How do you express $sin (\frac{π}{6}) \cdot cos (\frac{15 π}{12})$ $sin(pi/ 6 ) * cos( ( 15 pi) / 12 )$ without using products of trigonometric functions?

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How do you express $sin (\frac{π}{6}) \cdot cos (\frac{15 π}{12})$ $sin(pi/ 6 ) * cos( ( 15 pi) / 12 )$ without using products of trigonometric functions?

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Answer 1 · 2016-03-26T21:52:58

$- \frac{\sqrt{2}}{4}$ $-sqrt2/4$

Explanation:

$P = sin (\frac{π}{6}) cos (\frac{15 π}{12}) .$ $P = sin (pi/6)cos ((15pi)/12).$
Trig table --> $\frac{sin π}{6} = \frac{1}{2}$ $sin pi/6 = 1/2$
$cos (\frac{15 π}{12}) = cos (\frac{- 9 π}{12} + 2 π) = cos (\frac{- 3 π}{4}) =$ $cos ((15pi)/12) = cos ((-9pi)/12 + 2pi) = cos ((-3pi)/4) =$
$= cos (\frac{3 π}{4}) = - \frac{\sqrt{2}}{2} .$ $= cos ((3pi)/4) = -sqrt2/2.$
$P = (\frac{1}{2}) (- \frac{\sqrt{2}}{2}) = - \frac{\sqrt{2}}{4}$ $P = (1/2)(-sqrt2/2) = -sqrt2/4$

Answer 2 · 2016-04-04T19:42:38

$sin (\frac{π}{6}) cos (\frac{15 π}{12}) = \frac{1}{2} sin (\frac{17 π}{12}) - \frac{1}{2} sin (\frac{13 π}{12})$ $sin (pi/6) cos ((15pi)/12)=1/2 sin ((17pi)/12) - 1/2sin ((13pi)/12)$

Explanation:

$2 sin A cos B = sin (A + B) + sin (A - B)$ $2 sin A cos B=sin (A+B) + sin (A-B)$
$sin A cos B = \frac{1}{2} (sin (A + B) + sin (A - B))$ $sin A cos B=1/2 (sin (A+B) + sin (A-B))$
$A = \frac{π}{6}, B = \frac{15 π}{12}$ $A=pi/6, B=(15pi)/12$
$sin (\frac{π}{6}) cos (\frac{15 π}{12}) = \frac{1}{2} (sin (\frac{π}{6} + \frac{15 π}{12}) + sin (\frac{π}{6} - \frac{15 π}{12}))$ $sin (pi/6) cos ((15pi)/12)=1/2 (sin (pi/6+(15pi)/12) + sin (pi/6-(15pi)/12))$
$= \frac{1}{2} (sin (\frac{17 π}{12}) + sin (- \frac{13 π}{12}))$ $=1/2 (sin ((17pi)/12) + sin (-(13pi)/12))$
$= \frac{1}{2} sin (\frac{17 π}{12}) - \frac{1}{2} sin (\frac{13 π}{12})$ $=1/2 sin ((17pi)/12) - 1/2sin ((13pi)/12)$

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How do you express $sin (\frac{π}{6}) \cdot cos (\frac{15 π}{12})$ $sin(pi/ 6 ) * cos( ( 15 pi) / 12 )$ without using products of trigonometric functions?

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