103 Trigonometry Problems

152 103 Trigonometry Problems
The addition and subtraction formulas give
and so
sin C = sin(180 ◦ − B − A) = sin(135 ◦ − A)
√
2(sin A + cos A)
= sin 135 ◦ cos A − cos 135 ◦ sin A =
,
2
√ √
2 2
sin C −
2 = (sin A + cos A − 1).
2
Plugging the last equation into equation (∗) yields
1 − (sin A + cos A − 1) 2 = 2
(√
2 − 1
) [ sin A +
Expanding both sides of the last equation gives
or
√
2
2 (sin A + cos A − 1) ]
.
1 − (sin A + cos A) 2 + 2(sin A + cos A) − 1
(√ )(
= 2 − 1 2 + √ ) (
2 sin A + 2 − √ )
2 (cos A − 1)
sin 2 A + cos 2 A + 2 sin A cos A =
(
2 − √ )
2 sin A + √ (
2 cos A + 2 − √ )
2 .
Consequently, we have
(
2 sin A cos A − 2 − √ )
2 sin A − √ (√ )
2 cos A + 2 − 1 = 0;
that is,
(√ )(√ √ )
2 sin A − 1 2 cos A − 2 + 1 = 0.
√
2
This implies that sin A =
problem is
√
2
sin A = or sin A =
2
√
2 or cos A = 1 − 2
2
√
√
1 − cos 2 A =
. Therefore, the answer to the
4 √ 2 − 2
.
2