Animated Solution for Mathematics - Trigonometry: Given $A={\sin }^2\theta +{\cos }^4\theta$ then for all real values of $\theta$

The expression provided is $A={\sin }^2\theta +{\cos }^4\theta$. The immediate challenge here is the mix of sine and cosine functions.

In trigonometry, as in life, complexity often arises from mixing different "languages." Our goal is to translate this into a single, unified language using the fundamental identity ${\cos }^2\theta =1-{\sin }^2\theta$.

By substituting this identity, we transform the expression into:

Now, let us expand this expression carefully using the identity $(a-b{)}^2=a^2-2ab+b^2$. Applying this to the term $(1-{\sin }^2\theta {)}^2$, we obtain $1-2{\sin }^2\theta +{\sin }^4\theta$.

Adding the initial ${\sin }^2\theta$ back into the mix, our expression becomes:

Combining the like terms, ${\sin }^2\theta -2{\sin }^2\theta$ simplifies to $-{\sin }^2\theta$. Thus, we arrive at the elegant quadratic form:

Domain Constraints

Here is where many students stumble. We define a new variable $x={\sin }^2\theta$, turning our expression into the function $f(x)=x^2-x+1$.

We must strictly define the domain of $x$. Since $\sin \theta$ ranges from $-1$ to $1$, its square, ${\sin }^2\theta$, must lie within the interval $[0,1]$.

This is the crucial constraint that defines our physical reality. By respecting this interval, we ensure our solution remains valid within the trigonometric context.

Vertex and Boundaries

We now analyze the parabola $f(x)=x^2-x+1$ on the interval $x\in [0,1]$. To find the minimum, we complete the square:

The vertex is located at $x=\frac{1}{2}$. Since $\frac{1}{2}$ is within our domain $[0,1]$, the minimum value is $f(\frac{1}{2})=\frac{3}{4}$.

For a parabola, the maximum on a closed interval must occur at the boundaries. Checking the endpoints:

Thus, the maximum value is $1$. We have successfully navigated the trap, and the range of $A$ is $[\frac{3}{4},1]$.