What is the #sin^(-1)(sqrt(3)/2)#?

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Alan P. Share
Sep 7, 2015

Answer:

#sin^(-1)(sqrt(3)/2) = pi/3# (radians)

Explanation:

#sin^(-1)(sqrt(3)/2) = theta#
#color(white)("XX")#means
#sin(theta)=sqrt(3)/2#
#color(white)("XXXX")#with the restriction that #theta in [-pi/2,pi/2]#

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

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Write your answer here...
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Answer

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Answer:

Explanation

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3
May 10, 2016

Answer:

#theta# = 60° or 120°

Explanation:

It helps if one recognises the lengths 3 and # 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 #sqrt3#. The angle is therefore 60°.

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

Therefore #theta# = 60° or 120°

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