Can you prove the definite value of $f(x)$?

[Albert1]

$f(x)=\dfrac {sin \,3x}{sin\, x}-\dfrac {cos\, 3x}{cos \,x}$

prove the value of $f(x)$ is always definite

 

[Albert1]

$f(x)=\dfrac {sin \,3x}{sin\, x}-\dfrac {cos\, 3x}{cos \,x}$

prove the value of $f(x)$ is always definite

of course :$sin\,x\,\times cos\,x \neq 0$
if $sin\,x\times \,cos\,x=0$ then $f(x)$ is meaningless
hint :$sin\,3x=?\,\,\,$ $cos\,3x=?$

 

[kaliprasad]

$f(x)=\dfrac {sin \,3x}{sin\, x}-\dfrac {cos\, 3x}{cos \,x}$

prove the value of $f(x)$ is always definite

I do not know what you mean by definiten but if you mean constant then

Spoiler

$\dfrac {sin \,3x}{sin\, x}-\dfrac {cos\, 3x}{cos \,x}$
= $\dfrac {sin \,3x\cos\, x-cos\, 3x\sin\, x}{cos \,x\sin\,x }$
= $\dfrac {sin \,2x}{cos \,x\sin\,x }$
= $\dfrac {2 sin \,2x}{2 cos \,x\sin\,x }$
= $\dfrac {2 sin \,2x}{\sin\,2x }$
= 2
which is constant

 

[Albert1]

I do not know what you mean by definiten but if you mean constant then

Spoiler

$\dfrac {sin \,3x}{sin\, x}-\dfrac {cos\, 3x}{cos \,x}$
= $\dfrac {sin \,3x\cos\, x-cos\, 3x\sin\, x}{cos \,x\sin\,x }$
= $\dfrac {sin \,2x}{cos \,x\sin\,x }$
= $\dfrac {2 sin \,2x}{2 cos \,x\sin\,x }$
= $\dfrac {2 sin \,2x}{\sin\,2x }$
= 2
which is constant

yes the definite value (constant) of f(x) is 2

 

[MarkFL]

$f(x)=\dfrac {sin \,3x}{sin\, x}-\dfrac {cos\, 3x}{cos \,x}$

prove the value of $f(x)$ is always definite

Here's another way to proceed:

Spoiler

We are given:

\(\displaystyle f(x)=\frac{\sin(3x)}{\sin(x)}-\frac{\cos(3x)}{\cos(x)}\)

Combine terms:

\(\displaystyle f(x)=\frac{\sin(3x)\cos(x)-\cos(3x)\sin(x)}{\sin(x)\cos(x)}\)

In the numerator, apply angle-difference identity for sine and in the denominator apply double-angle identity for sine:

\(\displaystyle f(x)=\frac{\sin(3x-x)}{\dfrac{1}{2}\sin(2x)}=2\)

 

[Albert1]

another solution:

Spoiler

$\dfrac {sin\,3x}{sin\,x}-\dfrac {cos\,3x}{cos\,x}$

$=\dfrac {3sin\,x-4sin^3x}{sin\,x}-\dfrac {4cos^3\,x -3cos\,x}{cosx}$

$=6-4(sin^2x+cos^2x)=6-4=2$