How do you express $cos (\frac{5 π}{6}) \cdot cos (\frac{5 π}{12})$ $cos( (5 pi)/6 ) * cos (( 5 pi) /12 )$ without using products of trigonometric functions?

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How do you express $cos (\frac{5 π}{6}) \cdot cos (\frac{5 π}{12})$ $cos( (5 pi)/6 ) * cos (( 5 pi) /12 )$ without using products of trigonometric functions?

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Answer 1 · 2016-03-18T00:36:16

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

Explanation:

$P = cos (\frac{5 π}{6}) . cos (\frac{5 π}{12})$ $P = cos ((5pi)/6).cos ((5pi)/12)$
Trig table --> $cos (\frac{5 π}{6}) = - \frac{\sqrt{3}}{2}$ $cos ((5pi)/6) = -sqrt3/2$
Find $cos (\frac{5 π}{12})$ $cos ((5pi)/12)$ by the identity: $cos 2 a = 2 {cos}^{2} a - 1$ $cos 2a = 2cos^2a - 1$
$cos (\frac{5 π}{6}) = - \frac{\sqrt{3}}{2} = 2 {cos}^{2} (\frac{5 π}{12}) - 1$ $cos ((5pi)/6) = -sqrt3/2 = 2cos^2 ((5pi)/12) - 1$
$2 {cos}^{2} (\frac{5 π}{12}) = 1 - \frac{\sqrt{3}}{2} = \frac{2 - \sqrt{3}}{2}$ $2cos^2 ((5pi)/12) = 1 - sqrt3/2 = (2 - sqrt3)/2$
${cos}^{2} (\frac{5 π}{12}) = \frac{2 - \sqrt{3}}{4}$ $cos^2 ((5pi)/12) = (2 - sqrt3)/4$
$cos (\frac{5 π}{12}) = \frac{\sqrt{2 - \sqrt{3}}}{2}$ $cos ((5pi)/12) = sqrt(2 - sqrt3)/2$ --> $cos (\frac{5 π}{12})$ $cos ((5pi)/12)$ is positive.
Finally,
$P = (- \frac{\sqrt{3}}{2}) (\frac{\sqrt{2 - \sqrt{3}}}{2}) = - (\frac{\sqrt{3}}{4}) (\sqrt{2 - \sqrt{3}})$ $P = (-sqrt3/2)(sqrt(2 - sqrt3)/2) = - (sqrt3/4)(sqrt(2 - sqrt3))$

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How do you express $cos (\frac{5 π}{6}) \cdot cos (\frac{5 π}{12})$ $cos( (5 pi)/6 ) * cos (( 5 pi) /12 )$ without using products of trigonometric functions?

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Explanation:

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How do you express cos(5π6)⋅cos(5π12)cos( (5 pi)/6 ) * cos (( 5 pi) /12 ) without using products of trigonometric functions?

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How do you express $cos (\frac{5 π}{6}) \cdot cos (\frac{5 π}{12})$ $cos( (5 pi)/6 ) * cos (( 5 pi) /12 )$ without using products of trigonometric functions?