How do you express $sin(pi/ 4 ) * sin( ( 15 pi) / 8 )$ without using products of trigonometric functions?

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Answer 1 · 2016-03-10T10:26:19

$sin((5pi)/4)sin((15pi)/8)=1/2(cos((5pi)/8)+cos(pi/8))$

For this one uses the identity

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

Subtracting upper identity from lower, one gets

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

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

Hence $sin((5pi)/4)sin((15pi)/8)$

= $1/2(cos((15pi)/8-(5pi)/4)-cos((15pi)/8+(5pi)/4))$

= $1/2(cos((5pi)/8)-cos((25pi)/8))$

= $1/2(cos((5pi)/8)-cos(3pi+pi/8))$

= $1/2(cos((5pi)/8)+cos(pi/8))$

How do you express sin(pi/ 4 ) * sin( ( 15 pi) / 8 ) sin(pi/ 4 ) * sin( ( 15 pi) / 8 ) without using products of trigonometric functions?