# 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, sin^{2}A + sin^{2}B = sin(A + B).

We will make use of the following trigonometric identities:

sin(/2 − X) = cos X | (1) |

sin^{2}X + cos^{2}X = 1 | (2) |

sin(X + Y) = sin X cos Y + cos X sin Y | (3) |

## Right-angled triangle sin^{2}A + sin^{2}B = 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 sin^{2}A + sin^{2}B = sin^{2}A + cos^{2}A = 1. (Using (1) and (2).)

Also, sin(A + B) = sin(/2) = 1.

Therefore, the triangle is right-angled sin^{2}A + sin^{2}B = sin(A + B).

## sin^{2}A + sin^{2}B = sin(A + B) right-angled triangle

Now suppose sin^{2}A + sin^{2}B | = 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 sin^{2}A + cos^{2}A > sin^{2}B + cos^{2}B.

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, sin^{2}A + sin^{2}B = sin(A + B).

Source: Original

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