Question

If θ = 30°, verify that: 1 - sin 2θ = (sinθ - cosθ)² 

Answer 1

Given: θ = 30°
1 - sin2θ
= 1 - sin2 x 30°
= 1  - sin60°

= ${\displaystyle 1-\frac{\sqrt{3}}{2}}$

= ${\displaystyle \frac{2-\sqrt{3}}{2}}$
(sinθ - cosθ)² 
= sin²θ + cos²θ - 2sinθ cosθ
= 1 - 2 x sin30° x cos30

= ${\displaystyle 1-2\times \frac{1}{2}\times \frac{\sqrt{3}}{2}}$

= ${\displaystyle 1-\frac{\sqrt{3}}{2}}$

= ${\displaystyle \frac{2-\sqrt{3}}{2}}$
⇒ 1 - sin2θ = (sinθ - cosθ)².