Show that $(1 – \cos θ – \sin θ )^2 – 2(1 – \sin θ )(1 – \cos θ ) = 0$.

Question

Show that $(1 – \cos θ – \sin θ )^2 – 2(1 – \sin θ )(1 – \cos θ ) = 0$.

What kind of formulas should I use?

Answer 1

Observe \begin{align} (1-\cos\theta-\sin\theta)^2 & =(1-\cos\theta)^2-2(1-\cos\theta)\sin\theta+\sin^2\theta \\ & = 1 - 2\cos\theta+\color{red}{\cos^2\theta}-2(1-\cos\theta)\sin\theta+\color{red}{\sin^2\theta} \\ & = 1+\color{red}{1}-2\cos\theta -2(1-\cos\theta)\sin\theta\\ & =2(1-\cos\theta)-2(1-\cos\theta)\sin\theta\\ & =2(1-\cos\theta)(1-\sin\theta) \end{align}

Answer 2

LHS = $1 - 2\cos\theta - 2\sin\theta + 2\cos(\theta)\sin\theta + \cos^2\theta + \sin^2\theta -2 + 2\cos\theta + 2\sin\theta -2\cos(\theta)\sin\theta = 0$ = RHS