How do you express $sin (\frac{π}{6}) \cdot cos (\frac{3 π}{8})$ $sin(pi/ 6 ) * cos( (3 pi) / 8 )$ without using products of trigonometric functions?

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How do you express $sin (\frac{π}{6}) \cdot cos (\frac{3 π}{8})$ $sin(pi/ 6 ) * cos( (3 pi) / 8 )$ without using products of trigonometric functions?

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Answer 1 · 2016-03-17T00:55:06

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

Explanation:

$P = sin (\frac{π}{6}) . cos (\frac{3 π}{8}) .$ $P = sin (pi/6).cos ((3pi)/8).$
Trig table --> $\frac{sin π}{6} = \frac{1}{2}$ $sin pi/6 = 1/2$
Find $cos (\frac{3 π}{8})$ $cos ((3pi)/8)$ by the identity: $cos 2 a = 2 {cos}^{2} a - 1$ $cos 2a = 2cos^2 a - 1$
$cos (\frac{6 π}{8}) = 2 {cos}^{2} (\frac{3 π}{8}) - 1 .$ $cos ((6pi)/8) = 2cos^2 ((3pi)/8) - 1.$
$cos (\frac{3 π}{4}) = - \frac{\sqrt{2}}{2} = 2 {cos}^{2} (\frac{3 π}{8}) - 1$ $cos ((3pi)/4) = -sqrt2/2 = 2cos^2 ((3pi)/8) - 1$
$2 {cos}^{2} (\frac{3 π}{8}) = 1 - \frac{\sqrt{2}}{2} = \frac{2 - \sqrt{2}}{2}$ $2cos^2 ((3pi)/8) = 1 - sqrt2/2 = (2 - sqrt2)/2$
${cos}^{2} (\frac{3 π}{8}) = \frac{2 - \sqrt{2}}{4}$ $cos^2 ((3pi)/8) = (2 - sqrt2)/4$
$cos (\frac{3 π}{8}) = \frac{\sqrt{2 - \sqrt{2}}}{2}$ $cos ((3pi)/8) = sqrt(2 - sqrt2)/2$ (since $cos (\frac{3 π}{8})$ $cos ((3pi)/8)$ is positive.
Finally,

$P = (\frac{1}{2}) (\frac{\sqrt{2 - \sqrt{2}}}{2}) = (\frac{\sqrt{2 - \sqrt{2}}}{4})$ $P = (1/2)(sqrt(2 - sqrt2)/2) = (sqrt(2 - sqrt2)/4)$

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How do you express $sin (\frac{π}{6}) \cdot cos (\frac{3 π}{8})$ $sin(pi/ 6 ) * cos( (3 pi) / 8 )$ without using products of trigonometric functions?

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Explanation:

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How do you express $sin (\frac{π}{6}) \cdot cos (\frac{3 π}{8})$ $sin(pi/ 6 ) * cos( (3 pi) / 8 )$ without using products of trigonometric functions?