三角函數公式證明

Let AD = a

In ΔADB :
sin β = DB/a_ =>_ DB = a sin β
cos β = AB/a =>_AAB = a cos β

In ΔADQ :
sin α = DQ/a_ =>_ DQ = a sin α
cos α = AQ/a_ =>_ AQ = a cos α

In ΔABC :
sin (α - β) = BC/AB = BC/ a cosβ_ =>_ BC = a sin (α - β) cos β
cos (α - β) = AC/AB = AC/ a cosβ_ =>_ AC = a cos (α - β) cos β

∠ABP = ∠BAQ = α - β (alt. ∠s, AQ // PB)
Hence, ∠PBD = 90° - (α - β) and ∠BDP = α - β

In ΔBDP :
sin (α - β) = BP/DB = BP/ a sinβ_ =>_ BP = a sin (α - β) sin β
cos (α - β) = DP/DB = DP/ a sinβ_ =>_ DP = a cos (α - β) sin β

DQ = DP + PQ
DQ = DP + BC
a sin α = a cos (α - β) sin β + a sin (α - β) cos β
cos (α - β) sin β + sin (α - β) cos β = sin α ...... [1]

AC = AQ + QC
AC = AQ + BP
a cos (α - β) cos β = a cos α + a sin (α - β) sin β
cos (α - β) cos β - sin (α - β) sin β = cos α ...... [2]

[1] x cos β :
cos (α - β) sin β cos β + sin (α - β) cos² β = sin α cosβ ...... [3]

[2] x sin β :
cos (α - β) sin β cos β - sin (α - β) sin² β = cos α sin β...... [4]

[3] - [4] :
sin (α - β) (sin² β + cos² β) = sin α cos β - cos α sinβ
Hence, sin (α - β) = sin α cos β - cos α sinβ

[1] x sin β :
cos (α - β) sin² β + sin (α - β) sin β cos β = sin α sin β ...... [5]

[2] x cos β :
cos (α - β) cos² β - sin (α - β) sin β cos β = cos α cos β...... [6]

[5] + [6] :
cos (α - β) (sin² β + cos² β) = sin α sin β + cos α cos β
Hence, cos (α - β) = sin αsin β + cos α cos β

∠QPC = 90˚

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