Question

If acosθ + bsinθ = c have roots α and β. Then, what will be the value of sinα + sinβ?

(a) 2bc/(a^2 + b^2)

(b) 0

(c) 1

(d) (c^2 + a^2)/(a^2 + b^2)

I had been asked this question in quiz.

My enquiry is from Quadratic Equations in division Complex Numbers and Quadratic Equations of Mathematics – Class 11

Answer 1

Correct choice is (a) 2bc/(a^2 + b^2)

Explanation: Given, acosθ + bsinθ = c

So, this implies acosθ = c – bsinθ

Now squaring both the sides we get,

(acosθ)^2 = (c – bsinθ)^2

a^2 cos^2 θ = c^2 + b^2 sin^2 θ – 2b c sinθ

a^2 (1- sin^2 θ) = c^2 + b^2 sin^2 θ – 2b c sinθ

a^2 – a^2 sin^2 θ = c^2 + b^2 sin^2 θ – 2b c sinθ

Now rearranging the elements,

(a^2 + b^2) sin^2 θ – 2b c sinθ  +( c^2 – a^2) = θ

So, as sum of the roots are in the form –b/a if there is a quadratic equation ax^2 + bx + c = 0

Now, we can conclude that

sinα + sinβ = 2bc/(a^2 + b^2).