How do you express $sin (\frac{π}{4}) \cdot sin (\frac{5 π}{12})$ $sin(pi/ 4 ) * sin( ( 5 pi) / 12 )$ without using products of trigonometric functions?

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How do you express $sin (\frac{π}{4}) \cdot sin (\frac{5 π}{12})$ $sin(pi/ 4 ) * sin( ( 5 pi) / 12 )$ without using products of trigonometric functions?

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Answer 1 · 2016-06-24T03:24:50

$(\frac{\sqrt{2}}{4}) \sqrt{2 + \sqrt{3}}$ $(sqrt2/4)sqrt(2 + sqrt3)$

Explanation:

Product $P = sin (\frac{π}{4}) . sin (\frac{5 π}{12})$ $P = sin (pi/4).sin ((5pi)/12)$
Trig table -->
$sin (\frac{π}{4}) = \frac{\sqrt{2}}{2}$ $sin (pi/4) = sqrt2/2$
P can be expressed as:
$(\frac{\sqrt{2}}{2}) . sin (\frac{5 π}{12}) .$ $(sqrt2/2).sin ((5pi)/12).$
We can evaluate $sin (\frac{5 π}{12})$ $sin ((5pi)/12)$ by using the trig identity:
$cos 2 a = 1 - 2 {sin}^{2} a$ $cos 2a = 1 - 2sin^2 a$
$cos (\frac{10 π}{12}) = cos (\frac{5 π}{6}) = - \frac{\sqrt{3}}{2} = 1 - 2 {sin}^{2} (\frac{5 π}{12})$ $cos ((10pi)/12) = cos ((5pi)/6) = -sqrt3/2 = 1 - 2sin^2 ((5pi)/12)$
$2 {sin}^{2} (\frac{5 π}{12} = 1 + \frac{\sqrt{3}}{2} = \frac{2 + \sqrt{3}}{2}$ $2sin^2 ((5pi)/12 = 1 + sqrt3/2 = (2 + sqrt3)/2$
${sin}^{2} (\frac{5 π}{12}) = \frac{2 + \sqrt{3}}{4}$ $sin^2 ((5pi)/12) = (2 + sqrt3)/4$
$sin (\frac{5 π}{12}) = \frac{\sqrt{2 + \sqrt{3}}}{2}$ $sin ((5pi)/12) = sqrt(2 + sqrt3)/2$ (note: sin ((5pi)/12) is positive)
Finally,
$P = (\frac{\sqrt{2}}{4}) (\sqrt{2 + \sqrt{3}})$ $P = (sqrt2/4)(sqrt(2 + sqrt3))$

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How do you express $sin (\frac{π}{4}) \cdot sin (\frac{5 π}{12})$ $sin(pi/ 4 ) * sin( ( 5 pi) / 12 )$ without using products of trigonometric functions?

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Explanation:

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How do you express $sin (\frac{π}{4}) \cdot sin (\frac{5 π}{12})$ $sin(pi/ 4 ) * sin( ( 5 pi) / 12 )$ without using products of trigonometric functions?