Langleyの問題を三角関数で解いてみた。(3)

　ネットで見つけたラングレーの問題を三角関数で解いてみました。
　問題の設定や図は、次の通りです。ここでは、xも与えられていて、証明問題になっています。(^_^;

　a = 38°, b = 46°, c = 22°, d = 48°; x = 18°?

　このとき、e = 180°-(46+22+48)°= 64°, f = 180°-(38+46+22)°= 74°
　前回と同様に、x,yの連立方程式を立てて、変数部分と定数部分P,Qに分けると、
$x+y=46+22=68=P \cdots(1)$
$\frac{\sin(y^{\circ})}{\sin(x^{\circ})}=\frac{\sin(46^{\circ})\sin(48^{\circ})\sin(74^{\circ})}{\sin(38^{\circ})\sin(22^{\circ})\sin(64^{\circ})}=Q \cdots(2)$
　x=18°のとき、(1)式から、y=68-18=50°
　これらを(2)式に代入して、両辺の分母を払って、
$\sin(18^{\circ})\sin(46^{\circ})\sin(48^{\circ})\sin(74^{\circ})=\sin(22^{\circ})\sin(38^{\circ})\sin(50^{\circ})\sin(64^{\circ}) \cdots(3)$
　結局、この等式の左辺lhsと右辺rhsの差D=0を示せばよい。
　ここで、三角関数の積和の公式を数回連続で使って導いた次の公式を使います。(^_^;

　sin(a)cos(b)             = (1/2) {sin(a+b)+sin(a-b)}
　sin(a)cos(b)cos(c)       = (1/4) {sin(a+b+c)+sin(a+b-c)+sin(a-b+c)+sin(a-b-c)}
　sin(a)cos(b)cos(c)cos(d) = (1/8) {sin(a+b+c+d)+sin(a+b+c-d)+sin(a+b-c+d)+sin(a+b-c-d)
                                   +sin(a-b+c+d)+sin(a-b+c-d)+sin(a-b-c+d)+sin(a-b-c-d)}

$8lhs=8\sin(18^{\circ})\cos(90-46^{\circ})\cos(90-48^{\circ})\cos(90-74^{\circ})$
　 $=8\sin(18^{\circ})\cos(44^{\circ})\cos(42^{\circ})\cos(16^{\circ})$
　 $=\sin(120^{\circ})+\sin(88^{\circ})+\sin(36^{\circ})+\sin(4^{\circ})+\sin(32^{\circ})+\sin(0^{\circ})-\sin(52^{\circ})-\sin(84^{\circ})$
　 $=\sin(60^{\circ})+\sin(88^{\circ})+\sin(36^{\circ})+\sin(4^{\circ})+\sin(32^{\circ})-\sin(52^{\circ})-\sin(84^{\circ})$

$8rhs=8\sin(22^{\circ})\cos(90-38^{\circ})\cos(90-50^{\circ})\cos(90-64^{\circ})$
　 $=8\sin(22^{\circ})\cos(52^{\circ})\cos(40^{\circ})\cos(26^{\circ})$
　 $=\sin(140^{\circ})+\sin(88^{\circ})+\sin(60^{\circ})+\sin(8^{\circ})+\sin(36^{\circ})-\sin(16^{\circ})-\sin(44^{\circ})-\sin(96^{\circ})$
　 $=\sin(40^{\circ})+\sin(88^{\circ})+\sin(60^{\circ})+\sin(8^{\circ})+\sin(36^{\circ})-\sin(16^{\circ})-\sin(44^{\circ})-\sin(84^{\circ})$

