How do you express $sin (\frac{π}{4}) \cdot sin (\frac{7 π}{8})$ $sin(pi/ 4 ) * sin( ( 7 pi) / 8 )$ without using products of trigonometric functions?

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How do you express $sin (\frac{π}{4}) \cdot sin (\frac{7 π}{8})$ $sin(pi/ 4 ) * sin( ( 7 pi) / 8 )$ without using products of trigonometric functions?

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Answer 1 · 2016-04-07T14:02:47

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

Explanation:

$P = sin (\frac{π}{4}) . sin (\frac{7 π}{8})$ $P = sin (pi/4).sin ((7pi)/8)$
Trig table --> sin (pi/8) = sqrt2/2
Trig unit circle -->
$sin (\frac{7 π}{8}) = sin (\frac{π}{8})$ $sin ((7pi)/8) = sin (pi/8)$
Evaluate sin (pi/8) by using the thig identity:
$cos 2 a = 1 - 2 {sin}^{2} a$ $cos 2a = 1 - 2sin^2 a$
$cos (\frac{π}{4}) = \frac{\sqrt{2}}{2} = 1 - 2 {sin}^{2} (\frac{π}{8})$ $cos (pi/4) = sqrt2/2 = 1 - 2sin^2 (pi/8)$
$2 {sin}^{2} (\frac{π}{8}) = 1 - \frac{\sqrt{2}}{2} = \frac{2 - \sqrt{2}}{2}$ $2sin^2 (pi/8) = 1 - sqrt2/2 = (2 - sqrt2)/2$
${sin}^{2} (\frac{π}{8}) = \frac{2 - \sqrt{2}}{4}$ $sin^2 (pi/8) = (2 - sqrt2)/4$
$sin (\frac{π}{8}) = \frac{\sqrt{2 - \sqrt{2}}}{2}$ $sin (pi/8) = sqrt(2 - sqrt2)/2$ --> $sin (\frac{π}{8})$ $sin (pi/8)$ is positive.
Finally:
$P = (\sqrt{2}) (\frac{\sqrt{2 - \sqrt{2}}}{4})$ $P = (sqrt2)((sqrt(2 - sqrt2)]/4)$

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How do you express $sin (\frac{π}{4}) \cdot sin (\frac{7 π}{8})$ $sin(pi/ 4 ) * sin( ( 7 pi) / 8 )$ without using products of trigonometric functions?

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Explanation:

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How do you express $sin (\frac{π}{4}) \cdot sin (\frac{7 π}{8})$ $sin(pi/ 4 ) * sin( ( 7 pi) / 8 )$ without using products of trigonometric functions?