How do you find the value for ${cos}^{- 1} (- \frac{\sqrt{3}}{2})$ $cos^ -1 (- sqrt 3/2)$ ?

Question

How do you find the value for ${cos}^{- 1} (- \frac{\sqrt{3}}{2})$ $cos^ -1 (- sqrt 3/2)$ ?

1 Answer

Impact of this question

83378 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-09-23T12:51:16

$5 \frac{π}{6}$ $5pi/6$ radians or $150^{o}$ $150^o$

Explanation:

as $- \frac{\sqrt{3}}{2} i s N e g a t i v e$ $-sqrt3/2 is Negative$
and cos function is negative in 2nd and 3rd quadrant
$- \frac{\sqrt{3}}{2} = cos (π - \frac{π}{6}) or cos (π + \frac{π}{6})$ $-sqrt3/2=cos(pi-pi/6)or cos(pi+pi/6)$
as $arccos x$ $arccosx$ range is in $[0, π]$ $[0 , pi]$ $cos (π + \frac{π}{6})$ $cos(pi+pi/6)$ is ruled out
$- \frac{\sqrt{3}}{2} = cos (π - \frac{π}{6})$ $-sqrt3/2=cos(pi-pi/6)$ $\Rightarrow - \frac{\sqrt{3}}{2} = arccos cos (π - \frac{π}{3}) = π - \frac{π}{6} = 5 \frac{π}{6} r a d i a n s = \frac{5}{6} \cdot 180^{o} = 150^{o}$ $=>-sqrt3/2= arccoscos(pi-pi/3)=pi-pi/6=5pi/6radians =5/6*180^o=150^o$

Science

Math

Humanities

... and beyond

How do you find the value for ${cos}^{- 1} (- \frac{\sqrt{3}}{2})$ $cos^ -1 (- sqrt 3/2)$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you find the value for cos−1(−√32)cos^ -1 (- sqrt 3/2)?

1 Answer

Explanation:

Related questions

Impact of this question

How do you find the value for ${cos}^{- 1} (- \frac{\sqrt{3}}{2})$ $cos^ -1 (- sqrt 3/2)$ ?