103 Trigonometry Problems

144 103 Trigonometry Problems
Let x = ̸ PCB (in degrees). Then ̸ PBC = 80 ◦ − x.Bythelaw of sines or
by Ceva’s theorem,
1 = PA
PB · PB
PC · PC
PA = sin ̸ PBA
sin ̸ PAB · sin ̸ PCB
sin ̸ PBC · sin ̸ PAC
sin ̸ PCA
sin 20 ◦ sin x sin 40 ◦
=
sin 10 ◦ sin(80 ◦ − x)sin 30 ◦ = 4 sin x sin 40◦ cos 10 ◦
sin(80 ◦ .
− x)
The product-to-sum formulas yield
and so
1 = 2 sin x(sin 30◦ + sin 50 ◦ )
sin(80 ◦ − x)
= sin x(1 + 2 cos 40◦ )
sin(80 ◦ ,
− x)
2 sin x cos 40 ◦ = sin(80 ◦ − x) − sin x = 2 sin(40 ◦ − x)cos 40 ◦ ,
̸ ̸
by the difference-to-product formulas. We conclude that x = 40 ◦ − x, or
x = 20 ◦ . It follows that ACB = 50 ◦ = BAC, and so triangle ABC is
isosceles.
22. Let a 0 = √ 2 + √ 3 + √ 6, and let a n+1 = a2 n −5
2(a n +2)
for integers n>0. Prove
that
( 2 n−3 )
π
a n = cot − 2
3
for all n.
Solution: By either the double-angle or the half-angle formulas, we obtain
cot π 24 = cos 24
π
sin
24
π = 2 cos2 24
π
2 sin
24 π cos 24
π = 1 + cos 12
π
sin
12
π
= 1 + cos ( π
3
− π )
4
sin ( π
3
− π ) = 1 + cos π 3 cos π 4 + sin π 3 sin π 4
sin π 4
3 cos π 4 − cos π 3 sin π 4
Hence a n = cot
√
4 + 6
4
√ √
6
4 − 2
4
= 1 + √
2
= 4 + √ 6 + √ 2
√
6 −
√
2
= 4(√ 6 + √ 2) + ( √ 6 + √ 2) 2
( √ 6 − √ 2)( √ 6 + √ 2)
= 2 + √ 2 + √ 3 + √ 6 = a 0 + 2.
)
− 2 is true for n = 0.
(
2 n−3 π
3
= 4(√ 6 + √ 2) + 8 + 4 √ 3
4