12.1 Depth Analysis of Amino Acids in Proteins 331
Since |θ| < |ψ| if 0 <ψ≤
π
2
, −ψ<θ<ψand 0 <ψ+ θ<2ψ ≤ π,
we get sgn (sin ψ) = sgn [sin(ψ + θ)]. If −
π
2
≤ ψ<0, ψ<θ<−ψ and
2ψ<ψ+ θ<0, we also find that sgn (sin ψ) = sgn [sin(ψ + θ)] holds. Thus,
the theorem is proved.
Properties of a Plane Turning Around a Point
Lemma 5 suggests that the inner product changes if a plane is turning around
a line. To discuss this in detail, we introduce the following notations.
Let b, b
,andr be three arbitrary vectors whose inner products are b, r,
b
, r.Letθ, ψ be the angle between b and b
, r, respectively, in which 0 ≤ θ,
ψ ≤ π.
Lemma 6. If θ and ψ satisfy either of the following conditions, then
sgn (b, r)=sgn(b
, r) . (12.5)
Condition 1 If 0 ≤ ψ ≤ π/2,and 0 ≤ θ ≤ π/2 −ψ ≤ π/2.
Condition 2 If π/2 ≤ ψ ≤ π,and 0 ≤ θ ≤ ψ − π/2 ≤ π/2.
Proof. For the sake of simplicity, in a rectangular coordinate system, we take
b =(0, 0, 1) , r =(0, sin ψ, cos ψ) , b
=(cosφ sin θ, sin φ sin θ, cos θ) ,
in which, φ is from (0, 2π). Then
sgn (b, r)=sgn[cosψ]=
+1 , if 0 ≤ ψ ≤ π/2 ,
−1 , if π/2 <ψ≤ π
We calculate the value of sgn (b
, r). Here
b
, r =sinψ sin φ sin θ +cosψ cos θ. (12.6)
Tofinishtheproofofthelemma,wediscussthefollowingtwosteps:
1. If 0 ≤ ψ ≤ π/2, 0 ≤ θ ≤ π/2 − ψ ≤ π/2 (Fig. 12.4a), then cos ψ ≥ 0.
Therefore, we need only prove b
, r≥0. Following from (12.6), we have
b
, r≥cos ψ cos θ − sin ψ sin θ =cos(ψ + θ) ≥ 0
and since sin ψ, sin θ ≥ 0, sin φ ≥−1, 0 ≤ ψ + θ ≤ π/2. This implies that
b
, r≥0 holds. Thus the lemma is true under condition 1.
2. If π/2 ≤ ψ ≤ π, 0 ≤ θ ≤ ψ − π/2 ≤ π/2 (Fig. 12.4b), then cos ψ ≤ 0, so
we need only prove b
, r≤0. Following from (12.6), we have
b
, r≤cos ψ cos θ +sinψ sin θ =cos(ψ − θ) ≤ 0 .
Since sin ψ, sin θ ≥ 0, sin φ ≤ 1,π/2 ≤ ψ − θ ≤ π, we find that b
, r≤0
holds. Additionally, cos ψ = 0 implies that cos(ψ − θ)=0.Thusthe
lemma is true under condition two, and the lemma is shown to be true in
its entirety.