How do you prove $- sin (\frac{π}{3}) = cos (\frac{5 π}{6})$ $-sin( pi/3) = cos ((5pi)/6)$ ?

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How do you prove $- sin (\frac{π}{3}) = cos (\frac{5 π}{6})$ $-sin( pi/3) = cos ((5pi)/6)$ ?

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Answer 1 · 2016-04-04T16:45:28

$- sin (\frac{π}{3}) = sin (π + \frac{π}{3}) = cos (\frac{π}{2} - (π + \frac{π}{3})) = cos (- 5 \frac{π}{6}) = cos (5 \frac{π}{6})$ $-sin(pi/3) = sin (pi+pi/3)= cos (pi/2-(pi+pi/3))=cos(-5pi/6)=cos(5pi/6)$

Explanation:

$- sin x = sin (π - x) = = cos (\frac{π}{2} - (π - x)) = cos (- (x + \frac{π}{2})) = cos (x + \frac{π}{2})$ $-sinx=sin (pi-x)==cos(pi/2-(pi-x))=cos(-(x+pi/2))=cos(x+pi/2)$
Here, $x = \frac{π}{3}$ $x=pi/3$ .

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How do you prove $- sin (\frac{π}{3}) = cos (\frac{5 π}{6})$ $-sin( pi/3) = cos ((5pi)/6)$ ?

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How do you prove -sin( pi/3) = cos ((5pi)/6)−sin(π3)=cos(5π6)-sin( pi/3) = cos ((5pi)/6)?

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How do you prove $- sin (\frac{π}{3}) = cos (\frac{5 π}{6})$ $-sin( pi/3) = cos ((5pi)/6)$ ?