How do you express $sin (\frac{π}{8}) \cdot cos ((\frac{2 π}{3})$ $sin(pi/ 8 ) * cos(( ( 2 pi) / 3 )$ without using products of trigonometric functions?

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How do you express $sin (\frac{π}{8}) \cdot cos ((\frac{2 π}{3})$ $sin(pi/ 8 ) * cos(( ( 2 pi) / 3 )$ without using products of trigonometric functions?

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Answer 1 · 2016-10-31T12:54:08

$\frac{\sqrt{2 - \sqrt{2}}}{2}$ $sqrt(2 - sqrt2)/2$

Explanation:

$P = sin (\frac{π}{8}) . cos (\frac{2 π}{3})$ $P = sin (pi/8).cos ((2pi)/3)$
Trig table --> $cos (\frac{2 π}{3}) = - \frac{1}{2}$ $cos ((2pi)/3) = -1/2$ , then P can be expressed as:
$P = - (\frac{1}{2}) sin (\frac{π}{8})$ $P = - (1/2)sin (pi/8)$
We can evaluate the value of sin (pi)/8 by the trig identity:
$2 {sin}^{2} (\frac{π}{8}) = 1 - cos (\frac{2 π}{8}) = 1 - cos (\frac{π}{4}) = 1 - \frac{\sqrt{2}}{2}$ $2sin^2 (pi/8) = 1 - cos ((2pi)/8) = 1 - cos (pi/4) = 1 - sqrt2/2$
${sin}^{2} (\frac{π}{8}) = \frac{2 - \sqrt{2}}{4}$ $sin^2 (pi/8) = (2 - sqrt2)/4$
$sin (\frac{π}{8}) = \frac{\sqrt{2 - \sqrt{2}}}{2}$ $sin (pi/8) = sqrt(2 - sqrt2)/2$
Only the positive value is accepted because $sin (\frac{π}{8})$ $sin (pi/8)$ is positive
Finally,
$P = - (\frac{1}{2}) sin (\frac{π}{8}) = - \frac{\sqrt{2 - \sqrt{2}}}{4}$ $P = - (1/2)sin (pi/8) = - (sqrt(2 - sqrt2))/4$

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How do you express $sin (\frac{π}{8}) \cdot cos ((\frac{2 π}{3})$ $sin(pi/ 8 ) * cos(( ( 2 pi) / 3 )$ without using products of trigonometric functions?

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Explanation:

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How do you express $sin (\frac{π}{8}) \cdot cos ((\frac{2 π}{3})$ $sin(pi/ 8 ) * cos(( ( 2 pi) / 3 )$ without using products of trigonometric functions?