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Question: How do you find the value of \[\csc ( - 210)?\]...

How do you find the value of csc(210)?\csc ( - 210)?

Explanation

Solution

We know that cosec(x)=cosecx\cos ec( - x) = - \cos ecx .
Again the function y=cosecxy = \cos ecx has a period of 2π2\pi or 360{360^ \circ } , i.e., the value of cosecx\cos ecx repeats after an interval of 2π2\pi or 360{360^ \circ } . Therefore we calculate the given period using the intervals. After that we use the trigonometric formulas and we get the required value.

Complete step by step answer:
We know that the function y=cosecxy = \cos ecx has a period of 2π2\pi or 360{360^ \circ } , i.e., the value of cosecx\cos ecx repeats after an interval of 2π2\pi or 360{360^ \circ } .
First we draw the graph of y=cosecxy = \cos ecx

Now the given data 120=90+30{120^ \circ } = {90^ \circ } + {30^ \circ }
Therefore csc(120)\csc ( - 120)
Using the property csc(x)=cscx\csc ( - x) = - \csc x , we get
csc(120)- \csc (120)
We can write the above statement as,
=csc(90+30)= - \csc (90 + 30)
We know that cscx=1sinx\csc x = \dfrac{1}{{\sin x}} , use this property and we get
=1sin(90+30)= - \dfrac{1}{{\sin (90 + 30)}}
We know that the property of trigonometric that is sin(90+x)=cosx\sin (90 + x) = \cos x
Use this in the above function and we get
=1cos30= - \dfrac{1}{{\cos 30}}
From the value table we know the value of cos30=32=sin60\cos 30 = \dfrac{{\sqrt 3 }}{2} = \sin 60 , put this in above equation and we get
=132= - \dfrac{1}{{\tfrac{{\sqrt 3 }}{2}}}
Simplifying and we get
=23= - \dfrac{2}{{\sqrt 3 }}
Therefore our required answer is csc(120)=23\csc ( - 120) = - \dfrac{2}{{\sqrt 3 }} .

Note: Note that formulas which is very important and try to remember all the times
1.sinx=1cscx,cosx=1secx,tanx=1cotx1.\sin x = \dfrac{1}{{\csc x}},\cos x = \dfrac{1}{{\sec x}},\tan x = \dfrac{1}{{\cot x}}
2.sin2x+cos2x=12.{\sin ^2}x + {\cos ^2}x = 1
3.sec2xtan2x=13.{\sec ^2}x - {\tan ^2}x = 1
4.csc2xcot2x=14.{\csc ^2}x - {\cot ^2}x = 1
5.sin(x)=sinx5.\sin ( - x) = - \sin x
6.cos(x)=cosx6.\cos ( - x) = \cos x
7.tan(x)=tanx7.\tan ( - x) = - \tan x
8.sin(2nπ±x)=sinx8.\sin (2n\pi \pm x) = \sin x , period 2π2\pi
9.cos(2nπ±x)=cosx9.\cos (2n\pi \pm x) = \cos x , period 2π2\pi
10.tan(nπ±x)=tanx10.\tan (n\pi \pm x) = \tan x , period π\pi
WE also have to know about the sign conversion ,
In the first quadrant we know all the trigonometric functions are positive. In the second quadrant sine function is positive and others are negative. The third quadrant tangent is positive and in the fourth quadrant cos is positive. At least we must know about the trigonometric function value table. We have all the basic values of all trigonometric functions.