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Question

If A,b,c,d are the smallest positive angles in ascending order of magnitude their sines equal to the positive quantity K , then 4sinA2+3sinB2+2sinC2+sinD2=21+k.

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Solution

sinA=sinB=sinC=sinD=K

B=πA=B2=π2A2

C=2π+A=C2=π+A2

D=3πA=D2=3π2A2

Now 4sinA2+3sinB2+2sinC2+sinD2

4sinA2+3cosA2+2sinA2cosA2

2sinA2+2cosA2

2(sinA2+cosA2)

2(sinA2+cosA2)2

2sin2A2+cos2A2+sinA

21+sinA

21+K


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