103 Trigonometry Problems

100 103 Trigonometry Problems
By the law of cosines,
cos A = b2 + c 2 − a 2
.
2bc
Hence, by the half-angle formulas,wehave
sin 2 A 2 = 1 − cos A = 1 2 2 − b2 + c 2 − a 2
= a2 − (b 2 + c 2 − 2bc)
4bc
4bc
= a2 − (b − c) 2 (a − b + c)(a + b − c)
=
=
4bc
(2s − 2b)(2s − 2c)
4bc
=
4bc
(s − b)(s − c)
,
bc
where 2s = a + b + c is the perimeter of triangle ABC. It follows that
sin A √
(s − b)(s − c)
2 = ,
bc
and the analogous formulas for sin B 2 and sin C 2 . Hence
sin A 2 sin B 2 sin C 2
by Heron’s formula. It follows that
(s − a)(s − b)(s − c)
=
abc
s(s − a)(s − b)(s − c)
=
sabc
sin A 2 sin B 2 sin C 2 = [ABC]
s
· [ABC]
abc
= r ·
= [ABC]2
sabc
1
4R ,
from which (d) follows.
Now we prove (e). By the extended law of sines, we have a cos A = 2R sin A·
cos A = R sin 2A. Likewise, b cos B = R sin 2B and c cos C = R sin 2C.By
(a) and (b), we have
It suffices to show that
which is Problem 24(a).
4R sin A sin B sin C = abc
2R 2 .
sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C,