Prove that for all values of θ, |sin θ cos θ sin 2θ; sin(θ + 2π/3) cos(θ + 2π/3) sin(2θ + 4π/3); sin

Q: Why can't I just expand the determinant along the first row? It seems like a lot of work, but it would eventually give the answer, right?

While technically true, expanding a $3 \times 3$ determinant with trigonometric functions is a recipe for algebraic disaster. You would end up with a massive expression involving products of sines and cosines, and you would have to spend twenty minutes simplifying it using identities. In JEE Advanced, the goal is to find the 'elegant' path. The row operation method takes seconds and avoids the high probability of a calculation error.

Q: How did you know to add $R_2$ and $R_3$? Why not subtract them or add $R_1$?

Great question! It comes down to pattern recognition. I looked at the angles $\theta + \frac{2\pi}{3}$ and $\theta - \frac{2\pi}{3}$. When you add these, the $\frac{2\pi}{3}$ terms cancel out, leaving you with $2\theta$. This structure is designed to trigger the sum-to-product identities. If you had subtracted them, you would have gotten terms like $\sin(\frac{4\pi}{3})$, which is also solvable, but adding them directly aligns the result with the first row, making the linear dependence immediately visible.

Q: I keep getting confused with the signs of $\cos(\frac{2\pi}{3})$ and $\cos(\frac{4\pi}{3})$. How do I stop making this mistake?

Always visualize the unit circle. $\frac{2\pi}{3}$ is $120^\circ$, which is in the second quadrant where cosine is negative. $\frac{4\pi}{3}$ is $240^\circ$, which is in the third quadrant where cosine is also negative. Both have a reference angle of $60^\circ$ (or $\frac{\pi}{3}$), and $\cos(60^\circ) = \frac{1}{2}$. Therefore, both values must be $-\frac{1}{2}$. Drawing a quick sketch of the unit circle takes five seconds and saves you from losing marks on a sign error.

Question

Prove that for all values of $	heta$, $\begin{vmatrix} \sin 	heta & \cos 	heta & \sin 2	heta \ \sin \left( 	heta + \frac{2\pi}{3} ight) & \cos \left( 	heta + \frac{2\pi}{3} ight) & \sin \left( 2	heta + \frac{4\pi}{3} ight) \ \sin \left( 	heta - \frac{2\pi}{3} ight) & \cos \left( 	heta - \frac{2\pi}{3} ight) & \sin \left( 2	heta - \frac{4\pi}{3} ight) \end{vmatrix} = 0$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

When you first look at this matrix, your instinct might be to expand it along the first row. Stop. Breathe. In JEE Advanced, brute force is the enemy of elegance.

If you try to expand this directly, you will drown in a sea of trigonometric products. Instead, let us look at the structure. We have $\sin 	heta$, $\cos 	heta$, and $\sin 2	heta$ in the first row.

In the second and third rows, we see these phase shifts of $\pm \frac{2\pi}{3}$ and $\pm \frac{4\pi}{3}$. This is not random noise; it is a harmonic pattern.

### The Master Equation

The key to unlocking this problem lies in the sum-to-product identities. Specifically, we need to recall the following identities:

$$2\sin A \cos B = \sin(A+B) + \sin(A-B)$$

$$2\cos A \cos B = \cos(A+B) + \cos(A-B)$$

Why does this matter? Because if we perform the row operation $R_2 	o R_2 + R_3$, we are essentially adding these phase-shifted terms together.

### Executing the Transformation

Let us execute this operation. When we add the first elements of $R_2$ and $R_3$, we get $\sin(	heta + \frac{2\pi}{3}) + \sin(	heta - \frac{2\pi}{3})$.

Applying our identity, this becomes $2\sin 	heta \cos(\frac{2\pi}{3})$. Since $\cos(\frac{2\pi}{3}) = -\frac{1}{2}$, the term simplifies to $-\sin 	heta$.

Do you see the magic? The second row is becoming the negative of the first row. We repeat this for the second column, where $\cos(	heta + \frac{2\pi}{3}) + \cos(	heta - \frac{2\pi}{3})$ becomes $2\cos 	heta \cos(\frac{2\pi}{3})$, which simplifies to $-\cos 	heta$.

Finally, for the third column, $\sin(2	heta + \frac{4\pi}{3}) + \sin(2	heta - \frac{4\pi}{3})$ becomes $2\sin 2	heta \cos(\frac{4\pi}{3})$, which is $-\sin 2	heta$.

### Conclusion

We have arrived at a row that is exactly $-1$ times the first row. In the language of linear algebra, these rows are linearly dependent.

The moment you have linearly dependent rows, the determinant collapses to **zero**. This is the beauty of mathematics—what seemed complex is revealed to be perfectly balanced.

Remember, whenever you see complex trigonometric arguments in a determinant, look for the symmetry. The problem is not asking you to calculate; it is asking you to observe.

Animated Solution for Mathematics - Matrices and Determinants: Prove that for all values of $θ$ , $sin θ sin (θ + \frac{2 π}{3}) sin (θ - \frac{2 π}{3}) cos θ cos (θ + \frac{2 π}{3}) cos (θ - \frac{2 π}{3}) sin 2 θ sin (2 θ + \frac{4 π}{3}) sin (2 θ - \frac{4 π}{3}) = 0$ .

Visualized Solution

Analyzing the Determinant $D$

The Sum-to-Product Identities

Applying Row Operation $R_{2} \to R_{2} + R_{3}$

Simplifying the First Element $R_{21}$

Simplifying the Second Element $R_{22}$

Simplifying the Third Element $R_{23}$

Evaluating Trigonometric Constants

Finding the Cosine Values

Substituting the Constants into $R_{2}$

Conclusion: Determinant $D = 0$

The Sigma Insight: Properties of Determinants

Analyzing the Setup

The Master Equation

Executing the Transformation

Conclusion

Similar Questions

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Visualized Solution

Analyzing the Determinant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D

The Sum-to-Product Identities

Applying Row Operation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R2​→R2​+R3​

Simplifying the First Element .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R21​

Simplifying the Second Element .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R22​

Simplifying the Third Element .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R23​