How do you express $sin (\frac{π}{12}) \cdot cos (\frac{7 π}{12})$ $sin(pi/12) * cos(( 7 pi)/12 )$ without products of trigonometric functions?

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How do you express $sin (\frac{π}{12}) \cdot cos (\frac{7 π}{12})$ $sin(pi/12) * cos(( 7 pi)/12 )$ without products of trigonometric functions?

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Answer 1 · 2018-07-01T09:08:53

$- \frac{1}{2} + \frac{\sqrt{3}}{4}$ $-1/2+sqrt(3)/4$

Explanation:

$sin (\frac{π}{12}) = \frac{\sqrt{3} - 1}{2 \sqrt{2}}$ $sin(pi/12)=(sqrt(3)-1)/(2sqrt(2))$
$cos (\frac{7 π}{12}) = - \frac{\sqrt{3} - 1}{2 \sqrt{2}}$ $cos((7pi)/12)=-(sqrt(3)-1)/(2sqrt(2))$

$sin (\frac{π}{12}) \cdot cos (\frac{7 π}{12}) = - {(\frac{\sqrt{3} - 1}{2 \sqrt{2}})}^{2}$ $sin(pi/12) * cos(( 7 pi)/12 )=-((sqrt(3)-1)/(2sqrt(2)))^2$

$- {(\frac{\sqrt{3} - 1}{2 \sqrt{2}})}^{2} = - \frac{1}{8} {(\sqrt{3} - 1)}^{2} = - \frac{3}{8} + \frac{\sqrt{3}}{4} - \frac{1}{8} = - \frac{1}{2} + \frac{\sqrt{3}}{4}$ $-((sqrt(3)-1)/(2sqrt(2)))^2=-1/8(sqrt(3)-1)^2=-3/8+sqrt(3)/4-1/8=-1/2+sqrt(3)/4$

Answer 2 · 2018-07-01T09:36:10

$\Rightarrow \frac{\sqrt{3} - 2}{4}$ $color(indigo)(=> (sqrt3 - 2) / 4$

Explanation:

$sin (\frac{π}{12}) \cdot cos (\frac{7 π}{12})$ $sin (pi/12) * cos ((7pi) / 12)$

$\Rightarrow (\frac{1}{2}) (sin (\frac{π}{12} + \frac{7 π}{12}) + (sin (\frac{π}{12}) - (\frac{7 π}{12}))$ $=> (1/2) (sin (pi/12 + (7pi)/12) + (sin (pi/12) - ((7pi)/12))$

$\Rightarrow (\frac{1}{2}) (sin (\frac{2 π}{3}) + sin (- (\frac{π}{2}))$ $=> (1/2) (sin ((2pi)/3) + sin (-(pi/2))$

$\Rightarrow (\frac{1}{2}) (sin (\frac{π}{3}) - sin (\frac{π}{2}))$ $=> (1/2) (sin (pi/3) - sin (pi/2))$

$\Rightarrow (\frac{1}{2}) (\frac{\sqrt{3}}{2} - 1)$ $=> (1/2) (sqrt3 /2 - 1)$

$\Rightarrow \frac{\sqrt{3} - 2}{4}$ $color(indigo)(=> (sqrt3 - 2) / 4$

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How do you express $sin (\frac{π}{12}) \cdot cos (\frac{7 π}{12})$ $sin(pi/12) * cos(( 7 pi)/12 )$ without products of trigonometric functions?

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How do you express $sin (\frac{π}{12}) \cdot cos (\frac{7 π}{12})$ $sin(pi/12) * cos(( 7 pi)/12 )$ without products of trigonometric functions?