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Solution to puzzle 119: Three sines

Skip restatement of puzzle.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).


We will make use of the following trigonometric identities:

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

Right-angled triangle implies 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 = pi/2.
Hence sin B = sin(pi/2 − A) = cos A, and so sin2A + sin2B = sin2A + cos2A = 1.  (Using (1) and (2).)
Also, sin(A + B) = sin(pi/2) = 1.
Therefore, the triangle is right-angled implies sin2A + sin2B = sin(A + B).

sin2A + sin2B = sin(A + B) implies 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(pi/2 − B) implies A = pi/2 − B, which does indeed satisfy (4).
Hence A + B = pi/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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