103 Trigonometry Problems

4. Solutions to Introductory Problems 113
But we have not shown that indeed, cos α sin β can obtain all values in the
interval [ − 1 2 , 1 2]
. To do this, we consider
(cos α sin β) 2 = (1 − sin 2 α)(1 − cos 2 β)
= 1 − (sin 2 α + cos 2 β) + sin 2 α cos 2 β
= 5 4 − (sin2 α + cos 2 β)
= 5 4 − (sin α + cos β)2 + 2 sin α cos β
= 1 4 − (sin α + cos β)2 .
Let x = sin α and y = cos β. Then −1 ≤ x,y,≤ 1 and xy =−2 1 . Consider
the range of the sum s = sin α + cos β = x + y.Ifxy =−2 1 and x + y = s,
then x and y are the roots of the quadratic equation
Thus, {x,y} =
s+ √ s 2 +2
u 2 − su − 1 2 = 0.
(∗)
{ √ }
s+ s 2 +2
2
, s−√ s 2 +2
2
. By checking the boundary condition
2
≤ 1, we obtain s ≤
2 1 . By checking similar boundary conditions, we
conclude that the equation (∗) has a pair of solutions x and y with −1 ≤ x,y ≤
1 for all −
2 1 ≤ s ≤ 2 1 . Because both the sine and cosine functions are surjective
functions [ from R to the interval [−1, 1], the range of s = sin α + cos β is
−
1
2
,
2
1 ]
for sin α cos β =−
1
2
. Thus, the range of s 2 is [ 0,
2
1 ]
. Thus the range
of (cos α sin β) 2 is [ 0, 4] 1 [
, and so the range of cos α sin β is −
1
2
, 2] 1 .
45. Let a,b,c be real numbers. Prove that
(ab + bc + ca − 1) 2 ≤ (a 2 + 1)(b 2 + 1)(c 2 + 1).
Solution: Let a = tan x, b = tan y, c = tan z with − π 2 < x,y,z < π 2 .
Then a 2 + 1 = sec 2 x, b 2 + 1 = sec 2 y, and c 2 + 1 = sec 2 z. Multiplying by
cos 2 x cos 2 y cos 2 z on both sides of the desired inequality gives
Note that
[(ab + bc + ca − 1) cos x cos y cos z] 2 ≤ 1.
(ab + bc) cos x cos y cos z = sin x sin y cos z + sin y sin z cos x
= sin y sin(x + z)