##### #9 Precalculus questions

 Mathematics Tutor: None Selected Time limit: 1 Day

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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May 15th, 2015
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May 15th, 2015
Dec 7th, 2016
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