If and denotes the angles of a triangle such that , then triangle is a/an triangle.
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Using the identity sin ( x ) sin ( y ) = 2 1 ( cos ( x − y ) − cos ( x + y ) ) we see that
S = sin ( 2 A ) sin ( 2 B ) sin ( 2 C ) = 8 1
⟹ 2 1 sin ( 2 A ) ( cos ( 2 B − C ) − cos ( 2 B + C ) ) − 8 1 = 0 , (equation (i)).
Now since A = π − ( B + C ) we see that cos ( 2 B + C ) = cos ( 2 π − 2 A ) = sin ( 2 A ) .
With θ = 2 B − C , equation (i) then becomes
4 sin 2 ( 2 A ) − 4 cos ( θ ) sin ( 2 A ) + 1 = 0 ,
which is a quadratic in sin ( 2 A ) , and thus
sin ( 2 A ) = 8 4 cos ( θ ) ± 1 6 cos 2 ( θ ) − 1 6 = 2 cos ( θ ) ± cos 2 ( θ ) − 1 .
Now sin ( 2 A ) is real, so we must have that cos 2 ( θ ) − 1 ≥ 0 . But as the range of the cosine function is [ − 1 , 1 ] we must have either cos ( θ ) = − 1 ⟹ θ = π or cos ( θ ) = 1 ⟹ θ = 0 .
But as θ = 2 B − C the only possible option is that θ = 0 , i.e., that B = C . This in turn implies that sin ( 2 A ) = 2 cos ( 0 ) = 2 1 ⟹ A = 3 π radians.
But then since A + B + C = A + 2 B = π we see that B = C = 3 π radians, and thus Δ A B C is an equilateral triangle.