Question

If \( 0^{\circ}<\theta<90^{\circ} \) and \( (\sin \theta+\cos \theta)=\sqrt{2} \), then \( \theta= \) ?

(a) \( \frac{\pi}{3} \) 

(b) \( \frac{\pi}{4} \) 

(c) \( \frac{\pi}{6} \) 

(d) \( \pi \)

Answer 1

Given that 0° < θ < 90°

Also sin θ + cos θ = \(\sqrt2\)

⇒ sin²θ + cos²θ + 2sinθ cosθ= 2

By squaring both side

⇒ 1 + sin 2θ = 2 (\(\because\) sin² θ + cos²θ = 1 and 2 sinθ cosθ = sin 2θ)

⇒ sin 2θ = 2 - 1 = 1 = sin π/2 (\(\because\) if 0° < θ < 90° ⇒ 0° < 2 θ < 180°)

⇒ 2 θ = sin π/2

⇒ θ = π/4

Alternative method:

⇒ sin θ + cos θ = \(\sqrt2\) (Given)

⇒ sin θ + cos θ = 1/√2 + 1/√2 (\(\because\) 1/√2 + 1/√2 = 2/√2 = √2)

⇒ sin θ + cos θ  = sin \(\frac{\pi}4\) + cos \(\frac{\pi}4\) 

\(\therefore\) θ = \(\frac{\pi}4\)