Step-by-step explanation:
We need to explain why sin²θ + cos²θ = 1
Consider a right angled triangle, we have base, perpendicular and hypotenuse.
Refer the figure
[tex]sin\theta =\frac{\texttt{perpendicular}}{\texttt{hypotenuse}}\\\\cos\theta =\frac{\texttt{base}}{\texttt{hypotenuse}}[/tex]
Taking sin²θ + cos²θ
[tex]sin^2\theta =\frac{\texttt{perpendicular}^2}{\texttt{hypotenuse}^2}\\\\cos^2\theta =\frac{\texttt{base}^2}{\texttt{hypotenuse}^2}\\\\sin^2\theta+cos^2\theta=\frac{\texttt{perpendicular}^2}{\texttt{hypotenuse}^2}+\frac{\texttt{base}^2}{\texttt{hypotenuse}^2}\\\\sin^2\theta+cos^2\theta=\frac{\texttt{perpendicular}^2+\texttt{base}^2}{\texttt{hypotenuse}^2}[/tex]
We know Pythagoras theorem
perpendicular² + base² = hypotenuse²
[tex]sin^2\theta+cos^2\theta=\frac{\texttt{hypotenuse}^2}{\texttt{hypotenuse}^2}\\\\sin^2\theta+cos^2\theta=1[/tex]
Hence sin²θ + cos²θ = 1
