How do you express $sin (\frac{π}{8}) \cdot cos ((\frac{π}{4})$ $sin(pi/ 8 ) * cos(( pi / 4 )$ without using products of trigonometric functions?

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How do you express $sin (\frac{π}{8}) \cdot cos ((\frac{π}{4})$ $sin(pi/ 8 ) * cos(( pi / 4 )$ without using products of trigonometric functions?

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Answer 1 · 2016-12-31T13:15:56

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

Explanation:

$P = sin (\frac{π}{8}) . cos (\frac{π}{4})$ $P = sin (pi/8).cos (pi/4)$
Trig table gives $cos (\frac{π}{4}) = \frac{\sqrt{2}}{2}$ $cos (pi/4) = sqrt2/2$ , then P can be expressed as:
$P = (\frac{\sqrt{2}}{2}) sin (\frac{π}{8})$ $P = (sqrt2/2)sin ((pi)/8)$ .
We can evaluate $sin (\frac{π}{8})$ $sin (pi/8)$ by applying the trig identity:
$2 {sin}^{2} a - 1 - cos 2 a$ $2sin^2 a - 1 - cos 2a$
$2 {sin}^{2} (\frac{π}{8}) = 1 - cos (\frac{π}{4}) = 1 - \frac{\sqrt{2}}{2} = \frac{2 - \sqrt{2}}{2}$ $2sin^2 (pi/8) = 1 - cos (pi/4) = 1 - sqrt2/2 = (2 - sqrt2)/2$
${sin}^{2} (\frac{π}{8}) = \frac{2 - \sqrt{2}}{4}$ $sin^2 (pi/8) = (2 - sqrt2)/4$
$sin (\frac{π}{8}) = \pm (\frac{\sqrt{2 - \sqrt{2}}}{2})$ $sin (pi/8) = +- (sqrt(2 - sqrt2)/2)$
Since $sin (\frac{π}{8})$ $sin (pi/8)$ is positive, take the positive value.
Finally:
$P = (\frac{\sqrt{2}}{2}) (\frac{\sqrt{2 - \sqrt{2}}}{2})$ $P = ((sqrt2)/2)(sqrt(2 - sqrt2)/2)$

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How do you express $sin (\frac{π}{8}) \cdot cos ((\frac{π}{4})$ $sin(pi/ 8 ) * cos(( pi / 4 )$ without using products of trigonometric functions?

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Explanation:

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How do you express $sin (\frac{π}{8}) \cdot cos ((\frac{π}{4})$ $sin(pi/ 8 ) * cos(( pi / 4 )$ without using products of trigonometric functions?