What does $arcsin (cos (\frac{5 π}{6}))$ $arcsin(cos ((5pi)/6))$ equal?

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What does $arcsin (cos (\frac{5 π}{6}))$ $arcsin(cos ((5pi)/6))$ equal?

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Answer 1 · 2016-03-13T15:27:59

$= - \frac{π}{3}$ $=-pi/3$

Explanation:

"principal value" of the arcsin function means it's between
$- \frac{π}{2} \leq θ \leq + \frac{π}{2}$ $-pi/2<=theta<=+pi/2$

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

for least positve value
$arcsin (cos (5 \frac{π}{6})) = arcsin (cos (\frac{π}{2} + \frac{π}{3})) = arcsin (- sin (\frac{π}{3})) = arcsin sin (π + \frac{π}{3}) = 4 \frac{π}{3}$ $arcsin(cos(5pi/6))=arcsin(cos(pi/2+pi/3))=arcsin(-sin(pi/3))=arcsinsin(pi+pi/3)=4pi/3$

Answer 2 · 2016-03-18T15:30:50

$\frac{4 π}{3}, \frac{5 π}{3}$ $(4pi)/3, (5pi)/3$

Explanation:

Trig table gives -->
$cos (\frac{5 π}{6}) = - \frac{\sqrt{3}}{2}$ $cos ((5pi)/6) = - sqrt3/2$
Find $arcsin (- \frac{\sqrt{3}}{2})$ $arcsin(-sqrt3/2)$
Trig unit circle, and trig table give -->
$sin x = - \frac{\sqrt{3}}{2}$ $sin x = -sqrt3/2$ -->2 solutions -->
arc $x = - \frac{π}{3}$ $x = - pi/3$ and arc $x = \frac{4 π}{3}$ $x = (4pi)/3$
Answers: $(\frac{4 π}{3})$ $((4pi)/3)$ and $(\frac{5 π}{3})$ $((5pi)/3)$ --> (co-terminal with $(- \frac{π}{3})$ $(-pi/3)$

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What does $arcsin (cos (\frac{5 π}{6}))$ $arcsin(cos ((5pi)/6))$ equal?

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