How do you express sin(pi/ 4 ) * sin( ( pi) / 3 ) sin(π4)sin(π3) without using products of trigonometric functions?

1 Answer
May 20, 2016

sin(pi/4)sin(pi/3)=1/2[cos(pi/12)+cos((5pi)/12)]sin(π4)sin(π3)=12[cos(π12)+cos(5π12)]

Explanation:

We know that cos(A+B)=cosAcosB-sinAsinBcos(A+B)=cosAcosBsinAsinB ..............(1)

and cos(A-B)=cosAcosB+sinAsinBcos(AB)=cosAcosB+sinAsinB ..............(2)

Now subtracting (2) from (1)

2sinAsinB=cos(A-B)-cos(A+B)2sinAsinB=cos(AB)cos(A+B) or

sinAsinB=1/2cos(A-B)-1/2cos(A+B)sinAsinB=12cos(AB)12cos(A+B)

Hence, sin(pi/4)sin(pi/3)=1/2cos((pi/4)-(pi/3))-1/2cos((pi/4)+(pi/3))sin(π4)sin(π3)=12cos((π4)(π3))12cos((π4)+(π3))

= 1/2cos(-pi/12)-1/2cos((7pi)/12)12cos(π12)12cos(7π12),

but cos(-x)=cosxcos(x)=cosx and cos(pi-x)=-cosxcos(πx)=cosx,

hence above is equal to

1/2cos(pi/12)-1/2cos(pi-(5pi)/12)12cos(π12)12cos(π5π12)

= 1/2cos(pi/12)+1/2cos((5pi)/12)=1/2[cos(pi/12)+cos((5pi)/12)]12cos(π12)+12cos(5π12)=12[cos(π12)+cos(5π12)]