Q.

If 2cosA=cosB+cos3Band2sinA=sinBsin3B, then sin(AB)=

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a

±14

b

±13

c

±12

d

±1

answer is C.

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Detailed Solution

2cosA=cosB+cos3B(i)

2sinA=sinBsin3B(ii)

2sinAcosB2cosAsinB=(sinBsin3B)cosB(cosB+cos3B)sinB

2sin(AB)=sinBcosB

sin(AB)=122sin2B

Now squaring and adding equations (i) and (ii), we get

2=cos2B+sin2B+cos6B+sin6B+2(cos4Bsin4B)

=1+(13cos2Bsin2B)+2cos2B

2=234sin22B+2cos2B

3cos22B+8cos2B3=0

cos2B=13sin2B=±223

sin(AB)=±13.

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