Question

if alpha and bita are solution of a cost thita +bsin thita, then prove that cos(alpha-bita)=a2-b2÷a2+b2​

Answer 1

Answer:

Step-by-step explanation:

If alpha and beta are solution of the  

acosθ + bsinθ = c

Then

acosα + bsinα = c

and

acosβ+bsinβ = c

acosα + bsinα =acosβ+bsinβ

acosα - acosβ = bsinβ  - bsinα

a( cosα - cosβ) = b(sinβ - sinα)

sinβ - sinα/ cosα - cosβ = a/b

[sinβ - sinα/ cosα - cosβ]² = (a/b)²

Apply componendo dividendo rule and solve left hand side. Combine the terms and take common. You will be left with desired result.

(sinβ - sinα) ² -  (cosα - cosβ)²/ (sinβ - sinα) ² + (cosα - cosβ)² = a² - b²/a² + b²

LHS

sin²β + sin²α - 2sinαsinβ - cos²α - cos²β+2cosαCosβ / sin²β + sin²α - 2sinαsinβ + cos²α + cos²β - 2cosαCosβ

=> - cos2β - cos2α + 2cos(α+β)/ 1 + 1  - 2cos(α+β)

=> -2cos(α+β)cos(α-β)+2cos(α+β)/2 - 2cos(α+β)

=> cos(α+β)

=> cos(α+β) = a² - b²/a² + b²