Kathy P. answered 02/24/21
Mechanical Engineer with 10+ years of teaching and tutoring experience
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(1.) ArcCos [ sin (5pi/6) ]
sin(5pi/6) = 1/2 Note: 5pi/6 radians = 150 degrees
ArcCos [1/2] = The angle whose cosine is 1/2
ArcCos [ 1/2 ] = pi/3 or 5*pi/3
Note: On the unit circle, there are two angles who have a cosine = 1/2
Since the cosine is positive, those angles will be in Q1 and Q4
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(2.) sin [ ArcCos [ 1/3 ]
Construct a right-triangle that has a cosine of 1/3
The adjacent side = 1 and hypotenuse = 3
So, the Cosine = 1/3
Use Pythagorean Theorem to calculate the opposite side
Opposite Side = sqrt( 3^2 - 1^1)
Opposite Side = sqrt( 8 )
Opposite Side = 2*sqrt(2)
So,
Sine = Opposite / hypotenuse
Sine = 2*sqrt(2) / 3
Therefore,
sin [ ArcCos [ 1/3 ] = 2*sqrt(2) / 3
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(3.) ArcTan [ tan(2pi/3 )]
These are inverse, or opposite, functions.
Therefore,
ArcTan [ tan(2pi/3 )] = 2pi/3