∣\begin{vmatrix}a & b\\c & d\end{vmatrix} the value is ad−bcad - bc. Here a=cos⁡x, b=sin⁡x, c=−cos⁡x, d=sin⁡xa = \cos x,\ b = \sin x,\ c = -\cos x,\ d = \sin x, so the value is (cos⁡x)(sin⁡x)−(sin⁡x)(−cos⁡x)=2sin⁡xcos⁡x=sin⁡2x(\cos x)(\sin x) - (\sin x)(-\cos x) = 2\sin x \cos x = \sin 2x.

What are all solutions of sin⁡2x=1\sin 2x = 1?

sin⁡2x=1 when 2x=π2+2nπ\sin 2x = 1 \text{ when } 2x = \frac{\pi}{2} + 2n\pi for integer n, i.e. x=π4+nπx = \frac{\pi}{4} + n\pi. Among the given options only x=π4 satisfies this\text{only } x = \frac{\pi}{4} \text{ satisfies this}.