Question

Prove that, ${\sin }^{2}A+{\cos }^{2}A=1$

Answer 1

Step 1. Find out $\sin A\mathrm{and}\cos A$

Let $∆ABC$ be right-angle triangle where $AB=p,BC=qandAC=r$.

In the right-angle triangle $∆ABC$ ,

$$\begin{gathered} \sin A=\frac{\mathrm{Perpendicular}}{\mathrm{Hypotenuse}} \\ \Rightarrow \sin A=\frac{BC}{AC} \\ \Rightarrow \sin A=\frac{q}{r} \end{gathered}$$

And,

$$\begin{gathered} \cos A=\frac{\mathrm{Base}}{\mathrm{Hypotenuse}} \\ \Rightarrow \cos A=\frac{AB}{AC} \\ \Rightarrow \cos A=\frac{p}{r} \end{gathered}$$

Step 2. Proving ${\sin }^{2}A+{\cos }^{2}A=1$

Consider LHS:

$\begin{array}{rcl}\mathrm{LHS}& =& {\sin }^{2}A+{\cos }^{2}A\\ & =& {\left(\frac{q}{r}\right)}^2+{\left(\frac{p}{r}\right)}^2\\ & =& \frac{p^2+q^2}{r^2}\left[\mathrm{by}\mathrm{pythagoras}\mathrm{theorem},p^2+q^2=r^2\right]\\ & =& \frac{r^2}{r^2}\\ & =& 1\\ & =& \mathrm{RHS}\end{array}$

Thus,$\mathrm{LHS}=\mathrm{RHS}$

Hence,${\sin }^{2}A+{\cos }^{2}A=1$ is proved.