4.2 Trigonometry 233Remark.It is interesting to know that Leonardo da Vinci’s manuscripts contain drawings
of such decompositions. Later, however, Leonardo himself realized that the decompo-
sitions were impossible, and the drawings were mere optical illusions. Note also that
Dehn’s invariant mentioned in the first chapter provides an obstruction to the decompo-
sition.
We conclude the introduction with a problem by the second author of the book.Example.Leta 0 =
√
2 +
√
3 +
√
6 and letan+ 1 = a(^2) n− 5
2 (an+ 2 )forn≥0. Prove that
an=cot
(
2 n−^3 π
3)
−2 for alln.Solution.We have
cot
π
24=
cos
π
24
sinπ
24=
2 cos^2
π
24
2 sinπ
24cosπ
24=
1 +cos
π
12
sinπ
12=
1 +cos(π
3−
π
4)
sin(π
3−
π
4).
Using the subtraction formulas for sine and cosine we find that this is equal to
1 +√
2
4 +√
6
√^4
6
4 −√
2
4=
4 +
√
6 +
√
2
√
6 −
√
2
=
4 (
√
6 +
√
2 )+(
√
6 +
√
2 )^2
6 − 2
=
4 (
√
6 +
√
2 )+ 8 + 4
√
3
4
= 2 +
√
2 +
√
3 +
√
6 =a 0 + 2.Hence the equalityan=cot(^2
n− (^3) π
3 )−2 is true at least forn=0.
To verify it in general, it suffices to prove thatbn=cot(^2
n− (^3) π
3 ), wherebn=an+2,
n≥1. The recurrence relation becomes
bn+ 1 − 2 =
(bn− 2 )^2 − 5
2 bn
,
orbn+ 1 =b
(^2) n− 1
2 bn. Assuming inductively thatbk=cotck, whereck=
2 k−^3 π
3 , and using the
double-angle formula, we obtain
bk+ 1 =
cot^2 ck− 1
2 cotck
=cot( 2 ck)=cotck+ 1.
This completes the proof.
656.Prove that
sin 70◦cos 50◦+sin 260◦cos 280◦=