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Question: If a = $A_{\pi/2}$ (α, β, γ), b = $A_{\pi/3}$ (α, β, γ). Which of the following is true...

If a = Aπ/2A_{\pi/2} (α, β, γ), b = Aπ/3A_{\pi/3} (α, β, γ). Which of the following is true

A

a = b

B

a < b

C

a > b

D

2a = b

Answer

a = b

Explanation

Solution

We are given Aθ(α,β,γ)=∣cos⁡(α+θ)sin⁡(α+θ)1cos⁡(β+θ)sin⁡(β+θ)1cos⁡(γ+θ)sin⁡(γ+θ)1∣A_{\theta}(\alpha,\beta,\gamma)= \begin{vmatrix} \cos(\alpha+\theta) & \sin(\alpha+\theta) & 1 \\ \cos(\beta+\theta) & \sin(\beta+\theta) & 1 \\ \cos(\gamma+\theta) & \sin(\gamma+\theta) & 1 \end{vmatrix}.

Key idea: Notice that if we set x=α+θ,y=β+θ,z=γ+θx=\alpha+\theta, y=\beta+\theta, z=\gamma+\theta, then the determinant becomes Aθ(α,β,γ)=∣cos⁡xsin⁡x1cos⁡ysin⁡y1cos⁡zsin⁡z1∣A_{\theta}(\alpha,\beta,\gamma)=\begin{vmatrix}\cos x & \sin x & 1 \\\cos y & \sin y & 1 \\\cos z & \sin z & 1\end{vmatrix}.

It is a known fact (which one can verify by direct expansion) that for any three angles x,y,zx,y,z, ∣cos⁡xsin⁡x1cos⁡ysin⁡y1cos⁡zsin⁡z1∣=−4sin⁡(y−x2)sin⁡(z−x2)sin⁡(z−y2)\begin{vmatrix}\cos x & \sin x & 1 \\\cos y & \sin y & 1 \\\cos z & \sin z & 1\end{vmatrix} =-4\sin(\frac{y-x}{2})\sin(\frac{z-x}{2})\sin(\frac{z-y}{2}).

Thus, we have Aθ(α,β,γ)=−4sin⁡(β−α2)sin⁡(γ−α2)sin⁡(γ−β2)A_{\theta}(\alpha,\beta,\gamma)=-4\sin(\frac{\beta-\alpha}{2})\sin(\frac{\gamma-\alpha}{2})\sin(\frac{\gamma-\beta}{2}).

The important observation is that the final expression is independent of θ\theta.

Let K=−4sin⁡(β−α2)sin⁡(γ−α2)sin⁡(γ−β2)K=-4\sin(\frac{\beta-\alpha}{2})\sin(\frac{\gamma-\alpha}{2})\sin(\frac{\gamma-\beta}{2}). Then for every θ\theta, Aθ(α,β,γ)=KA_\theta(\alpha,\beta,\gamma) = K.

Given: a=Aπ/2(α,β,γ)=K,b=Aπ/3(α,β,γ)=Ka=A_{\pi/2}(\alpha,\beta,\gamma)=K, b=A_{\pi/3}(\alpha,\beta,\gamma)=K. Therefore, a=ba=b.