Thanks for Alexander Bogomolny's kind reply to my
previous note. 1)cosine proof:With the usual diagrams and notations.
In TRIANGLE ABC,angle C=90deg.
side a is opposite Angle A
side b is opp angle B
side c is opp angle C
From angle C drop a perpendicular to line AB at D.
BD=a.cos.B.. DA=c-a.cos.B
cos.A=b/c=(c-a.cos.B)/b....and cos.B=a/c
:: b.b=c.c-ac.cos.B =c.c-ac.a/c
:: b.b=c.c-a.a ::b2 +a2=c2
2)sine proof:BD=a.sin.A ...angle DCB is size A
angle CDB is 90.deg.
DA=BA-BD...Angle ACB is of sizeB.
sin.B=DA/AC=(c-a.sin.A)/b=b/c
sin.A=a/c ::cc-ca.a/c=bb ::cc-aa=bb
these steps are derived from similar triangles.
Thanks to readers