Mathematics - Trigonometric Ratios & Identities Question with Solution | TestHub

The function is given by $f(x)=({\sin }^{2}x+{\cos }^{2}x{)}^3-3{\sin }^{2}x{\cos }^{2}x({\sin }^{2}x+{\cos }^{2}x)$.

Using the identity ${\sin }^{2}x+{\cos }^{2}x=1$, we can simplify this to:

$$f(x)=(1{)}^3-3{\sin }^{2}x{\cos }^{2}x(1)=1-3{\sin }^{2}x{\cos }^{2}x$$

We know that $\sin (2x)=2\sin x\cos x$, so ${\sin }^2(2x)=4{\sin }^{2}x{\cos }^{2}x$.

Therefore, ${\sin }^{2}x{\cos }^{2}x=\frac{1}{4}{\sin }^2(2x)$.

Substituting this back into the expression for $f(x)$:

$$f(x)=1-3\left(\frac{1}{4}{\sin }^2(2x)\right)=1-\frac{3}{4}{\sin }^2(2x)$$

The range of ${\sin }^2(2x)$ is $[0,1]$.

To find the range of $f(x)$, we consider the minimum and maximum values of ${\sin }^2(2x)$:

When ${\sin }^2(2x)=0$, $f(x)=1-\frac{3}{4}(0)=1$.

When ${\sin }^2(2x)=1$, $f(x)=1-\frac{3}{4}(1)=1-\frac{3}{4}=\frac{1}{4}$.

Thus, the range of $f(x)$ is $\left[\frac{1}{4},1\right]$.

Note: While derivatives can be used to determine extrema, some questions are more efficiently solved by algebraic manipulation and understanding the range of trigonometric functions.