Question

In triangle \(ABC\), if \(\cos A \cos B + \sin A \sin B \sin C = 1\), then \(\sin A + \sin B + \sin C =\)

• A. \( \dfrac{2+\sqrt{3}}{2} \)
• B. \( 1+\sqrt{2} \)
• C. \( \dfrac{2\sqrt{3}-1}{2} \)
• D. \( \dfrac{3+\sqrt{3}}{2} \)

Hint:

When a trigonometric equation in a triangle leads to a sum like \(\cos A \cos B + \sin A \sin B \sin C = 1\), try expressing in terms of a single angle using triangle angle sum properties and trigonometric identities. Often, symmetry (\(A = B\)) or right angles (\(C = 90^\circ\)) are involved.

Answer 1

The Correct Option is B

Given: \[ \cos A \cos B + \sin A \sin B \sin C = 1 \] Recall the identity: \[ \cos A \cos B + \sin A \sin B = \cos(A - B) \] So, \[ \cos A \cos B + \sin A \sin B \sin C = \cos(A - B) + \sin A \sin B (\sin C - 1) \] But from the equation, for the sum to be 1, \(\cos(A - B)\) must be maximized, i.e., \(\cos(A - B) = 1\), so \(A = B\). Now, in a triangle, \(A = B\) and \(A + B + C = \pi\), so \(2A + C = \pi\), \(C = \pi - 2A\). Now, substitute \(A = B\) and \(C = \pi - 2A\) into the original equation: \[ \cos^2 A + \sin^2 A \sin(\pi - 2A) = 1 \] But \(\sin(\pi - 2A) = \sin 2A\), so: \[ \cos^2 A + \sin^2 A \cdot \sin 2A = 1 \] But \(\cos^2 A + \sin^2 A = 1\), so for the equation to hold, \(\sin 2A = 1\), so \(2A = \dfrac{\pi}{2}\), \(A = \dfrac{\pi}{4}\). Now, \[ C = \pi - 2A = \pi - \dfrac{\pi}{2} = \dfrac{\pi}{2} \] Now, calculate \(\sin A + \sin B + \sin C\): Since \(A = B = \dfrac{\pi}{4}\), \(C = \dfrac{\pi}{2}\): \[ \sin A + \sin B + \sin C = 2\sin \dfrac{\pi}{4} + \sin \dfrac{\pi}{2} \] \[ = 2 \cdot \dfrac{1}{\sqrt{2}} + 1 = \sqrt{2} + 1 \]