Question

Prove identity ${\sin }^2\theta +{\cos }^2\theta =1$, geometrically.

Answer 1

In right $△ABC$, $\angle B={90}^o$

By Pythagorus theorem,

$A{C}^2=A{B}^2+B{C}^2$
Dividing both sides by $A{C}^2$
$\Rightarrow \frac{A{C}^2}{A{C}^2}=\frac{A{B}^2}{A{C}^2}+\frac{B{C}^2}{A{C}^2}$
$\Rightarrow 1={\left(\frac{AB}{AC}\right)}^2+{\left(\frac{BC}{AC}\right)}^2$

$\Rightarrow 1=(\cos A{)}^2+(\sin A{)}^2$

$\Rightarrow {\sin }^{2}A+{\cos }^{2}A=1$

Hence, proved.