How do you express $sin (\frac{π}{8}) \cdot cos ((\frac{π}{3})$ $sin(pi/ 8 ) * cos(( ( pi) / 3 )$ without using products of trigonometric functions?

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Answer 1 · 2016-04-11T04:16:34

$\frac{sin (\frac{11 π}{24}) - sin (\frac{5 π}{24})}{2}$ $(sin((11pi)/24)-sin((5pi)/24))/2$

Use
$sin 2 A - sin 2 B = 2 cos (A + B) sin (A - B)$ $sin 2A - sin 2B = 2 cos(A+B)sin(A-B)$ .

Here, $A + B = \frac{π}{3} and A - B = \frac{π}{8}$ $A+B=pi/3 and A-B=pi/8$ .

So, $2 A = 11 \frac{π}{24} and 2 B = 5 \frac{π}{24}$ $2A=11pi/24 and 2B=5pi/24$ .

Answer: $\frac{sin (\frac{11 π}{24}) - sin (\frac{5 π}{24})}{2}$ $(sin((11pi)/24)-sin((5pi)/24))/2$

How do you express sin(pi/ 8 ) * cos(( ( pi) / 3 ) sin(π8)⋅cos((π3)sin(pi/ 8 ) * cos(( ( pi) / 3 ) without using products of trigonometric functions?