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Question
If cos(θ−α)=a and sin(θ−β)=b, then cos2(α−β)+2absin(α−β) is equal to
If cos(θ−α)=a and sin(θ−β)=b, then cos2(α−β)+2absin(α−β) is equal to
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Solution
The correct option is A a2+b2
sin((θ−β)−(θ−α))=sin(α−β)=sin(θ−β)cos(θ−α)−cos(θ−β)sin(θ−α)
sin(α−β)=ba−√1−a2√1−b2
=ab−√1−a2√1−b2
cos((θ−β)−(θ−α))=cos(α−β)[∵cos(A−B)=cosAcosB+sinAsinB]
=a√1−b2+b√1−a2
Therefore, cos2(α−β)+2absin(α−β)
=a2−a2b2+b2−b2a2+2ab√a−b2√1−a2+2a2b2−2ab√1−a2√1−b2
=2a2b2−a2b2−b2a2+a2+b2
=a2+b2
sin((θ−β)−(θ−α))=sin(α−β)=sin(θ−β)cos(θ−α)−cos(θ−β)sin(θ−α)
sin(α−β)=ba−√1−a2√1−b2
=ab−√1−a2√1−b2
cos((θ−β)−(θ−α))=cos(α−β)[∵cos(A−B)=cosAcosB+sinAsinB]
=a√1−b2+b√1−a2
Therefore, cos2(α−β)+2absin(α−β)
=a2−a2b2+b2−b2a2+2ab√a−b2√1−a2+2a2b2−2ab√1−a2√1−b2
=2a2b2−a2b2−b2a2+a2+b2
=a2+b2
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