103 Trigonometry Problems

174 103 Trigonometry Problems
Similarly,
tan y + tan z =
sin(y + z) sin(x + w)
=
cos y cos z cos y cos z ,
because x + w = 180 ◦ − (y + z). It follows that
(tan x + tan w)(tan y + tan z)
sin 2 (x + w)
=
cos x cos y cos z cos w = 1 − cos2 (x + w)
cos x cos y cos z cos w
[1 − cos(x + w)][1 + cos(x + w)]
=
cos x cos y cos z cos w
[1 + cos(y + z)][1 + cos(x + w)]
= = st.
cos x cos y cos z cos w
The desired inequality becomes s+t ≤ st,or(1−s)(1−t) = 1−s−t+st ≥ 1.
It suffices to show that 1 − s ≥ 1 and 1 − t ≥ 1. By symmetry, we have only
to show that 1 − s ≥ 1; that is,
1 + cos(x + w)
cos x cos w ≥ 2.
Multiplying both sides of the inequality by cos x cos w and applying the addition
and subtraction formulas gives
1 + cos x cos w − sin x sin w ≥ 2 cos x cos w,
or 1 ≥ cos x cos w + sin x sin w = cos(x − w), which is evident. Equality
holds if and only if x = w. Therefore, inequality (∗) is true, with equality if
and only if x = w and y = z, which happens precisely when AD ‖ BC and
|AB| =|CD|, as was to be shown.
Second Solution: We maintain the same notation as in the first solution.
Applying the law of cosines to triangles ADI and BCI gives
and
|AI| 2 +|DI| 2 = 2 cos(x + w)|AI|·|DI|+|AD| 2
|BI| 2 +|CI| 2 = 2 cos(y + z)|BI|·|CI|+|BC| 2 .
Adding the last two equations and completing squares gives
(|AI|+|DI|) 2 + (|BI|+|CI|) 2 + 2|AD|·|BC|
= 2 cos(x + w)|AI|·|DI|+2 cos(y + z)|BI|·|CI|
+ 2|AI|·|DI|+2|BI|·|CI|+(|AD|+|BC|) 2 .