Question

If cos 0 +sin0 = V2 cose, then prove that
cose -o sin 0 = V2 sin ​

Answer 1

Correct Question

If cos∅ + sin∅ = √2 cos∅, then prove that cos∅ - sin∅ = √2 sin∅.

Proof

→ cos∅ + sin∅ = √2 cos∅

→ sin∅ = √2 cos∅ - cos∅

→ sin∅ = cos∅(√2 - 1)

→ sin∅/(√2 - 1) = cos∅

On rationalizing we get,

→ sin∅/(√2 - 1) × (√2 + 1)/(√2 + 1) = cos∅

→ sin∅(√2 + 1)/(2 - 1) = cos∅

→ √2 sin∅ + sin∅ = cos∅

→ √2 sin∅ = cos∅ - sin∅

Or cos∅ - sin∅ = √2 sin∅

Hence Proved

Answer 2

Correct Question :-

$\sf{ If \: cos \theta + sin \theta = \sqrt{2} cos \theta}$

$\textsf{ Then prove that }$

$\sf{ cos\theta - sin\theta = \sqrt{2} sin\theta }$

Solution :-

Given :-

$cos\theta + sin\theta = \sqrt{2}cos\theta$

Now by taking like terms on same side :-

$\rightarrow sin\theta = \sqrt{2}cos\theta - cos\theta$

$\rightarrow sin\theta = cos\theta( \sqrt{2} - 1 )$

$\rightarrow \dfrac{sin\theta}{(\sqrt{2} - 1)} = cos\theta$

Now we will rationalize the denominator :-

$\rightarrow \dfrac{sin\theta}{(\sqrt{2} - 1)} \times \dfrac{( \sqrt{2} + 1)}{( \sqrt{2} + 1)} = cos\theta$

$\rightarrow \dfrac{ sin\theta(\sqrt{2} +1)}{(\sqrt{2})^2 - 1^2 } = cos\theta$

$\rightarrow \dfrac{ sin\theta(\sqrt{2} +1)}{2 - 1 } = cos\theta$

$\rightarrow \dfrac{ sin\theta(\sqrt{2} +1)}{1 } = cos\theta$

$\rightarrow \sqrt{2}sin\theta + sin\theta = cos\theta$

$\rightarrow cos\theta - sin\theta = \sqrt{2}sin\theta$

Hence Proved