How do you express #sin(pi/ 6 ) * cos(( ( pi) / 2 ) # without using products of trigonometric functions?

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Answer 1 · 2016-04-13T12:54:23

#sin(pi/6)*cos(pi/2)=0#

As #sin(A+B)=sinAcosB+cosAsinB# and

#sin(A-B)=sinAcosB-cosAsinB#,

Now adding two and moving RHS to LHS, we get

#2sinAcosB=sin(A+B)+sin(A-B)# or

#sinAcosB=1/2sin(A+B)+1/2sin(A-B)#

Hence, #sin(pi/6)*cos(pi/2)=1/2sin(pi/6+pi/2)+1/2sin(pi/6-pi/2)#

= #1/2[sin(pi/2+pi/6)+sin-(pi/2-pi/6)]# or

= #1/2[cos(pi/6)-cos(pi/6)]=0#