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Question

If 4+8-32+768=a2cos11πb, where a and b are natural numbers then b-a is divisible by:


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Solution

Step 1: Simplify the given expression to required form

Given that 4+8-32+768=a2cos11πb

Let 768=32cosθ

768=32cosθ163=32cosθcosθ=32θ=cos-132θ=π6

768=32cosπ61

Consider 4+8-32+768 ….2

Now, substitute 1 in 2, we get,

4+8-32+768=4+8-32+32cosπ6

=4+8-321+cosπ6

=4+8-322cos2π121+cosx=2cos2x2

=4+8-8cosπ12=4+81-cosπ12=4+82sin2π241-cosx=2sin2x2=4+4sinπ24=4+4cosπ2-π24sinx=cosπ2-x=4+4cos11π24=41+cos11π24=42cos211π48=22cos11π48

Step 2: Solve for the required value

4+8-32+768=a2cos11πb

22cos11π48=a2cos11πb

On comparing, we get, a=2 and b=48

b-a=48-2=46

We know that factors of 46 are 1,2,23,46

Hence the value of b-a is divisible by 1,2,23,46


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