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

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

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Answer 1 · 2017-02-17T01:04:29

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

Explanation:

$P = sin (\frac{π}{6}) . cos (\frac{π}{8})$ $P = sin (pi/6).cos (pi/8)$
Trig table gives --> sin (pi/6) = 1/2
$P = (\frac{1}{2}) cos (\frac{π}{8})$ $P = (1/2)cos (pi/8)$ .
We can evaluate $cos (\frac{π}{8})$ $cos (pi/8)$ by using trig identity:
$2 {cos}^{2} a = 1 + cos 2 a$ $2cos^2 a = 1 + cos 2a$
In this case:
$2 {cos}^{2} (\frac{π}{8}) = 1 + cos (\frac{π}{4}) = 1 + \frac{\sqrt{2}}{2} = \frac{2 + \sqrt{2}}{2}$ $2cos^2 (pi/8) = 1 + cos (pi/4) = 1 + sqrt2/2 = (2 + sqrt2)/2$
${cos}^{2} (\frac{π}{8}) = \frac{2 + \sqrt{2}}{4}$ $cos^2 (pi/8) = (2 + sqrt2)/4$
$cos (\frac{π}{8}) = \pm \frac{\sqrt{2 + \sqrt{2}}}{2}$ $cos (pi/8) = +- sqrt(2 + sqrt2)/2$ .
Since $cos (\frac{π}{8})$ $cos (pi/8)$ is positive, take the positive answer.
Finally,
$P = (\frac{1}{2}) cos (\frac{π}{8}) = \frac{\sqrt{2 + \sqrt{2}}}{4}$ $P = (1/2)cos (pi/8) = sqrt(2 + sqrt2)/4$

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

1 Answer

Explanation:

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