Question

The points $A(0,0),B(\cos \alpha ,\sin \alpha )$, and $C(\cos \beta ,\sin \beta )$ are the vertices of a right-angled triangle if

• A. ${\sin }^2\frac{\alpha -\beta }{2}=\frac{1}{2\sqrt{2}}$
• B. $\cos \frac{\alpha -\beta }{2}=-\frac{1}{\sqrt{2}}$
• C. ${\cos }^2\frac{\alpha -\beta }{2}=\frac{1}{2\sqrt{2}}$
• D. $\sin \frac{\alpha -\beta }{2}=-\frac{1}{\sqrt{2}}$

Answer 1

The correct option is A ${\sin }^2\frac{\alpha -\beta }{2}=\frac{1}{2\sqrt{2}}$

$\begin{array}{c}ByusingPythagoroustheorem\\ Then,A{C}^2+A{B}^2=B{C}^2\\ \Rightarrow \sqrt{\left({\cos }^2\beta +{\sin }^2\beta \right)+\left({\cos }^2\alpha +{\sin }^2\alpha \right)}={\left(\cos \beta -\cos \alpha \right)}^2+{\left(\sin \beta -\sin \alpha \right)}^2\\ \Rightarrow \sqrt{1+1}=\left({\cos }^2\beta +{\sin }^2\beta +{\sin }^2\alpha +{\cos }^2\alpha -2\cos \beta .\cos \alpha -2sin\alpha sin\beta \right)\\ \Rightarrow \sqrt{2}=\left[2-2\left\{\sin \alpha .\sin \beta +\cos \beta .\cos \alpha \right\}\right]\\ \Rightarrow \sqrt{2}=2-2\cos \left(\alpha -\beta \right)\\ \Rightarrow \frac{\sqrt{2}}{2}=1-\cos \left(\alpha -\beta \right)\\ \Rightarrow \frac{1}{\sqrt{2}}=2{\sin }^2\frac{\left(\alpha -\beta \right)}{2}\\ \sqrt{\frac{1}{2\sqrt{2}}}=\sin \frac{\left(\alpha -\beta \right)}{2}\end{array}$