Animated Solution for Mathematics - Trigonometry: If the equation ${\cos }^4\theta +{\sin }^4\theta +\lambda =0$ has real solutions for $\theta$, then $\lambda$ lies in the interval

Imagine you are standing on a tightrope, balancing between the world of algebra and the world of trigonometry. We are given the equation ${\cos }^4\theta +{\sin }^4\theta +\lambda =0$ and asked to find the interval for $\lambda$ such that real solutions for $\theta$ exist.

First, let's isolate our parameter $\lambda$. By moving everything else to the other side, we get:

Let's define a new function $f(\theta )={\cos }^4\theta +{\sin }^4\theta$. Our entire mission now is to find the range of this function $f(\theta )$. Once we have that, finding the range of $\lambda$ will be a simple matter of negation.

Don't let those powers of four intimidate you. We can use the classic algebraic identity $a^2+b^2=(a+b{)}^2-2ab$.

If we let $a={\cos }^2\theta$ and $b={\sin }^2\theta$, our function $f(\theta )$ transforms beautifully into:

This is the core of the problem. We have taken a complex expression and reduced it to something much more manageable.

Look closely at the term $({\cos }^2\theta +{\sin }^2\theta )$. That is simply $1$, our most fundamental trigonometric identity!

So, the expression collapses to $f(\theta )=1-2{\sin }^2\theta {\cos }^2\theta$. We still have a product of sine and cosine, but we know the double angle formula: $\sin 2\theta =2\sin \theta \cos \theta$.

If we square both sides, we get ${\sin }^{2}2\theta =4{\sin }^2\theta {\cos }^2\theta$. Dividing by $2$, we find that:

Substituting this back into our function, we get:

Now, for the final act. We know that for any real $\theta$, the value of ${\sin }^{2}2\theta$ is bounded between $0$ and $1$.

When ${\sin }^{2}2\theta =0$, $f(\theta )$ reaches its maximum value of $1$. When ${\sin }^{2}2\theta =1$, $f(\theta )$ reaches its minimum value of:

Thus, the range of $f(\theta )$ is $[\frac{1}{2},1]$. Since $\lambda =-f(\theta )$, we multiply the entire inequality by $-1$, which flips the signs.

The final interval for the parameter is $\lambda \in [-1,-\frac{1}{2}]$.