How do you express $sin (\frac{π}{6}) \cdot cos ((\frac{π}{2})$ $sin(pi/ 6 ) * cos(( ( pi) / 2 )$ without using products of trigonometric functions?

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How do you express $sin (\frac{π}{6}) \cdot cos ((\frac{π}{2})$ $sin(pi/ 6 ) * cos(( ( pi) / 2 )$ without using products of trigonometric functions?

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Answer 1 · 2016-04-13T12:54:23

$sin (\frac{π}{6}) \cdot cos (\frac{π}{2}) = 0$ $sin(pi/6)*cos(pi/2)=0$

Explanation:

As $sin (A + B) = sin A cos B + cos A sin B$ $sin(A+B)=sinAcosB+cosAsinB$ and

$sin (A - B) = sin A cos B - cos A sin B$ $sin(A-B)=sinAcosB-cosAsinB$ ,

Now adding two and moving RHS to LHS, we get

$2 sin A cos B = sin (A + B) + sin (A - B)$ $2sinAcosB=sin(A+B)+sin(A-B)$ or

$sin A cos B = \frac{1}{2} sin (A + B) + \frac{1}{2} sin (A - B)$ $sinAcosB=1/2sin(A+B)+1/2sin(A-B)$

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

= $\frac{1}{2} [sin (\frac{π}{2} + \frac{π}{6}) + sin - (\frac{π}{2} - \frac{π}{6})]$ $1/2[sin(pi/2+pi/6)+sin-(pi/2-pi/6)]$ or

= $\frac{1}{2} [cos (\frac{π}{6}) - cos (\frac{π}{6})] = 0$ $1/2[cos(pi/6)-cos(pi/6)]=0$

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How do you express $sin (\frac{π}{6}) \cdot cos ((\frac{π}{2})$ $sin(pi/ 6 ) * cos(( ( pi) / 2 )$ without using products of trigonometric functions?

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Explanation:

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