Question

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

Answer 1

It is known that if g and h are two continuous functions, then
(i)$\frac{h(x)}{g(x)}$, g(x)≠0 is continues
(ii)$\frac{1}{g(x)}$,g(x)$$\neq 0 is continues
(iii)$\frac{1}{h(x)}$,$h(x)\neq0$ is continues
It has to be proved first that g(x)=sinx and h(x)=cosx are continuous functions.
Let g(x)=sinx It is evident that g(x)=sinx is defined for every real number.
Let c be a real number. Put x=c+h
If x$\rightarrow$c,then h$\rightarrow$0
g(c)=sin c
$\lim_{x\rightarrow c}$g(x)=$\lim_{x\rightarrow c}$sin x
=$\lim_{h\rightarrow 0}$sin(c+h)
=$\lim_{h\rightarrow 0}$[sin c cos h+cos c sin h]
=$\lim_{h\rightarrow 0}$(sin c cos h)+$\lim_{h\rightarrow 0}$(cos c sin h)
=sin0 cos0+cos c sin0
=sin c+0
=sin c
∴$\lim_{x\rightarrow c}$g(x)=g(c)
Therefore,g is a continuous function.
Let h(x)=cos x It is evident that h(x)=cos x is defined for every real number.
Let c be a real number.Put x=c+h
If x$\rightarrow$c, then h$\rightarrow$0
h(c)=cos c
$\lim_{x\rightarrow c}$h(x)=$\lim_{x\rightarrow c}$cosx
=$\lim_{h\rightarrow 0}$cos(c+h)
=$\lim_{h\rightarrow 0}$[cos c cos h-sin c sin h]
=$\lim_{h\rightarrow 0}$cos c cos h-$\lim_{h\rightarrow 0}$ sin c sin h
=cos c cos 0-sin c sin 0
=cos c$\times$1-sinc$\times$0
=cos c
∴$\lim_{x\rightarrow c}$h(x)=h(c)
Therefore,h(x)=cosx is a continuous function.
It can be concluded that,
cosec x=$\frac{1}{sin\,x}$, sinx≠0 is continues
⇒cosec x,x≠nπ(n∈Z) is continues
Therefore,cosecant is continuous except at x=np, nÎZ
secx=$\frac{1}{cos\,x}$,cos x≠0 is continuous
⇒sec x, x≠(2n+1)$\frac{\pi}{2}$(n∈Z) is continues
Therefore,secant is continuous except at x=(2n+1)$\frac{\pi}{2}$(n∈Z)
cotx=$\frac{cos\,x}{sin\,x}$,sinx≠0 is continuous

⇒cotx, x≠nπ(n∈Z) is continues
Therefore, cotangent is continuous except at x=np,nÎZ