How do you express $sin (\frac{π}{6}) \cdot cos (\frac{3 π}{4})$ $sin(pi/ 6 ) * cos( ( 3 pi) / 4 )$ without using products of trigonometric functions?

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How do you express $sin (\frac{π}{6}) \cdot cos (\frac{3 π}{4})$ $sin(pi/ 6 ) * cos( ( 3 pi) / 4 )$ without using products of trigonometric functions?

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Answer 1 · 2016-02-16T08:07:02

Using Sum or Difference
$\frac{1}{2} \cdot sin (\frac{11 π}{12}) - \frac{1}{2} \cdot sin (\frac{7 π}{12})$ $1/2*sin ((11pi)/12)-1/2*sin ((7pi)/12)$

Explanation:

From "Sum and Difference Formulas"

$sin (x + y) = sin x cos y + cos x sin y$ $sin (x+y)=sin x cos y + cos x sin y$
$sin (x - y) = sin x cos y - cos x sin y$ $sin (x-y)=sin x cos y - cos x sin y$

Add the equations to obtain

$sin (x + y) + sin (x - y) = 2 \cdot sin x cos y$ $sin (x+y)+sin(x-y)=2*sin x cos y$

so that, after dividing both sides by 2

$sin x cos y = \frac{1}{2} \cdot sin (x + y) + \frac{1}{2} \cdot sin (x - y)$ $sin x cos y=1/2*sin(x+y)+1/2*sin(x-y)$

at this point , Let $x = \frac{π}{6}$ $x=pi/6$ and $y = \frac{3 π}{4}$ $y=(3pi)/4$

$sin x cos y = \frac{1}{2} \cdot sin (x + y) + \frac{1}{2} \cdot sin (x - y)$ $sin x cos y=1/2*sin(x+y)+1/2*sin(x-y)$

$sin (\frac{π}{6}) cos (\frac{3 π}{4}) = \frac{1}{2} \cdot sin (\frac{π}{6} + \frac{3 π}{4}) + \frac{1}{2} \cdot sin (\frac{π}{6} - \frac{3 π}{4})$ $sin (pi/6) cos ((3pi)/4)=1/2*sin(pi/6+(3pi)/4)+1/2*sin(pi/6-(3pi)/4)$

$sin (\frac{π}{6}) cos (\frac{3 π}{4}) = \frac{1}{2} \cdot sin (\frac{11 π}{12}) + \frac{1}{2} \cdot sin (\frac{- 7 π}{12})$ $sin (pi/6) cos ((3pi)/4)=1/2*sin((11pi)/12)+1/2*sin((-7pi)/12)$

but $sin (\frac{- 7 π}{12}) = - sin (\frac{7 π}{12})$ $sin((-7pi)/12)=-sin((7pi)/12)$

therefore

$sin (\frac{π}{6}) cos (\frac{3 π}{4}) = \frac{1}{2} \cdot sin (\frac{11 π}{12}) - \frac{1}{2} \cdot sin (\frac{7 π}{12})$ $sin (pi/6) cos ((3pi)/4)=1/2*sin((11pi)/12)-1/2*sin((7pi)/12)$

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How do you express $sin (\frac{π}{6}) \cdot cos (\frac{3 π}{4})$ $sin(pi/ 6 ) * cos( ( 3 pi) / 4 )$ without using products of trigonometric functions?

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Explanation:

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