Using the identity

$$ \sin^2 A-\sin^2 B=\sin(A+B)\sin(A-B), $$

we get

$$ f(x)=\sin^2(7x)-\sin^2(5x) $$

$$ =\sin(12x)\sin(2x). $$

Now check the options.

For Option B, if

$$ 0

then

$$ 0<12x<\frac{\pi}{4} $$

and

$$ 0<2x<\frac{\pi}{24}. $$

Both $\sin(12x)$ and $\sin(2x)$ are positive. So

$$ f(x)=\sin(12x)\sin(2x)>0. $$

So Option B is true.

For Option C,

$$ f\left(x+\frac{\pi}{2}\right) =\sin\left(12x+6\pi\right)\sin(2x+\pi). $$

Now,

$$ \sin(12x+6\pi)=\sin(12x) $$

and

$$ \sin(2x+\pi)=-\sin(2x). $$

$$ f\left(x+\frac{\pi}{2}\right) =-\sin(12x)\sin(2x) =-f(x). $$

Hence,

$$ f\left(x+\frac{\pi}{2}\right)+f(x)=0. $$

So Option C is true.

For Option D,

$$ f\left(\frac{\pi}{12}\right) =\sin\left(12\cdot \frac{\pi}{12}\right)\sin\left(2\cdot \frac{\pi}{12}\right) $$

$$ =\sin \pi \cdot \sin\frac{\pi}{6} =0. $$

So Option D is true.

Now check Option A.

Option A says that $f$ is increasing on

$$ \left(\frac{3\pi}{2},2\pi\right). $$

Take

$$ x=\frac{5\pi}{3}. $$

Then

$$ f\left(\frac{5\pi}{3}\right) =\sin(20\pi)\sin\left(\frac{10\pi}{3}\right) =0. $$

Now take a slightly bigger value,

$$ x=\frac{17\pi}{10}. $$

Clearly,

$$ \frac{5\pi}{3}<\frac{17\pi}{10}<2\pi. $$

Now,

$$ f\left(\frac{17\pi}{10}\right) =\sin\left(\frac{102\pi}{5}\right)\sin\left(\frac{17\pi}{5}\right). $$

Here,

$$ \sin\left(\frac{102\pi}{5}\right)=\sin\left(20\pi+\frac{2\pi}{5}\right)>0 $$

and

$$ \sin\left(\frac{17\pi}{5}\right)=\sin\left(3\pi+\frac{2\pi}{5}\right)<0. $$

$$ f\left(\frac{17\pi}{10}\right)<0. $$

Thus, for a larger value of $x$, the value of $f(x)$ becomes smaller:

$$ f\left(\frac{17\pi}{10}\right) < f\left(\frac{5\pi}{3}\right). $$

So $f$ is not increasing on $\left(\frac{3\pi}{2},2\pi\right)$.

Hence, the statement which is NOT true is

$$ \boxed{\text{Option A}} $$

Trigonometric Ratios And Identities

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