Science

Math

Humanities

... and beyond

Socratic Meta
Featured Answers

Topics

How do you prove ${cos}^{- 1} (- sin ((\frac{2}{3}) π)$ $cos^-1(-sin((2/3) pi)$ ?

2 Answers

Jun 22, 2016

Note there is nothing here to prove.
If the intended question was to evaluate
$XXX {cos}^{- 1} (- sin (\frac{2 π}{3})) = - \frac{π}{6}$ $color(white)("XXX")cos^(-1)(-sin((2pi)/3))=color(green)(-pi/6$

Explanation:

$\frac{2 π}{3}$ $(2pi)/3$ is equivalent to a reference angle of $\frac{π}{3}$ $pi/3$ in Quadrant II.
In Quadrant II the $sin$ $sin$ of the reference angle is equal to the $sin$ $sin$ of the actual angle.

$\frac{π}{3}$ $pi/3$ is a standard angle with $sin (\frac{π}{3}) = \frac{\sqrt{3}}{2}$ $sin(pi/3)=sqrt(3)/2$

So
$XXX {cos}^{1} (- sin (\frac{2 π}{3}))$ $color(white)("XXX")cos^(1)(-sin((2pi)/3))$

$XXXXXX = {cos}^{- 1} (- \frac{\sqrt{3}}{2})$ $color(white)("XXXXXX")=cos^(-1)(-sqrt(3)/2)$

${cos}^{- 1}$ $cos^(-1)$ or ( $arccos$ $arccos$ ) using the standard function definitions is restricted to the range $(- \frac{π}{2}, + \frac{π}{2}]$ $(-pi/2,+pi/2]$

Within this interval only
$XXX cos (- \frac{π}{6}) = - \frac{\sqrt{3}}{2}$ $color(white)("XXX")cos(-pi/6)=-sqrt(3)/2$
(again using standard trigonometric triangles)

So
$XXX {cos}^{- 1} (- \frac{\sqrt{3}}{2}) = - \frac{π}{6}$ $color(white)("XXX")cos^(-1)(-sqrt(3)/2)=-pi/6$

A. S. Adikesavan

Jun 23, 2016

$\frac{5 π}{6}$ $(5pi)/6$ in $[0 . π]$ $[0. pi]$ .

Explanation:

Let us use ${cos}^{- 1} (cos x) = x$ $cos^(-1)( cos x )= x$

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

Now, the given expression is ${cos}^{- 1} (cos (\frac{5 π}{6})) = \frac{5 π}{6}$ $cos^(-1) (cos ((5pi)/6)) =(5pi)/6$ .

Related questions

See all questions in Basic Inverse Trigonometric Functions

Impact of this question

2947 views around the world

You can reuse this answer
Creative Commons License