Use identity to simplify the expression? 2sin(π/9-π/2)cos(π/2-π/9)

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Answer 1 · 2018-04-02T04:22:59

#2sin(pi/9-pi/2)cos(pi/2-pi/9)=-sin((2pi)/9)#

We use #sin(-A)=-sinA#, #sin(pi/2-A)=cosA#, #cos(pi/2-A)=sinA# and #2sinAcosA=sin2A#

#sin(pi/9-pi/2)=sin(-(pi/2-pi/9))#

= #-sin(pi/2-pi/9)=-cos(pi/9)#

Similarly #cos(pi/2-pi/9)=sin(pi/9)#

Hence #2sin(pi/9-pi/2)cos(pi/2-pi/9)#

= #-2cos(pi/9)*sin(pi/9)#

= #-2sin(pi/9)cos(pi/9)#

= #-sin((2pi)/9)#