103 Trigonometry Problems

96 103 Trigonometry Problems
then there is a triangle ABC such that x = sin A 2 , y = sin B 2 , and z = sin C 2 .
Solution: Solving the second given equation as a quadratic in x gives
x = −2yz + √ 4y 2 z 2 − 4(y 2 + z 2 − 1)
2
√
=−yz + (1 − y 2 )(1 − z 2 ).
We make the trigonometric substitution y = sin u and z = sin v, where
0 ◦