How do you express $sin (\frac{π}{6}) \cdot cos (\frac{9 π}{8})$ $sin(pi/ 6 ) * cos( ( 9 pi) / 8 )$ without using products of trigonometric functions?

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How do you express $sin (\frac{π}{6}) \cdot cos (\frac{9 π}{8})$ $sin(pi/ 6 ) * cos( ( 9 pi) / 8 )$ without using products of trigonometric functions?

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Answer 1 · 2016-03-25T12:38:27

$= - \frac{1}{2} \sqrt{\frac{1}{2} \cdot (1 + \frac{1}{\sqrt{2}})}$ $=-1/2sqrt(1/2*(1+1/sqrt2))$

Explanation:

$sin (\frac{π}{6}) \cdot cos (π + \frac{π}{8}) = - \frac{1}{2} \cdot cos (\frac{π}{8})$ $sin(pi/6)*cos(pi+pi/8)= -1/2*cos(pi/8)$
$= - \frac{1}{2} \sqrt{\frac{1}{2} \cdot (1 + cos (\frac{π}{4}))} = - \frac{1}{2} \sqrt{\frac{1}{2} \cdot (1 + \frac{1}{\sqrt{2}})}$ $= -1/2sqrt(1/2*(1+cos(pi/4))) =-1/2sqrt(1/2*(1+1/sqrt2))$

It may also be expressed as sum of two trigonometric functions
using formula
$sin A cos B = \frac{1}{2} \cdot (sin (A + B) + sin (A - B))$ $sinAcosB=1/2*(sin(A+B)+sin(A-B))$

$sin (\frac{π}{6}) \cdot cos (π + \frac{π}{8}) = - sin (\frac{π}{6}) \cdot cos (\frac{π}{8})$ $sin(pi/6)*cos(pi+pi/8)= -sin(pi/6)*cos(pi/8)$

$= - \frac{1}{2} \cdot (sin (\frac{π}{6} + \frac{π}{8}) + sin (\frac{π}{6} - \frac{π}{8}))$ $=-1/2*(sin(pi/6+pi/8)+sin(pi/6-pi/8))$

$= - \frac{1}{2} \cdot (sin (7 \frac{π}{24}) + sin (\frac{π}{24})$ $=-1/2*(sin(7pi/24)+sin(pi/24)$

Answer 2 · 2016-03-25T12:57:29

$- (\frac{1}{2}) (\frac{cos π}{8})$ $- (1/2)(cos pi/8)$

Explanation:

$P = sin (\frac{π}{6}) . cos (\frac{9 π}{8})$ $P = sin (pi/6).cos ((9pi)/8)$
Trig table --> $sin (\frac{π}{6}) = \frac{1}{2}$ $sin (pi/6) = 1/2$
$cos (\frac{9 π}{8}) = cos (\frac{π}{8} + π) = - cos (\frac{π}{8})$ $cos ((9pi)/8) = cos (pi/8 + pi) = - cos (pi/8)$
The product P can be expressed as:
$P = - (\frac{1}{2}) (\frac{cos π}{8})$ $P = - (1/2) (cos pi/8)$

If being asked, we can find P by evaluating $cos (\frac{π}{8})$ $cos (pi/8)$ , using the trig identity: $cos 2 a = 2 {cos}^{2} a - 1 .$ $cos 2a =2cos^2a - 1.$

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How do you express $sin (\frac{π}{6}) \cdot cos (\frac{9 π}{8})$ $sin(pi/ 6 ) * cos( ( 9 pi) / 8 )$ without using products of trigonometric functions?

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How do you express $sin (\frac{π}{6}) \cdot cos (\frac{9 π}{8})$ $sin(pi/ 6 ) * cos( ( 9 pi) / 8 )$ without using products of trigonometric functions?