#9 Precalculus questions
Mathematics

Tutor: None Selected  Time limit: 1 Day 
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
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