How do you express $sin (\frac{π}{6}) \cdot cos (\frac{5 π}{12})$ $sin(pi/ 6 ) * cos( ( 5 pi) / 12 )$ without using products of trigonometric functions?

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How do you express $sin (\frac{π}{6}) \cdot cos (\frac{5 π}{12})$ $sin(pi/ 6 ) * cos( ( 5 pi) / 12 )$ without using products of trigonometric functions?

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Answer 1 · 2016-01-25T00:43:41

$(\frac{1}{4}) (\sqrt{2 - \sqrt{3}})$ $(1/4)(sqrt(2 - sqrt3))$

Explanation:

a. Find sin (pi/6).
Trig Table of special arcs --> $\frac{sin π}{6} = \frac{1}{2}$ $sin pi/6 = 1/2$
b. Find cos (5pi)/2
$\frac{cos (10 π)}{12} = cos (5 \frac{π}{6}) = - \frac{\sqrt{3}}{2}$ $cos (10pi)/12 = cos (5pi/6) = -sqrt3/2$
Use trig identity: $cos 2 x = 2 {cos}^{2} x - 1$ $cos 2x = 2cos^2 x - 1$
$cos (5 \frac{π}{6}) = - \frac{\sqrt{3}}{2} = 2 {cos}^{2} (5 \frac{π}{12}) - 1$ $cos (5pi/6) = -sqrt3/2 = 2cos^2 (5pi/12) - 1$
$2 {cos}^{2} (5 \frac{π}{12}) = 1 - \frac{\sqrt{3}}{2} = \frac{2 - \sqrt{3}}{2}$ $2cos^2 (5pi/12) = 1 - sqrt3/2 = (2 - sqrt3)/2$
$cos (5 \frac{π}{12}) = \frac{\sqrt{2 - \sqrt{3}}}{2}$ $cos (5pi/12) = sqrt(2 - sqrt3)/2$
Finally,
$\frac{sin π}{6} . cos (\frac{5 π}{12}) = (\frac{1}{4}) (\sqrt{2 - \sqrt{3}})$ $sin pi/6.cos ((5pi)/12) = (1/4)(sqrt(2 - sqrt3))$

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How do you express $sin (\frac{π}{6}) \cdot cos (\frac{5 π}{12})$ $sin(pi/ 6 ) * cos( ( 5 pi) / 12 )$ without using products of trigonometric functions?

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How do you express $sin (\frac{π}{6}) \cdot cos (\frac{5 π}{12})$ $sin(pi/ 6 ) * cos( ( 5 pi) / 12 )$ without using products of trigonometric functions?