How do you express sin(π6)cos((π2) without using products of trigonometric functions?

1 Answer
Apr 13, 2016

sin(π6)cos(π2)=0

Explanation:

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

sin(AB)=sinAcosBcosAsinB,

Now adding two and moving RHS to LHS, we get

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

sinAcosB=12sin(A+B)+12sin(AB)

Hence, sin(π6)cos(π2)=12sin(π6+π2)+12sin(π6π2)

= 12[sin(π2+π6)+sin(π2π6)] or

= 12[cos(π6)cos(π6)]=0