Question

Cosa- sina=1÷root 2 general solution​

Answer 1

Answer:

A = nπ / 2 + ( -1 )ⁿ ( π / 12 )

Step-by-step explanation:

Given---> CosA - SinA = 1 /√2

To find ---> General solution of given trigonometric equation

Solution---> ATQ,

CosA - SinA = 1 /√2

Squaring both sides , we get,

=> ( CosA - SinA )² = ( 1 / √2 )²

We have an identity,

( a - b )² = a² + b² - 2ab , applying it here , we get,

=> Cos²A + Sin²A - 2 SinA CosA = 1 / 2

We have an identity, Sin²θ + Cos²θ = 1 , applying it here , we get.

=> 1 - 2 SinA CosA = 1 / 2

=> - 2 SinA CosA = 1 /2 - 1

=> - 2 SinA CosA = - 1 / 2

=> 2 SinA CosA = 1 / 2

We know that, Sin2θ = 2 Sinθ Cosθ , applying it here , we get,

=> Sin2A = 1 / 2

=> Sin2A = Sin( π / 6 )

We know that general value of Sin is

{ nπ + ( -1 )ⁿ α } , applying it here , we get,

=> Sin2A = Sin { nπ + ( -1 )ⁿ π / 6 }

=> 2A = nπ + ( -1 )ⁿ π / 6

=> A = ( nπ / 2 ) + ( - 1 )ⁿ ( π / 12 )