Here clearly both
sinx and
cosx are defined in their domain.
Let's assume that g(x)=sinx and f(x)=cosx
=> Let's first prove that g(x) is continuous in it's domain.
Let c be a real number, put x=c+h
So if x⇒c, then it means that h⇒0
limx→c g(x) = limx→c sin(x)
Put x=h+c
And as mentioned above, when x⟶c then it means that h⟶0
Which gives us limh→0 sin(c+h)
Expanding sin(c+h) = sin(h)cos(c)+cos(h)sin(c)
Which gives us limh→0 sin(h)cos(c)+cos(h)sin(c)
=sin(c)cos(0)+cos(c)sin(0)
=sin(c)
So here we get,
limx→c g(x) = limx→c sin(x) = sin(c)=g(c)
And this proves that sin(x) is continuous all across its domain
=> Let's prove that f(x)=cos(x) is continuous in it's domain.
Let c be a real number, put x=c+h
So if x⇛c then it means that h⇒0
f(c)=cos(c)
limx→c f(x) = limx→c cos(x)
Put x=h+c
And as mentioned above, when x⇒c then it means that h⇒0
Which gives us limh→0 cos(c+h)
Expanding cos(h+c) = cos(h)cos(c)−sin(h)sin(c)
Which gives us limh→0 cos(h)cos(c)−sin(h)sin(c)
=cos(c)cos(0)−sin(c)sin(0)
=cos(c)
This gives us
limx→c f(x) = limx→c cos(x) = cos(c)=f(c)
And this proves that cos(x) is continuous all across its domain
So by theorem. If function f and function g are contnous then f+g is also continous.
∴ sin(x)+cos(x) is continious.