∴ $8D=8lhs-8rhs$
　 $=\{\sin(4^{\circ})+\sin(32^{\circ})-\sin(52^{\circ}) \}-\{\sin(40^{\circ})+\sin(8^{\circ})-\sin(16^{\circ})-\sin(44^{\circ}) \}$
　 $=\sin(4^{\circ})-\{\sin(40^{\circ})-\sin(32^{\circ}) \}-\{\sin(52^{\circ})-\sin(44^{\circ}) \}+\{\sin(16^{\circ})-\sin(8^{\circ}) \}$
　 $=\sin(4^{\circ})-2\cos(36^{\circ})\sin(4^{\circ})-2\cos(48^{\circ})\sin(4^{\circ})+2\cos(12^{\circ})\sin(4^{\circ})$
　 $=2\sin(4^{\circ})\{\frac{1}{2}-\cos(36^{\circ})-\cos(48^{\circ})+\cos(12^{\circ}) \}$
　 $=2\sin(4^{\circ})\{\frac{1}{2}-\sin(54^{\circ})-\sin(42^{\circ})+\sin(78^{\circ}) \}$
　 $=2\sin(4^{\circ})\{\frac{1}{2}-\sin(54^{\circ})+\sin(18^{\circ}) \}$
　 $=2\sin(4^{\circ})(\frac{1}{2}-\frac{\sqrt{5}+1}{4}+\frac{\sqrt{5}-1}{4} )=0$

　ちなみに、

$\sin(x)=\sin(x+60^{\circ})+\sin(x-60^{\circ})$
$\cos(x)=\cos(x+60^{\circ})+\cos(x-60^{\circ})=\cos(60^{\circ}+x)+\cos(60^{\circ}-x)$
$\sin(54^{\circ})-\sin(18^{\circ})=2\sin(54^{\circ})\sin(18^{\circ})=\frac{1}{2}$

P.S.
　8D=8lhs-8rhs以下の式変形が難しかったら、
　公式 sin(x)=sin(x+60ﾟ)+sin(x-60ﾟ)で、項数を5個まで減らせたら、
　公式 sin(x)=sin(x+36ﾟ)+sin(x-36ﾟ)-sin(x+72ﾟ)-sin(x-72ﾟ)を使うとよいです。
P.S.
[x]=sin(xﾟ)と略記すると、
8lhs=[60]+[88]+[36]+[4]+[32]-[52]-[84]
8rhs=[40]+[88]+[60]+[8]+[36]-[16]-[44]-[84]
∴8D=8lhs-8rhs
=[4]+[32]-[52]-[40]-[8]+[16]+[44]
=[4]+[32]-[40]+[16]+[44]-[68] (∵[8]=[68]-[52])
=[4]+[32]-[40]-[68]+[76] (∵[16]=[76]-[44])
=0 (∵[4]=[4+36]+[4-36]-[4+72]-[4-72]=[40]-[32]-[76]+[68])
　ちなみに、他も同様。
[32]=[32+36]+[32-36]-[32+72]-[32-72]=[68]-[4]-[76]+[40]

※参考URL
●[幾何大王からの挑戦状] 角度の問題#36
●Angular Angst - MathPages
●Langleyの問題をJavaで解いてみた。
●Langleyの問題を三角関数で解いてみた。
●sin(3°)系列
●sin(3°)系列 (2)

P.S.
　ちょっと、ネットで検索して見つけた1変数の式を使って解いてみます。以下、三角関数の単位は[°]で、省略します。

sin(a)sin(c)sin(b+c+d-x) = sin(a+b)sin(d)sin(x)

∴sin(38)sin(22)sin(46+22+48-x) = sin(38+46)sin(48)sin(x)
∴sin(38)sin(22)sin(116-x) = sin(84)sin(48)sin(x)
∴sin(38)sin(22)sin(116-18) = sin(84)sin(48)sin(18)
∴sin(22)sin(38)sin(82) = sin(18)sin(48)sin(84)

　結局、この等式の左辺lhsと右辺rhsの差D=0を示せばよい。
　ここで、今回は、積も和も全部+のcosだけの式になる三角関数の積和の公式を用いてみます。(^_^;

　cos(a)cos(b)             = (1/2) {cos(a+b)+cos(a-b)}
　cos(a)cos(b)cos(c)       = (1/4) {cos(a+b+c)+cos(a+b-c)+cos(a-b+c)+cos(a-b-c)}
　cos(a)cos(b)cos(c)cos(d) = (1/8) {cos(a+b+c+d)+cos(a+b+c-d)+cos(a+b-c+d)+cos(a+b-c-d)
                                   +cos(a-b+c+d)+cos(a-b+c-d)+cos(a-b-c+d)+cos(a-b-c-d)}

