Question

Solve. root3 sin theta-cos theta= root2​

Answer 1

Step-by-step explanation:

PLEASE...MARK...ME...AS

...... BRILLIANTLIST

We have,

→ cos θ + sin θ = √2cos θ .

[ Squaring both side, we get ] .

⇒ ( cos θ + sin θ )² = 2cos²θ .

⇒ cos²θ + sin²θ + 2cosθsinθ = 2cos² .

⇒ sin²θ + 2cosθsinθ = 2cos²θ - cos²θ .

⇒ sin²θ + 2cosθsinθ = cos²θ .

⇒ cos²θ - 2cosθsinθ = sin²θ .

[ Adding sin²θ both side, we get ] .

⇒ cos²θ - 2cosθsinθ + sin²θ = sin²θ + sin²θ .

⇒ ( cos θ - sin θ )² = 2sin²θ .

⇒ cos θ - sin θ = √( 2sin²θ ) .

∴ cos θ - sin θ = √2sin θ .

Answer 2

Answer:

Ф = n π + π / 6 + ( - 1 )ⁿ π / 4  where n € I

Step-by-step explanation:

Given :

√ 3 sin Ф - cos Ф = √ 2

r = √ ( √ 3 )² + 1² ) = 2

Adding whole given equation by 2 :

√ 3 / 2 sin Ф - 1 / 2 cos Ф = √ 2 / 2

sin Ф cos π / 6 - cos Ф sin π / 6 = 1 / √ 2

Using identity sin ( A - B ) = sin A cos B - cos A sin B

sin ( Ф - π / 6 ) = 1 / √ 2

We know :

sin π / 4 = 1 / √ 2

sin ( Ф - π / 6 ) = sin π / 4

On comparing we get :

Ф - π / 6 = π / 4

Ф = π / 6 + π / 4

We know :

if sin Ф = sin α

= > Ф = n π + ( - 1 )ⁿ α  where n € I

So , general equation of given equation is written as :

= > Ф = n π + π / 6 + ( - 1 )ⁿ π / 4  where n € I

Therefore we get answer.