Question

If θ < 90°, find the value of: sin²θ + cos²θ

Answer 1

Since θ <90°,
Consider θ = 45°
∴ sin² + cos²
= sin²45° + cos²45°

= `(1/sqrt(2))^2 + (1/sqrt(2))^2`

= `(1)/(2) + (1)/(2)`
= 1.