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

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Answer 1 · 2016-05-20T05:54:08

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

We know that #cos(A+B)=cosAcosB-sinAsinB# ..............(1)

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

Now subtracting (2) from (1)

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

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

Hence, #sin(pi/4)sin(pi/3)=1/2cos((pi/4)-(pi/3))-1/2cos((pi/4)+(pi/3))#

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

but #cos(-x)=cosx# and #cos(pi-x)=-cosx#,

hence above is equal to

#1/2cos(pi/12)-1/2cos(pi-(5pi)/12)#

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