How do you express $sin (\frac{π}{12}) \cdot cos (\frac{π}{4})$ $sin(pi/12) * cos(pi/4 )$ without products of trigonometric functions?

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How do you express $sin (\frac{π}{12}) \cdot cos (\frac{π}{4})$ $sin(pi/12) * cos(pi/4 )$ without products of trigonometric functions?

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Answer 1 · 2016-06-19T02:03:40

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

Explanation:

$P = sin (\frac{π}{12}) . cos (\frac{π}{4})$ $P = sin (pi/12).cos (pi/4)$
Trig table --> $cos (\frac{π}{4}) = \frac{\sqrt{2}}{2}$ $cos (pi/4) = sqrt2/2$ .
Then P can be expressed as: $P = (\frac{\sqrt{2}}{2}) sin (\frac{π}{12})$ $P = (sqrt2/2)sin (pi/12)$
We can evaluate $sin (\frac{π}{12})$ $sin (pi/12)$ , using the trig identity:
$cos 2 a = 1 - 2 {sin}^{2} a$ $cos 2a = 1 - 2sin^2 a$
$cos (\frac{π}{6}) = \frac{\sqrt{3}}{2} = 1 - 2 {sin}^{2} (\frac{π}{12})$ $cos (pi/6) = sqrt3/2 = 1 - 2sin^2 (pi/12)$
$2 {sin}^{2} (\frac{π}{12}) = 1 - \frac{\sqrt{3}}{2} = \frac{2 - \sqrt{3}}{2}$ $2sin^2 (pi/12) = 1 - sqrt3/2 = (2 - sqrt3)/2$
${sin}^{2} (\frac{π}{12}) = \frac{2 - \sqrt{3}}{4}$ $sin^2 (pi/12) = (2 - sqrt3)/4$
$sin (\frac{π}{12}) = \pm \frac{\sqrt{2 - \sqrt{3}}}{2}$ $sin (pi/12) = +- sqrt(2 - sqrt3)/2$
Take the positive value because $\frac{sin π}{12}$ $sin pi/12$ is positive
Finally,
$P = (\frac{\sqrt{2}}{4}) . \sqrt{2 - \sqrt{3}}$ $P = (sqrt2/4).sqrt(2 - sqrt3)$

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How do you express $sin (\frac{π}{12}) \cdot cos (\frac{π}{4})$ $sin(pi/12) * cos(pi/4 )$ without products of trigonometric functions?

1 Answer

Explanation:

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How do you express $sin (\frac{π}{12}) \cdot cos (\frac{π}{4})$ $sin(pi/12) * cos(pi/4 )$ without products of trigonometric functions?