How do you express $sin (\frac{π}{8}) \cdot cos (\frac{7 π}{6})$ $sin(pi/ 8 ) * cos(( 7 pi)/6 )$ without using products of trigonometric functions?

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How do you express $sin (\frac{π}{8}) \cdot cos (\frac{7 π}{6})$ $sin(pi/ 8 ) * cos(( 7 pi)/6 )$ without using products of trigonometric functions?

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Answer 1 · 2016-01-24T23:14:08

$- (\sqrt{3}) (\frac{\sqrt{2 - \sqrt{2}}}{4})$ $-(sqrt3)(sqrt(2 - sqrt2)/4)$

Explanation:

First, find $sin (\frac{π}{8})$ $sin (pi/8)$ and $cos (\frac{7 π}{6})$ $cos ((7pi)/6)$ separately.
Call $sin (\frac{π}{8}) = sin t$ $sin (pi/8) = sin t$
Use the trig identity: $cos (2 t) = 1 - 2 {sin}^{2} t$ $cos (2t) = 1 - 2sin^2 t$
$cos (\frac{π}{4}) = \frac{\sqrt{2}}{2} = 1 - {sin}^{2} t$ $cos (pi/4) = sqrt2/2 = 1 - sin^2 t$
$2 {sin}^{2} t = 1 - \sqrt{2} = \frac{2 - \sqrt{2}}{2}$ $2sin^2 t = 1 - sqrt2 = (2 - sqrt2)/2$
${sin}^{2} t = \frac{2 - \sqrt{2}}{4}$ $sin^2 t = (2 - sqrt2)/4$
$sin (\frac{π}{8}) = sin t = \frac{\sqrt{2 - \sqrt{2}}}{2}$ $sin (pi/8) = sin t = (sqrt(2 - sqrt2))/2$

$cos (\frac{7 π}{6}) = cos (\frac{π}{6} + π) = - cos (\frac{π}{6}) = - \frac{\sqrt{3}}{2}$ $cos ((7pi)/6) = cos (pi/6 + pi) = - cos (pi/6) = - sqrt3/2$

Finally, $sin (\frac{π}{8}) . cos (\frac{7 π}{6}) = - (\sqrt{3}) (\frac{\sqrt{2 - \sqrt{2}}}{4})$ $sin (pi/8).cos ((7pi)/6) = - (sqrt3)(sqrt(2 - sqrt2)/4)$

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How do you express $sin (\frac{π}{8}) \cdot cos (\frac{7 π}{6})$ $sin(pi/ 8 ) * cos(( 7 pi)/6 )$ without using products of trigonometric functions?

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How do you express $sin (\frac{π}{8}) \cdot cos (\frac{7 π}{6})$ $sin(pi/ 8 ) * cos(( 7 pi)/6 )$ without using products of trigonometric functions?