三角函數的題目!!拜託一下

三角形ABC中,已知角A-角C=90度,線段BC+線段AB=根號2的線段AC,求cos^2C=

sinA = sin(C + 90度) = cosC

a + c = √2b
sinA + sinC = √2sinB

cosC + sinC = √2sin(180度 - A - C) = √2sin(90度 - 2C) = √2cos2C = √2[(cosC)^2 - (sinC)^2]
1 = √2(cosC - sinC)
(cosC - sinC)^2 = 1/2
1 - sin2C = 1/2
C = 15度

(cosC)^2 = (2 + √3)/4

感謝thepiano!!我一直繞餘弦定理!都沒運用到正弦定理!!感恩