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Simplify cos[arccsc((2/3)/3)].
A)no solution
B)sqrt3/2
C)1/2

User Vijaykumar
by
7.5k points

1 Answer

1 vote

Answer:

1/2

Explanation:

We have
\cos(\arccsc((2√(3))/(3))).

Let
u=\arccsc((2√(3))/(3)).

This implies
\csc(u)=(2√(3))/(3).

Use that sine and cosecant are reciprocals.


\sin(u)=(3)/(2√(3))

Now I'm going to rationalize the denominator there by multiply numerator and denominator by
√(3):


\sin(u)=(3√(3))/(2(3))


\sin(u)=(3√(3))/(6)

Reduce the fraction:


\sin(u)=(√(3))/(2)

Now I'm going to use a Pythagorean Identity:
\cos^2(u)+\sin^2(u)=1.

This will give me the value of cos(u) which would give me the answer to my question if it exists.

Replace
\sin(u) with
(√(3))/(2) in:


\cos^2(u)+\sin^2(u)=1


\cos^2(u)+((√(3))/(2))^2=1


\cos^2(u)+(3)/(4)=1

Subtract 3/4 on both sides:


\cos^2(u)=(1)/(4)

Square root both sides:


\cos(u)=\pm (1)/(2) (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
\pi. Otherwise, it doesn't exist (unless you have a different definition for arc function).

So u led to this equation earlier:


\sin(u)=(√(3))/(2)

arcsin( ) only has outputs between
(-\pi)/(2) and
(\pi)/(2).

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
\arccsc((2√(3))/(3)).

So we have actually evaluated


\cos(\arccsc((2√(3))/(3))) without a calculator.

The value is 1/2.

User Bastl
by
9.2k points

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