4lhs=4cos(90-22)cos(90-38)cos(90-82)=4cos(68)cos(52)cos(8)
=+cos(68+52+8)+cos(68+52-8)+cos(68-52+8)+cos(68-52-8)
=+cos(128)+cos(112)+cos(24)+cos(8)
=-cos(52)-cos(68)+cos(24)+cos(8)

4rhs=4cos(90-18)cos(90-48)cos(90-84)=4cos(72)cos(42)cos(6)
=+cos(72+42+6)+cos(72+42-6)+cos(72-42+6)+cos(72-42-6)
=+cos(120)+cos(108)+cos(36)+cos(24)
=-cos(60)-cos(72)+cos(36)+cos(24)

∴4D=4lhs-4rhs
={-cos(52)-cos(68)+cos(24)+cos(8)}-{-cos(60)-cos(72)+cos(36)+cos(24)}
={-cos(52)-cos(68)+cos(8)}-{-cos(60)-cos(72)+cos(36)}
={-cos(8)+cos(8)}-{-cos(60)+cos(60)}=0

∵
cos(8)=cos(60+8)+cos(60-8)=cos(52)+cos(68)
cos(60)=cos(60+36)+cos(60-36)-cos(60+72)-cos(60-72)=cos(36)-cos(72)

※公式集（ただし、単位は、[°]）

$\cos(-x)=\cos(x), \cos(x)=-\cos(180^{\circ}-x)$
$\cos(x)=\cos(x+60^{\circ})+\cos(x-60^{\circ})=\cos(60^{\circ}+x)+\cos(60^{\circ}-x)$
$\sin(x)=\sin(x+60^{\circ})+\sin(x-60^{\circ})=\sin(60^{\circ}+x)-\sin(60^{\circ}-x)$
$\cos(x)=\cos(x+36^{\circ})+\cos(x-36^{\circ})-\cos(x+72^{\circ})-\cos(x-72^{\circ})$
$\sin(x)=\sin(x+36^{\circ})+\sin(x-36^{\circ})-\sin(x+72^{\circ})-\sin(x-72^{\circ})$
$\cos(60^{\circ})=\cos(60^{\circ}+36^{\circ})+\cos(60^{\circ}-36^{\circ})-\cos(60^{\circ}+72^{\circ})-\cos(60^{\circ}-72^{\circ})$
$=\cos(36^{\circ})-\cos(72^{\circ})=\frac{1}{2}$

P.S.
　証明は、たとえば、積和公式2sin(x)cos(y)=sin(x+y)+sin(x-y)で、y=60°とおく。
　また、積和公式で、y=36°とおいたものから、y=72°とおいたものを引いて、cos(36°)-cos(72°)=1/2を使う。
　ちなみに、cos(36°)-cos(72°)=2sin(18°)sin(54°)
=2sin(18°)sin(72°)*2sin(36°)sin(54°)/(2sin(36°)sin(72°))=1/2

（以下、単位は、[rad]）
$\cos(x)=\sum_{k=1}^{n}{(-1)^{k-1} \left(\cos\left(x+\frac{\pi k}{2n+1}\right)+\cos\left(x-\frac{\pi k}{2n+1}\right)\right)}$ 　(where n=1,2,3,...)
$\cos(x)=\sum_{k=1}^{2n}{(-1)^{k-1}\cos\left(x+\frac{\pi k}{2n+1}\right)}$
$\sin(x)=\sum_{k=1}^{2n}{(-1)^{k-1}\sin\left(x+\frac{\pi k}{2n+1}\right)}$

※参考URL
●kadai78