How do you express $cos (\frac{π}{3}) \cdot sin (\frac{π}{8})$ $cos(pi/ 3 ) * sin( ( pi) / 8 )$ without using products of trigonometric functions?

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How do you express $cos (\frac{π}{3}) \cdot sin (\frac{π}{8})$ $cos(pi/ 3 ) * sin( ( pi) / 8 )$ without using products of trigonometric functions?

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Answer 1 · 2016-07-11T17:54:25

$P = (\frac{1}{4}) \sqrt{2 - \sqrt{2}}$ $P = (1/4)sqrt(2 - sqrt2)$

Explanation:

$P = cos (\frac{π}{3}) . sin (\frac{π}{8})$ $P = cos (pi/3).sin (pi/8)$
Trig table --> cos (pi/3) = 1/2.
There for, P can be expressed as $P = (\frac{1}{2}) sin (\frac{π}{8})$ $P = (1/2)sin (pi/8)$
We can evaluate sin (pi/8) by using trig identity:
cos 2a = 1 - 2sin^2 a
$\frac{cos (2 π)}{8} = cos (\frac{π}{4}) = \frac{\sqrt{2}}{2} = 1 - 2 {sin}^{2} (\frac{π}{8})$ $cos (2pi)/8 = cos (pi/4) = sqrt2/2 = 1 - 2sin^2 (pi/8)$
$2 {sin}^{2} (\frac{π}{8}) = 1 - \frac{\sqrt{2}}{2} = \frac{2 - \sqrt{2}}{2}$ $2sin^2 (pi/8) = 1 - sqrt2/2 = (2 - sqrt2)/2$
${sin}^{2} (\frac{π}{8}) = \frac{2 - \sqrt{2}}{4}$ $sin^2 (pi/8) = (2 - sqrt2)/4$
$sin (\frac{π}{8}) = \pm \frac{\sqrt{2 - \sqrt{2}}}{2}$ $sin (pi/8) = +- sqrt(2 - sqrt2)/2$
Since $sin (\frac{π}{8})$ $sin (pi/8)$ is positive, then,
$P = (\frac{1}{4}) \sqrt{2 - \sqrt{2}}$ $P = (1/4)sqrt(2 - sqrt2)$

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How do you express $cos (\frac{π}{3}) \cdot sin (\frac{π}{8})$ $cos(pi/ 3 ) * sin( ( pi) / 8 )$ without using products of trigonometric functions?

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How do you express $cos (\frac{π}{3}) \cdot sin (\frac{π}{8})$ $cos(pi/ 3 ) * sin( ( pi) / 8 )$ without using products of trigonometric functions?