Respuesta :
Answer:
1/2
Step-by-step explanation:
We have [tex]\cos(\arccsc(\frac{2\sqrt{3}}{3}))[/tex].
Let [tex]u=\arccsc(\frac{2\sqrt{3}}{3})[/tex].
This implies [tex]\csc(u)=\frac{2\sqrt{3}}{3}[/tex].
Use that sine and cosecant are reciprocals.
[tex]\sin(u)=\frac{3}{2\sqrt{3}}[/tex]
Now I'm going to rationalize the denominator there by multiply numerator and denominator by [tex]\sqrt{3}[/tex]:
[tex]\sin(u)=\frac{3\sqrt{3}}{2(3)}[/tex]
[tex]\sin(u)=\frac{3\sqrt{3}}{6}[/tex]
Reduce the fraction:
[tex]\sin(u)=\frac{\sqrt{3}}{2}[/tex]
Now I'm going to use a Pythagorean Identity: [tex]\cos^2(u)+\sin^2(u)=1[/tex].
This will give me the value of cos(u) which would give me the answer to my question if it exists.
Replace [tex]\sin(u)[/tex] with [tex]\frac{\sqrt{3}}{2}[/tex] in:
[tex]\cos^2(u)+\sin^2(u)=1[/tex]
[tex]\cos^2(u)+(\frac{\sqrt{3}}{2})^2=1[/tex]
[tex]\cos^2(u)+\frac{3}{4}=1[/tex]
Subtract 3/4 on both sides:
[tex]\cos^2(u)=\frac{1}{4}[/tex]
Square root both sides:
[tex]\cos(u)=\pm \frac{1}{2}[/tex] (since 1/2*1/2=1/4 or -1/2*-1/2=1/4)
Now we must decide between the positive or the negative.
It depends where u lies. What quadrant? Hopefully it lays between 0 and [tex]\pi[/tex]. Otherwise, it doesn't exist (unless you have a different definition for arc function).
So u led to this equation earlier:
[tex]\sin(u)=\frac{\sqrt{3}}{2}[/tex]
arcsin( ) only has outputs between [tex]\frac{-\pi}{2}[/tex] and [tex]\frac{\pi}{2}[/tex].
This would have to be in the first quadrant because we have only positive sine values there.
So this means cos(u)=1/2 and not -1/2 because we are using that u is in the 1st quadrant.
Remember u was [tex]\arccsc(\frac{2\sqrt{3}}{3})[/tex].
So we have actually evaluated
[tex]\cos(\arccsc(\frac{2\sqrt{3}}{3}))[/tex] without a calculator.
The value is 1/2.
