Solution to puzzle 119: Three sines

A triangle has two acute angles, A and B.  Show that the triangle is right-angled if, and only if, sin²A + sin²B = sin(A + B).

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We will make use of the following trigonometric identities:

sin(/2 − X) = cos X	(1)
sin2X + cos2X = 1	(2)
sin(X + Y) = sin X cos Y + cos X sin Y	(3)

Right-angled triangle sin²A + sin²B = sin(A + B)

This direction of the if and only if statement is straightforward.
If the triangle is right-angled, then A + B = /2.
Hence sin B = sin(/2 − A) = cos A, and so sin²A + sin²B = sin²A + cos²A = 1.  (Using (1) and (2).)
Also, sin(A + B) = sin(/2) = 1.
Therefore, the triangle is right-angled sin²A + sin²B = sin(A + B).

sin²A + sin²B = sin(A + B) right-angled triangle

Now suppose sin2A + sin2B	= sin(A + B).(4)
	= sin A cos B + cos A sin B,  from (3).

Rearranging, we have sin A (sin A − cos B) = sin B (cos A − sin B).
Since sin A and sin B are positive, the two parenthesized factors must have the same sign: both positive, both negative, or both equal to zero.

Suppose both factors are positive, in which case sin A > cos B > 0 and cos A > sin B > 0.
Squaring the terms of each inequality, and adding, we obtain sin²A + cos²A > sin²B + cos²B.
This implies 1 > 1; a contradiction.
Similarly, if we suppose both factors are negative, in which case 0 < sin A < cos B and 0 < cos A < sin B, we arrive at 1 < 1; again a contradiction.

Therefore the only possible solution is where sin A = cos B and cos A = sin B.
But then sin A = sin(/2 − B) A = /2 − B, which does indeed satisfy (4).
Hence A + B = /2, and the triangle is right-angled.

Therefore, a triangle with two acute angles, A and B, is right-angled if, and only if, sin²A + sin²B = sin(A + B).

Source: Original

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