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Question: How do you evaluate \[\csc \left( {\dfrac{{5\pi }}{3}} \right)\] ?...

How do you evaluate csc⁡(5π3)\csc \left( {\dfrac{{5\pi }}{3}} \right) ?

Explanation

Solution

Hint : Trigonometric functions are those functions that tell us the relation between the three sides of a right-angled triangle. Sine, cosine, tangent, cosecant, secant and cotangent are the six types of trigonometric functions; sine, cosine and tangent are the main functions while cosecant, secant and cotangent are the reciprocal of sine, cosine and tangent respectively. Thus the given function can be converted in the form of sine easily. First we find the value of sine function then taking the reciprocal of sine we get the cosecant value. Also we need to know the supplementary angle of sine.

Complete step-by-step answer :
Given, csc⁡(5π3)\csc \left( {\dfrac{{5\pi }}{3}} \right) .
We know that the csc⁡(5π3)=1sin⁡(5π3)\csc \left( {\dfrac{{5\pi }}{3}} \right) = \dfrac{1}{{\sin \left( {\dfrac{{5\pi }}{3}} \right)}} .
Now we find the value of sin⁡(5π3)\sin \left( {\dfrac{{5\pi }}{3}} \right) .
We can write 5π3=2π−π3\dfrac{{5\pi }}{3} = 2\pi - \dfrac{\pi }{3} .
sin⁡(5π3)=sin⁡(2π−π3)\sin \left( {\dfrac{{5\pi }}{3}} \right) = \sin \left( {2\pi - \dfrac{\pi }{3}} \right)
We know that sin⁡(2π−θ)=−sin⁡θ\sin (2\pi - \theta ) = - \sin \theta . The negative sign is because sine lies in the fourth quadrant. In the fourth quadrant sine value is negative.
=−sin⁡(π3)= - \sin \left( {\dfrac{\pi }{3}} \right)
But we know the value of sin⁡(π3)\sin \left( {\dfrac{\pi }{3}} \right) , we have:
=−32= - \dfrac{{\sqrt 3 }}{2} .
Thus we have, sin⁡(5π3)=−32\sin \left( {\dfrac{{5\pi }}{3}} \right) = - \dfrac{{\sqrt 3 }}{2}
Now we have,
csc⁡(5π3)=1sin⁡(5π3)\csc \left( {\dfrac{{5\pi }}{3}} \right) = \dfrac{1}{{\sin \left( {\dfrac{{5\pi }}{3}} \right)}}
Substituting sin⁡(5π3)=−32\sin \left( {\dfrac{{5\pi }}{3}} \right) = - \dfrac{{\sqrt 3 }}{2} in above. We get,
=1(−32)= \dfrac{1}{{\left( { - \dfrac{{\sqrt 3 }}{2}} \right)}} .
=−23= - \dfrac{2}{{\sqrt 3 }} .
Thus we have csc⁡(5π3)=−23\csc \left( {\dfrac{{5\pi }}{3}} \right) = - \dfrac{2}{{\sqrt 3 }} .
So, the correct answer is “−23 - \dfrac{2}{{\sqrt 3 }}”.

Note : Remember A graph is divided into four quadrants, all the trigonometric functions are positive in the first quadrant, all the trigonometric functions are negative in the second quadrant except sine and cosine functions, tangent and cotangent are positive in the third quadrant while all others are negative and similarly all the trigonometric functions are negative in the fourth quadrant except cosine and secant. Depending on the problem we need to change the sign as we did in above.