Graphs of some difficult trigonometric functions -I

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Please understand the below functions carefully,

 1, f(x) = {\sin ^{ - 1}}\sin x

  • Periodic with period 2\pi  ( 'x' is directly operated by 'sin' and sin x is periodic with period '2pi' )
  • Domain = R ( domain of f(x) = sin x is R)
  • Range = [-pi/2,pi/2] ( Range of f(x) = arcsin(x) is [-pi/2,pi/2] ) 
And,

 2. f(x) = \sin {\sin ^{ - 1}}x
  • Non-periodic ( 'x' is directly operated by 'inverse of sin' and 'inverse of sin' is not periodic)
  • Domain = [-1,1] ( domain of f(x) = arcsin(x) is [-1,1] )
  • Range = [-1,1]  

From above arguments following statements are very clear, 

PERIODIC FUNCTIONS

1. f(x) = {\sin ^{ - 1}}\sin x , Period = 2\pi
2. f(x) = {\cos ^{ - 1}}\cos x, Period = 2\pi
3. f(x) = {\tan ^{ - 1}}\tan x, Period = \pi

NON-PERIODIC FUNCTIONS

1. f(x) = \sin {\sin ^{ - 1}}x
2. f(x) = \cos {\cos ^{ - 1}}x
3. f(x) = \tan {\tan ^{ - 1}}x

PERIODIC FUNCTIONS:


1. f(x) = {\sin ^{ - 1}}\sin x

Steps of drawing the graph of periodic function:

1. Draw the graph of function for one period
2. Repeat the graph for all values.



Since, range of arcsin(x) = [-pi/2, pi/2]

therefore, f(x) = {\sin ^{ - 1}}\sin x cannot go beyond this interval, also consider that 'sin x' is increasing in the interval [-pi/2,pi/2] and then decreasing in the interval [pi/2,3pi/2]

understand this carefully,

{\sin ^{ - 1}}\sin x = x ,  when  - \frac{\pi }{2} \le x \le \frac{\pi }{2}

                       = \pi - x, when  - \frac{\pi }{2} \le \pi - x \le \frac{\pi }{2}
                                           i.e.   \frac{\pi }{2} \le x \le \frac{{3\pi }}{2}  ( we have adjusted all the values such that they will always fall in the interval [-pi/2,pi/2]  )

 (* NOTE - As when we draw the graph between \frac{\pi }{2} \le x \le \frac{{3\pi }}{2} we have to adjusted it into interval [ - \pi /2,\pi /2]

 1st multiply with negative sign, we have

 - \frac{{3\pi }}{2} \le - x \le - \frac{\pi }{2}   ( inequality gets reversed)

 Add 'pi',

 \pi - \frac{{3\pi }}{2} \le \pi - x \le \pi - \frac{\pi }{2}

 or,  - \frac{\pi }{2} \le \pi - x \le \frac{\pi }{2}

 so, graph in interval \frac{\pi }{2} \le x \le \frac{{3\pi }}{2} is \pi - x )


Since, the period of f(x) = {\sin ^{ - 1}}\sin x is 2\pi

therefore, first we draw the graph for period 2\pi  i.e. [ - \pi /2,3\pi /2] and then repeat it for entire value of 'x'.



2. f(x) = {\cos ^{ - 1}}\cos x

 {\cos ^{ - 1}}\cos x = x, when {\cos ^{ - 1}}\cos x = x,0 \le x \le \pi

                      = 2\pi - x, when 0 \le 2\pi - x \le \pi
                                        i.e. \pi \le x \le 2\pi

Since, the period of f(x) = {\cos ^{ - 1}}\cos x is 2\pi

therefore, first we draw the graph for period 2\pi  i.e. 0 \le x \le 2\pi  and then repeat it for entire value of 'x'.


3. f(x) = {\tan ^{ - 1}}\tan x

 {\tan ^{ - 1}}\tan x = x, when  - \frac{\pi }{2} < x < \frac{\pi }{2}
Since, the period of f(x) = {\tan ^{ - 1}}\tan x is \pi

therefore, first we draw the graph for period \pi  i.e.  - \frac{\pi }{2} < x < \frac{\pi }{2} and then repeat it for entire value of 'x'.




NON-PERIODIC FUNCTIONS:

1. f(x) = \sin {\sin ^{ - 1}}x

\sin {\sin ^{ - 1}}x = x

Domain = [-1,1], (since domain of arcsin(x) is [-1,1])

Range = [-1,1]






2. f(x) = \cos {\cos ^{ - 1}}x

\cos {\cos ^{ - 1}}x = x

Domain = [-1,1], (since domain of arccos(x) is [-1,1])

Range = [-1,1]






3. f(x) = \tan {\tan ^{ - 1}}x

\tan {\tan ^{ - 1}}x = x

Domain = R (since domain of arctan(x) is R )

Range = R





Students are requested to sketch following graph and observe their domain, range and periodicity.

1. f(x) = \cot {\cot ^{ - 1}}x

2. f(x) = \cos ec\cos e{c^{ - 1}}x

3. f(x) = secse{c^{ - 1}}x

4. f(x) = {\cot ^{ - 1}}\cot x

5. f(x) = \cos e{c^{ - 1}}\cos ecx

6.. f(x) = se{c^{ - 1}}secx

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