# f(x)=Cos(x) g(x)=Csc(x)Tan(x)

• Jul 20th 2010, 03:23 PM
RosieLaird
f(x)=Cos(x) g(x)=Csc(x)Tan(x)
Use the given relations f and g to find (a) f · g (b) f/g (f over g) (c) f∘g (d)g∘f
f(x)=Cos(x) g(x)=Csc(x)Tan(x)

can i see the steps on how to on these?
• Jul 20th 2010, 03:36 PM
titos
first you have to fine the domain for f(x),g(x).Then give a name for each (example:h=f*g,d=f/g,l= f∘g,k=g∘f)then find the domain for the new functions(this very important to do),and for the end right the type for the new functions!!ok?
• Jul 20th 2010, 03:42 PM
RosieLaird
how do you find the domain and range for them?
• Jul 20th 2010, 03:46 PM
eumyang
I'll do (a)...
\displaystyle \begin{aligned} f \cdot g &= (cos \, x)(csc \, x)(tan \, x) \\ &= (cos \, x)\left( \frac{1}{sin \, x} \right) \left( \frac{sin \, x}{cos \, x} \right) \\ &= 1 \end{aligned}

But this is not the same as your linear function y = 1. $\displaystyle f \cdot g$ is going to look like a horizontal line with some "holes". The domain of f(x) = cos(x) is all real numbers, but the domain of g(x) = csc(x)tan(x) is all reals except multiples of pi/2. So the domain of $\displaystyle f \cdot g$ is the intersection of the two domains, or all reals except multiples of pi/2.
• Jul 20th 2010, 03:53 PM
titos
for the f(x) we see cos which is real in real numbers so domain of f is R
for g(x) it has y=css(x) wich means that the domain of that is R-{kπ},k in Z
and u=tan(x) has domain R-{kπ+π/2}k in Z
so domain for g(x) is R-{kπ+π/2,kπ},k in Z.ok now?
• Jul 20th 2010, 11:33 PM
Mathelogician
Quote:

Originally Posted by titos
for the f(x) we see cos which is real in real numbers so domain of f is R
for g(x) it has y=csc(x) wich means that the domain of that isR-{kπ},k in Z
and u=tan(x) has domain R-{kπ+π/2}k in Z
so domain for g(x) is R-{kπ+-π/2,kπ},k in Z.ok now?

That's Right!(Just do no forgot the -)
Now as eumyang said, the intersection of domains of f and g is the domain of f.g