How do you express sin(pi/ 6 ) * cos(( ( pi) / 2 ) sin(π6)cos((π2) without using products of trigonometric functions?

1 Answer
Apr 13, 2016

sin(pi/6)*cos(pi/2)=0sin(π6)cos(π2)=0

Explanation:

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

sin(A-B)=sinAcosB-cosAsinBsin(AB)=sinAcosBcosAsinB,

Now adding two and moving RHS to LHS, we get

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

sinAcosB=1/2sin(A+B)+1/2sin(A-B)sinAcosB=12sin(A+B)+12sin(AB)

Hence, sin(pi/6)*cos(pi/2)=1/2sin(pi/6+pi/2)+1/2sin(pi/6-pi/2)sin(π6)cos(π2)=12sin(π6+π2)+12sin(π6π2)

= 1/2[sin(pi/2+pi/6)+sin-(pi/2-pi/6)]12[sin(π2+π6)+sin(π2π6)] or

= 1/2[cos(pi/6)-cos(pi/6)]=012[cos(π6)cos(π6)]=0