What is the ${sin}^{- 1} (\frac{\sqrt{3}}{2})$ $sin^(-1)(sqrt(3)/2)$ ?

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What is the ${sin}^{- 1} (\frac{\sqrt{3}}{2})$ $sin^(-1)(sqrt(3)/2)$ ?

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Answer 1 · 2015-09-07T12:35:40

${sin}^{- 1} (\frac{\sqrt{3}}{2}) = \frac{π}{3}$ $sin^(-1)(sqrt(3)/2) = pi/3$ (radians)

Explanation:

${sin}^{- 1} (\frac{\sqrt{3}}{2}) = θ$ $sin^(-1)(sqrt(3)/2) = theta$
$XX$ $color(white)("XX")$ means
$sin (θ) = \frac{\sqrt{3}}{2}$ $sin(theta)=sqrt(3)/2$
$XXXX$ $color(white)("XXXX")$ with the restriction that $θ \in [- \frac{π}{2}, \frac{π}{2}]$ $theta in [-pi/2,pi/2]$

The possible triangles with this ratio of $y$ $y$ to $hypotenuse$ $"hypotenuse"$ (i.e. $sin$ $sin$ ) are pictured below:
enter image source here
...but note that only $\frac{π}{3}$ $pi/3$ falls within the restricted range for ${sin}^{- 1}$ $sin^(-1)$

Answer 2 · 2016-05-10T22:32:15

$θ$ $theta$ = 60° or 120°

Explanation:

It helps if one recognises the lengths 3 and $\sqrt{2}$ $sqrt2$ as
being two of the sides in one of the special triangle,
namely 30°, 60° 90°.

Sketch this by drawing an equilateral triangle of sides 2 units, and drawing in a line of symmetry to form two right-angled triangles.

The hypotenuse is of length 2 units and the opposite side
is of length $\sqrt{3}$ $sqrt3$ . The angle is therefore 60°.

However, the sign is positive. Sin is positive in the first and second quadrants.

Therefore $θ$ $theta$ = 60° or 120°

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What is the ${sin}^{- 1} (\frac{\sqrt{3}}{2})$ $sin^(-1)(sqrt(3)/2)$ ?

2 Answers

Explanation:

Explanation:

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