點解cos^2θ+sin^2θ=1

係一個直角3角形入面：
設
對邊 = a
鄰邊 = b
斜邊 = c
sinθ = a/c
sin2θ = a2/c2
cosθ = b/c
cos2θ = b2/c2
根據畢氏定理：
a2 + b2 = c2
將成條式除c2
a2/c2 + b2/c2 = 1
sin2θ + cos2θ = 1

設一 個直角三角形為ABC
設鄰邊為a 對邊為b 鈄邊為c

sinθ=b/c, sin^2θ=b^2/c^2
cosθ=a/c,cos^2θ=a^2/c^2

∴sin^2θ+cos^2θ=b^2/c^2+a^2/c^2
=b^2+a^2/c^2

根據畢氏定理
a^2+b^2=c^2

∴sin^2θ+cos^2θ=b^2+a^2/c^2
=c^2/c^2
=1

∴cos^2θ+sin^2θ=1

設三角形三條邊為3, 4, 5 的長度

斜邊是5
斜邊的對邊是3
餘下的一條邊是4

設5及4 之間的夾角為x

cosx=4/5
cos^2x=16/25

sinx=3/5
sin^2x=9/25

cos^2x+sin^2x=(16+9)/25=25/25=1

For a right-angled triangle, let A=adjacent side, O=opposite side and H=hypotenus.
By Pythagoras theorem:
O^2 + A^2 = H^2
All divided by H^2, we get O^2/H^2 + A^2/H^2 = H^2/H^2
that is sin^2 x + cos^2 x = 1.