Question

Discuss the continuity of the cosine, cosecant, secant and cotangent functions,

Answer 1

Consider the function g(x)=sinx.

Let k be any real number.

At x=k, g( k )=sink.

Consider the left hand limit,

LHL= lim x→ k − g( x ) lim x→ k − sinx= lim h→0 sin( k−h ) = lim h→0 sinkcosh−cosksinh =sink

Consider the right hand limit,

RHL= lim x→ k + g( x ) lim x→ k + sinx= lim h→0 sin( k+h ) = lim h→0 sinkcosh+cosksinh =sink

Here at x=k, LHL=RHL.

Hence, the function is continuous for all real numbers.

Consider the function h(x)=cosx.

Let k be any real number.

At x=k, h( k )=cosk.

Consider the left hand limit,

LHL= lim x→ k − h( x ) lim x→ k − cosx= lim h→0 cos( k−h ) = lim h→0 coskcosh+sinksinh =cosk

Consider the right hand limit,

RHL= lim x→ k + h( x ) lim x→ k + cosx= lim h→0 cos( k+h ) = lim h→0 coskcosh−sinksinh =cosk

Here, at x=k, LHL=RHL.

Hence, the function is continuous for all real numbers.

If g and h are two continuous functions, then the functions g h ;h≠0, 1 h ;h≠0 and 1 g ;g≠0 are also continuous functions.

The given function is f( x )=cosec x. Since cosecx= 1 sinx ;sinx≠0, then the function is continuous because 1 g ;g≠0 is continuous which means x≠nπ( n∈Z ) is continuous.

Hence, cosecx is continuous except x=nπ( n∈Z ).

The given function is f( x )=secx. Since secx= 1 cosx ;cosx≠0, hence the function is continuous for all x≠ ( 2n+1 )π 2 ( n∈Z ).

Therefore, secx is continuous except x= ( 2n+1 )π 2 ( n∈Z ).

The given function is f( x )=cotx. Since cotx= cosx sinx ;sinx≠0, hence the function is continuous for all x≠nπ( n∈Z ).

Therefore, cotx is a continuous except x=nπ( n∈Z ).

Thus, the functions cosine, cosecant, secant and cotangent are continuous in their domain.