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

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

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Answer 1 · 2016-04-04T04:40:26

1/2

Explanation:

$P = sin (\frac{π}{4}) cos (\frac{5 π}{4})$ $P = sin (pi/4)cos ((5pi)/4)$
Trig table --> $sin (\frac{π}{4}) = \frac{\sqrt{2}}{2} .$ $sin (pi/4) = sqrt2/2.$
$cos (\frac{5 π}{4} = cos (\frac{π}{4} + π) = - cos (\frac{π}{4}) = - \frac{\sqrt{2}}{2}$ $cos ((5pi)/4 = cos (pi/4 + pi) = -cos (pi/4) = - sqrt2/2$
$P = (- \frac{\sqrt{2}}{2}) (\frac{\sqrt{2}}{2}) = - \frac{1}{2}$ $P = (-sqrt2/2)(sqrt2/2) = - 1/2$

Answer 2 · 2016-04-04T17:10:36

$sin (\frac{π}{4}) cos (\frac{5 π}{4}) = \frac{1}{2} sin (\frac{3 π}{2}) - \frac{1}{2} sin (π) = - \frac{1}{2}$ $sin (pi/4) cos ((5pi)/4)=1/2 sin ((3pi)/2) -1/2 sin (pi) =-1/2$

Explanation:

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

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

$A = \frac{π}{4}, B = \frac{5 π}{4}$ $A=pi/4, B= (5pi)/4$

$sin (\frac{π}{4}) cos (\frac{5 π}{4}) = \frac{1}{2} (sin (\frac{π}{4} + \frac{5 π}{4}) + sin (\frac{π}{4} - \frac{5 π}{4}))$ $sin (pi/4) cos ((5pi)/4)=1/2 (sin (pi/4+(5pi)/4) + sin (pi/4-(5pi)/4))$

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

$= \frac{1}{2} (sin (\frac{3 π}{2}) + sin (- π))$ $=1/2 (sin ((3pi)/2) + sin (-pi))$

$= \frac{1}{2} (sin (\frac{3 π}{2}) - sin (π))$ $=1/2 (sin ((3pi)/2) -sin (pi))$

$= \frac{1}{2} sin (\frac{3 π}{2}) - \frac{1}{2} sin (π)$ $=1/2 sin ((3pi)/2) -1/2 sin (pi)$

$= \frac{1}{2} \cdot (- 1) - \frac{1}{2} (0)$ $=1/2 *(-1)-1/2 (0)$

$= - \frac{1}{2}$ $= -1/2$

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

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