Maximizing a trigonometric expression (IIT JEE prep question with cute solutions)

Found this cute problem in some IIT JEE prep material. The question as posed there, was a single answer multiple choice question, to find the maximum value. Anyway the puzzle here is:

For what $x \in [0, 2\pi]$ does the expression

$$ \cos x (\sin x + \sqrt{\sin^2 x + 1})$$

take the maximum value? What is the maximum value? No calculus allowed.

Scroll down for solutions.

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$\textbf{Method 1}$: One nice method is to use vector algebra. Consider the formula:

$$ u.v = |u | |v| \cos \theta$$

with $u = (\sin x, \cos x),  v= (\cos x, \sqrt{\sin^2 x + 1})$. 

The expression in question (say $f$) is $f = u.v$ with $|u| = 1$ and $|v| = \sqrt{2}$. Thus  $f \leq \sqrt{2}$. $f$ will be $ = \sqrt{2}$ iff $u$ and $v$ are parallel. And so we are looking for solutions of

$$\sin x \sqrt{\sin^2 x + 1} = \cos^2 x$$

Since $\cos x = 0$ cannot be a solution, we can divide by $\cos^2 x$ and rewrite as

$$\tan x \sqrt{\tan^2 x + \sec^2 x} = 1$$

If $t = \tan x$, then this is 

$$ t \sqrt{2t^2 + 1} = 1$$

Which is a quadractic in $t^2$ with the solution $ t  = \frac{1}{\sqrt{2}}$.

Thus $x$ must satisfy $\tan x = \frac{1}{\sqrt{2}}$. We can verify that it is indeed a solution.

$\textbf{Method 2}:$ This is also a nice solution.

Let $$f = \cos x (\sin x + \sqrt{\sin^2 x + 1})$$

Clearly $\cos x = 0$ cannot be it. So assume $\cos x \neq 0$.

Now this looks quite similar to formula for the root of a quadratic.

$$ f = \frac{ - ( -\sin x) + \sqrt{(-\sin x)^2 - 4 \left(\frac{\sec x}{2}\right) \left(\frac{-\cos x}{2}\right)}}{2 \times\frac{\sec x}{2}}$$

Thus $f$ is the root of $ax^2 + bx + c$ with $a = \frac{\sec x}{2}, b = -\sin x, c = \frac{-\cos x}{2}$.

Thus we have that

$$ \frac{\sec x}{2} f^2 - f \sin x - \frac{\cos x}{2} = 0$$

Multiplying by $2\sec x$ gives us

$$ \sec^2 x f^2 - 2f \tan x - 1 = 0$$

i.e

$$f^2 + \tan^2 x f^2 - 2f \tan x = 1$$

$$ f^2 + (f \tan x - 1)^2 = 2$$

Since $(f\tan x -1)^2 \ge 0$, we must have $f^2 \le 2$. For $f^2 = 2$ (and so $f = \sqrt{2}$) we need $f\tan x = 1$, ie $\tan x = \frac{1}{f} = \frac{1}{\sqrt{2}}$. This is easily verified.