Question

sin0 + cos0 = √2 (where 0 ° <0 <90 °) is the value of 0​

Answer 1

Correct question :-

If sinθ + cosθ = √2 (where 0° < θ < 90°) , then find the value of θ.

Solution :-

sinθ + cosθ = √2

Squaring on both sides

⇒ (sinθ + cosθ)² = (√2)²

⇒ sin²θ + cos²θ + 2sinθcosθ = 2

[ Because (x + y)² = x² + y² + 2xy ]

⇒ 1 + 2sinθcosθ = 2

[ Because sin²θ + cos²θ = 1 ]

⇒ 2sinθcosθ = 2 - 1

⇒ 2sinθcosθ = 1

⇒ sin 2θ = 1

[ Because 2sinθcosθ = sin 2θ ]

⇒ sin 2θ = sin 90°

[ Because sin90° = 1 ]

Comparing on both sides

⇒ 2θ = 90°

⇒ θ = 90°/2

⇒ θ = 45°

Therefore the value of θ is 45°.

Answer 2

HEY MATE YOUR ANSWER IS HERE...

LET THETA BE “ X ”

★ GIVEN ★

$sin \: x \: + \: cos \: x \: = \sqrt{2}$

NOW

BY HIT AND TRIAL METHOD

LET X = 45°

HENCE ,

$sin \: 45 \: = \frac{1}{ \sqrt{2} }$

$cos \: 45 = \frac{1}{ \sqrt{2} }$

NOW

$\frac{1}{ \sqrt{2} } + \frac{1}{ \sqrt{2} } = \sqrt{2}$

$\frac{2}{ \sqrt{2} } = \sqrt{2}$

HENCE

$\frac{ (\sqrt{2} ) ( \sqrt{2}) }{ \sqrt{2} } = \sqrt{2}$

NOW

$\sqrt{2} = \sqrt{2}$

HENCE VALUE OF “ X ” = 45°

OR VALUE OF THETA = 45°

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