Albert1
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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
prove the value of $f(x)$ is always definite
of course :$sin\,x\,\times cos\,x \neq 0$Albert said:$f(x)=\dfrac {sin \,3x}{sin\, x}-\dfrac {cos\, 3x}{cos \,x}$
prove the value of $f(x)$ is always definite
Albert said:$f(x)=\dfrac {sin \,3x}{sin\, x}-\dfrac {cos\, 3x}{cos \,x}$
prove the value of $f(x)$ is always definite
yes the definite value (constant) of f(x) is 2kaliprasad 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
Albert said:$f(x)=\dfrac {sin \,3x}{sin\, x}-\dfrac {cos\, 3x}{cos \,x}$
prove the value of $f(x)$ is always definite