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

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The function f(x) = (sin 3x/sin x) - (cos 3x/cos x) is discussed in terms of proving its definite value. It is established that f(x) is constant and equals 2, provided that sin x and cos x are not zero. The conversation emphasizes the need for sin x and cos x to be non-zero for f(x) to be meaningful. There is some confusion regarding the term "definite," with participants clarifying that it refers to a constant value. The discussion ultimately seeks to validate the constancy of f(x) across its domain.
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$f(x)=\dfrac {sin \,3x}{sin\, x}-\dfrac {cos\, 3x}{cos \,x}$

prove the value of $f(x)$ is always definite
 
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Albert said:
$f(x)=\dfrac {sin \,3x}{sin\, x}-\dfrac {cos\, 3x}{cos \,x}$

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 :$sin\,3x=?\,\,\,$ $cos\,3x=?$
 
Albert said:
$f(x)=\dfrac {sin \,3x}{sin\, x}-\dfrac {cos\, 3x}{cos \,x}$

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

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

$\dfrac {sin \,3x}{sin\, x}-\dfrac {cos\, 3x}{cos \,x}$
= $\dfrac {sin \,3x\cos\, x-cos\, 3x\sin\, x}{cos \,x\sin\,x }$
= $\dfrac {sin \,2x}{cos \,x\sin\,x }$
= $\dfrac {2 sin \,2x}{2 cos \,x\sin\,x }$
= $\dfrac {2 sin \,2x}{\sin\,2x }$
= 2
which is constant
 
kaliprasad said:
I do not know what you mean by definiten but if you mean constant then

$\dfrac {sin \,3x}{sin\, x}-\dfrac {cos\, 3x}{cos \,x}$
= $\dfrac {sin \,3x\cos\, x-cos\, 3x\sin\, x}{cos \,x\sin\,x }$
= $\dfrac {sin \,2x}{cos \,x\sin\,x }$
= $\dfrac {2 sin \,2x}{2 cos \,x\sin\,x }$
= $\dfrac {2 sin \,2x}{\sin\,2x }$
= 2
which is constant
yes the definite value (constant) of f(x) is 2
 
Albert said:
$f(x)=\dfrac {sin \,3x}{sin\, x}-\dfrac {cos\, 3x}{cos \,x}$

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

Here's another way to proceed:

We are given:

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

Combine terms:

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

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

$$f(x)=\frac{\sin(3x-x)}{\dfrac{1}{2}\sin(2x)}=2$$
 
another solution:
$\dfrac {sin\,3x}{sin\,x}-\dfrac {cos\,3x}{cos\,x}$

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

$=6-4(sin^2x+cos^2x)=6-4=2$
 
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