prove數學問題

Sol
For L.H.S.,
note that 2sinθcosθ=sin2θ
therefore, 1-2sinθcosθ=1-sin2θ
For R.H.S.,
(cosθ-sinθ)^2=cos^2θ-2cosθsinθ+sin^2θ
note that 2sinθcosθ=sin2θ and sin^2θ+cos^2θ = 1
=sin^2θ+cos^2θ - 2 sin θ cos θ = 1 - sin 2θ
=L.H. S.

Therefore,
1-2 sinθ cosθ =（cosθ- sinθ）^2

R.H.S.=（cosθ- sinθ）^2
=cos^2θ-2sinθcosθ+sin^2θ
=1-2sinθcosθ
=L.H.S.

R.H.S= (cosθ- sinθ)^2

=(cosθ)^2 +(sinθ)^2 -2sinθcosθ

=1- 2sinθcosθ

=L.H.S

1-2 sinθ cosθ =（cosθ- sinθ）^2
sol
1-2sinθ cosθ
=cos^2θ+sin^2θ-2sinθ cosθ
=cos^2θ-2sinθ cosθ+sin^2θ
=(cosθ- sinθ)^2

=(cosθ- sinθ)^2
=cos^2θ-2sinθ cosθ+sin^2θ
=cos^2θ+sin^2θ-2sinθ cosθ
=