If cos(θ-α)=a,sin(θ-β)=b, then a2-2absin(α-β)+b2 is equal to
sin2(α–β)
cos2(α–β)
1
0
Step 1. Find the value of a2-2absin(α-β)+b2:
Given, cos(θ-α)=a,sin(θ-β)=b
⇒α–β=(θ–β)–(θ–α)
Step 2. Take “sin” on both sides,
sin(α–β)=sin[(θ–β)–(θ–α)]=sin(θ–β)cos(θ–α)–cos(θ–β)sin(θ–α)=ba–(1–b2)(1–a2)
sin(α-β)=ab–(1–a2)(1–b2) ….(i)
Again,
α–β=(θ–β)–(θ–α)
Step 3. Take “cos” on both sides,
cos(α–β)=cos[(θ–β)–(θ–α)]=cos(θ–α)cos(θ–β)+sin(θ–α)sin(θ–β)=a(1–b2)+b(1–a2)
cos2(α–β)=a2(1–b2)+b2(1–a2)+2ab(1–a2)(1–b2) ….(ii)
∴a2–2absin(α–β)+b2=a2–2ab[ab–(1–a2)(1–b2)]+b2 {from (i)}
=a2–2a2b2+2ab(1–a2)(1–b2)+b2 …..(iii)
Step 4. By equating equation (ii) and (iii), we get
a2–2absin(α–β)+b2=cos2(α–β)
Hence, Option ‘B’ is Correct.
If cos(θ−α) = a, sin(θ−β) = b,
then cos2(α−β) + 2ab sin(α−β) is equal to
if alpha nad beta are the roots of acosx+bcosx=c show
that cos(alpha+beta)=(a2-b2) / (a2+b2)