Finding the maxima of a trigonometric function

Question

Show that the function $$\frac {\sin^2 x}{\sin(x+a) \sin(x+b)} \quad (0<a<b<\pi)$$ has an infinite number of minima equal to $0$ and maxima equal to $$-4 \sin a \sin b / \sin^2(a-b)$$

Using the quotient rule I found the minima quite easily. Just $(n \pi,0)$. I cannot find a nice way of deriving the expression for the maxima though. Any hints?

Answer 1

Let $f(x)$ be the function in the question. Notice that $$f(x)=\frac{1}{\sin(a)\sin(b)(\cot(x)+\cot(a))(\cot(x)+\cot(b))}$$

Differentiating both sides we obtain that $$f'(x) = \frac{\sin^2(x)}{\sin^4(x+a)\sin^4(x+b)}\sin(a)\sin(b)(2\cot(x)+\cot(a)+\cot(b))$$

From this we obtain that the extremum values of the function lies at when $\sin^2(x)=0$ or $2\cot(x)+\cot(a)+\cot(b)=0$. From the condition $2\cot(x)+\cot(a)+\cot(b)=0$ one can derive the required maxima.

Answer 2

Hint:

Let $y=\dfrac{\sin^2x}{\sin(x+a)\sin(x+b)}$

$$\iff\dfrac1y=(\cos a+\tan x\sin a)(\cos b+\tan x\sin b)$$ using $\dfrac{\sin(x+a)}{\sin x}=\dfrac{\sin x\cos a+\cos x\sin a}{\sin x}=?$

Now rearrange to form a Quadratic Equation in $\tan x$ which is real,

so the discriminant must be $\ge0$