The easiest trigonometry in the world2

 

abk

bk=bj½=0.0327190828217

sin1.875=0.0327190828217

ak=√(1-bk²)=√1-0.0327190828217²=0.9994645874763

cos1.875=0.9994645874763

abl ab1, al1

bk=0.0327190828217

ak=0.9994645874763

kl=1-ak=0.0005354125237

bl=√(kl²+bk²)=√0.0005354125237²+0.0462469961894²=0.0327234632528

abm

bm=bl½=0.0163617316264

sin0.9375=0.0163617316264

am=√(1-bm²)=0.9998661379095

cos0.9375=0.9998661379095

abn, ab1, an1

am=0.0163617316264

bm=0.9998661379095

mn=1-am=0.0001338620905

bn=√(mn²+bm²)=0.0163622792077

abo

bo=bn½=0.008181139603

sin0.46875=0.008181139603

an=√(1-bo²)=0.9999665339174

cos0.46875=0.9999665339174

 

 

bc×(ab-√(ab²-bc²))×½=be→be×(ab(1)-√(ab²-be²))×½=bf→bf×(ab-√(ab²-bf²))×½=bg→bg×(ab-√(ab²-bg²))×½=bh→bh×(ab-√(ab²-bh²))×½=bi→bi×(ab-√(ab²-bi²))×½=bj→bj×(ab-√(ab²-bj²(0.703125²)))×½=bk....

√(sinA²+(1-cosA)²)×½

bc(sin45)×(ab(1)-√(ab(1)-bc(sin45)²))×½=be(sin22.5)→be(sin22.5)×(ab(1)-√(ab(1)-be(sin22.5)²))×½=bf(sin11.25)→bf(sin11.5)×(ab(1)-√(ab(1)-bf(sin11.25)²))×½=bg(sin5.625)→bg(sin5.625)×(ab(1)-√(ab(1)-bg(sin5.625)²))×½=bh(sin2.8125)→bh(sin2.8125)×(ab(1)-√(ab(1)-bh(sin2.8125)²))×½=bi(sin1.40625)→bi(sin1.40625)×(ab(1)-√(ab(1)-bi(sin1.40625)²))×½=bj(sin0.703125)→bj(sin0.703125)×(ab(1)-√(ab(1)-bj(0.703125)²))×½=bk(sin0.3515625)....

√(sin30²+(1-cos30)²)×½