#9 Precalculus questions

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May 15th, 2015

1=simple solution:

That would be the same as the arcsin of 1 divided by the square root of 2 (if you multiply numerator and denominator by 1 over root 2). 
That would make it 45 degrees or pi/4 radians

details(optional just for understanding)

In order to find the value of arcsin(sqrt(2)/2) 

Let t equal that. 

t = arcsin(sqrt(2)/2) 

Take the sine of both sides, keeping in mind that sin and arcsin are inverses of each other. Just to show you in steps, 

sin(t) = sin [ arcsin(sqrt(2)/2) ] 
sin(t) = sqrt(2)/2 

Solve this equation (i.e. where on the unit circle is sine equal to sqrt(2)/2?) 

If we were to select the unit circle values that fall between 0 and 2pi, 
t = { pi/4, 3pi/4 } 

However, arcsin is a function, which means it is only supposed to return one value. 

arcsin(t) is only defined for values of t between -pi/2 to pi/2 inclusive. 
Since 3pi/4 falls out of that range, 
t = pi/4 

As a reference: 

t = arcsin(x) is the same as solving sin(t) = x, but for -pi/2 <= t <= pi/2 
t = arccos(x) is the same as solving cos(t) = x, but for 0 <= t <= pi 
t = arctan(x) is the same as solving tan(t) = x, but for -pi/2 < t < pi/2

May 15th, 2015

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