How do you express $cos (\frac{π}{3}) \cdot sin (\frac{9 π}{8})$ $cos(pi/ 3 ) * sin( ( 9 pi) / 8 )$ without using products of trigonometric functions?

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How do you express $cos (\frac{π}{3}) \cdot sin (\frac{9 π}{8})$ $cos(pi/ 3 ) * sin( ( 9 pi) / 8 )$ without using products of trigonometric functions?

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Answer 1 · 2016-06-24T03:36:26

$P = - (\frac{1}{2}) sin (\frac{π}{8})$ $P = - (1/2)sin (pi/8)$

Explanation:

Product $P = cos (\frac{π}{3}) . sin (\frac{9 π}{8})$ $P = cos (pi/3).sin ((9pi)/8)$
Trig table --> $cos (\frac{π}{3}) = \frac{1}{2}$ $cos (pi/3) = 1/2$
$sin (\frac{9 π}{8}) = sin (\frac{π}{8} + π) = - sin (\frac{π}{8})$ $sin ((9pi)/8) = sin (pi/8 + pi) = - sin (pi/8)$
P can be expressed as:
$P = - (\frac{1}{2}) sin (\frac{π}{8}) .$ $P = - (1/2)sin (pi/8).$
Note. We can evaluate $sin (\frac{π}{8})$ $sin (pi/8)$ by using the trig identity:
$cos 2 a = 1 - 2 {sin}^{2} a .$ $cos 2a = 1 - 2sin^2 a.$
$cos (\frac{2 π}{8}) = cos (\frac{π}{4}) = \frac{\sqrt{2}}{2} = 1 - 2 {sin}^{2} (\frac{π}{8})$ $cos ((2pi)/8) = cos (pi/4) = sqrt2/2 = 1 - 2sin^2 (pi/8)$
${sin}^{2} (\frac{π}{8}) = 1 - \sqrt{2} = \frac{2 - \sqrt{2}}{4}$ $sin^2 (pi/8) = 1 - sqrt2 = (2 - sqrt2)/4$
$sin (\frac{π}{8}) = \frac{\sqrt{2 - \sqrt{2}}}{2}$ $sin (pi/8) = sqrt(2 - sqrt2)/2$ .
Take the positive value since sin (pi/8) is positive.
$P = - (\frac{1}{4}) \sqrt{2 - \sqrt{2}}$ $P = - (1/4)sqrt(2 - sqrt2)$

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How do you express $cos (\frac{π}{3}) \cdot sin (\frac{9 π}{8})$ $cos(pi/ 3 ) * sin( ( 9 pi) / 8 )$ without using products of trigonometric functions?

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Explanation:

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How do you express $cos (\frac{π}{3}) \cdot sin (\frac{9 π}{8})$ $cos(pi/ 3 ) * sin( ( 9 pi) / 8 )$ without using products of trigonometric functions?