Question
If α, β, γ, δ are the smallest positive angles in ascending order of magnitude which have their sines equal to the positive
quantity k, then the value of 4 sin α2 + 3 sin β2+ 2 sin γ2 + sin δ2 is equal to
quantity k, then the value of 4 sin α2 + 3 sin β2+ 2 sin γ2 + sin δ2 is equal to
Answer: Option C
:
C
(c) Given α < β < γ < δ and sin α = sin β = sin γ = sin δ = k. Alsoα, β,γ,δ are smallest positive angles satisfying above two conditions.
∴ We can take β = π - α, γ = 2π + α, δ = 3π - α
Given expression
= 4 sin α2 + 3 sin(π2−α2) + 2 sin(π+α2) +3 sin(3π2−α2)
=4 sin α2 + 3cos α2 - 2sin α2 -cos α2 = 2 (sinα2+cosα2)
= 2 √(sin12α+cos12α)2 = 2 √1+sinα = 2√1+k.
Was this answer helpful ?
:
C
(c) Given α < β < γ < δ and sin α = sin β = sin γ = sin δ = k. Alsoα, β,γ,δ are smallest positive angles satisfying above two conditions.
∴ We can take β = π - α, γ = 2π + α, δ = 3π - α
Given expression
= 4 sin α2 + 3 sin(π2−α2) + 2 sin(π+α2) +3 sin(3π2−α2)
=4 sin α2 + 3cos α2 - 2sin α2 -cos α2 = 2 (sinα2+cosα2)
= 2 √(sin12α+cos12α)2 = 2 √1+sinα = 2√1+k.
Was this answer helpful ?
More Questions on This Topic :
Latest Videos
Latest Test Papers
Submit Solution