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# If $\alpha + \beta = c$ , where $\alpha ,\beta > 0$ each lying between 0 and $\dfrac{\pi }{2}$ and c is a constant, find the maximum or minimum value of A) $\sin \alpha + \sin \beta$B) $\sin \alpha .\sin \beta$C) $\tan \alpha + \tan \beta$

Last updated date: 13th Jun 2024
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Hint: We will use some formulas for trigonometric ratios to combine them somehow in the form $\alpha + \beta = c$ so that we get a constant term in the function and then vary the other one as per our requirements to get the maximum or minimum.

Complete step-by-step solution:
We will go through each of the function and part one by one:
Part A:
We have the function given by $\sin \alpha + \sin \beta$.
We know that we have a formula given by the following expression:-
$\Rightarrow \sin C + \sin D = 2\sin \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right)$
Applying this on the function we are given in part (a), we will get:-
$\Rightarrow \sin \alpha + \sin \beta = 2\sin \left( {\dfrac{{\alpha + \beta }}{2}} \right)\cos \left( {\dfrac{{\alpha - \beta }}{2}} \right)$
Now, since we are already given that $\alpha + \beta = c$. Therefore, we now have with us the following function:-
$\Rightarrow \sin \alpha + \sin \beta = 2\sin \left( {\dfrac{c}{2}} \right)\cos \left( {\dfrac{{\alpha - \beta }}{2}} \right)$
Since, c is a constant, therefore, $2\sin \left( {\dfrac{c}{2}} \right)$ is a constant.
Now, we know that the maximum value of a cosine function is when it takes the value 1.
Therefore, the maximum value of $\sin \alpha + \sin \beta = 2\sin \left( {\dfrac{c}{2}} \right)\cos \left( {\dfrac{{\alpha - \beta }}{2}} \right)$ will be $2\sin \left( {\dfrac{c}{2}} \right)$.
And, since $\alpha$ and $\beta$ both lie in the first quadrant. Therefore, $\alpha - \beta$ will either lie in first or in fourth quadrant in which in either case, its cosine can only be greater than or equal to 0.
Therefore, the minimum value of $\sin \alpha + \sin \beta = 2\sin \left( {\dfrac{c}{2}} \right)\cos \left( {\dfrac{{\alpha - \beta }}{2}} \right)$ is 0.
Part B:
We have the function given by $\sin \alpha .\sin \beta$.
We know that we have a formula given by the following expression:-
$\Rightarrow \sin C.\sin D = - \dfrac{1}{2}\left[ {\cos (C + D) - \cos (C - D)} \right]$
Applying this on the function we are given in part (a), we will get:-
$\Rightarrow \sin \alpha .\sin \beta = - \dfrac{1}{2}\left[ {\cos (\alpha + \beta ) - \cos (\alpha - \beta )} \right]$
Now, since we are already given that $\alpha + \beta = c$. Therefore, we now have with us the following function:-
$\Rightarrow \sin \alpha .\sin \beta = - \dfrac{1}{2}\left[ {\cos c - \cos (\alpha - \beta )} \right]$
Since, c is a constant, therefore, $- \dfrac{1}{2}\cos c$ is a constant.
Therefore, we are left with $\dfrac{1}{2}\cos (\alpha - \beta )$
Now, we know that the maximum value of a cosine function is when it takes the value 1.
Therefore, the maximum value of $\sin \alpha .\sin \beta = - \dfrac{{\cos c}}{2} + \dfrac{{\cos (\alpha - \beta )}}{2}$ will be $\dfrac{{1 - \cos c}}{2}$.
And, since $\alpha$ and $\beta$ both lie in the first quadrant. Therefore, $\alpha - \beta$ will either lie in first or in fourth quadrant in which in either case, its cosine can only be greater than or equal to 0.
Therefore, the minimum value of $\sin \alpha .\sin \beta = - \dfrac{{\cos c}}{2} + \dfrac{{\cos (\alpha - \beta )}}{2}$ is $- \dfrac{{\cos c}}{2}$.
Part C:
We have the function given by $\tan \alpha + \tan \beta$.
Now, since we are already given that $\alpha + \beta = c$. Therefore, we get:-
$\Rightarrow \beta = c - \alpha$
Applying this in the given function, we will get:-
$\Rightarrow f(\alpha ) = \tan \alpha + \tan \left( {c - \alpha } \right)$
Now, we will find its derivative using the fact that $\dfrac{d}{{dx}}(\tan x) = {\sec ^2}x$.
$\Rightarrow f'(\alpha ) = {\sec ^2}\alpha + {\sec ^2}\left( {c - \alpha } \right) \times ( - 1)$ (By chain rule)
$\Rightarrow f'(\alpha ) = {\sec ^2}\alpha - {\sec ^2}\left( {c - \alpha } \right)$
Now, for maximum or minimum, we need to put the derivative equal to 0:-
$\Rightarrow f'(\alpha ) = 0$
$\Rightarrow {\sec ^2}\alpha - {\sec ^2}\left( {c - \alpha } \right) = 0$
$\Rightarrow {\sec ^2}\alpha = {\sec ^2}\left( {c - \alpha } \right)$
Now, since $\sec A = \dfrac{1}{{\cos A}}$
$\Rightarrow {\cos ^2}\alpha = {\cos ^2}\left( {c - \alpha } \right)$
This is only possible if:
$\Rightarrow \alpha = 2n\pi \pm (c - \alpha )$
There arises two cases now:
Case 1: $\alpha = 2n\pi - (c - \alpha )$
$\Rightarrow \alpha = 2n\pi - c + \alpha$ which just eliminates $\alpha$ from both sides.
Case 2: $\alpha = 2n\pi + c - \alpha$
Taking $\alpha$ from RHS to LHS, we will get:-
$\Rightarrow 2\alpha = 2n\pi + c$
$\Rightarrow \alpha = n\pi + \dfrac{c}{2}$
Since, we are given that $\alpha ,\beta > 0$ each lies between 0 and $\dfrac{\pi }{2}$.
Therefore, n = 0.
$\Rightarrow \alpha = \dfrac{c}{2}$
Since $f(\alpha ) = \tan \alpha + \tan \left( {c - \alpha } \right)$
$\therefore f\left( {\dfrac{c}{2}} \right) = \tan \left( {\dfrac{c}{2}} \right) + \tan \left( {c - \dfrac{c}{2}} \right)$
$\Rightarrow f\left( {\dfrac{c}{2}} \right) = 2\tan \left( {\dfrac{c}{2}} \right)$ which is the maximum.

Note: The students must note that we checked for the slope that is the derivative of the function in the last part. Why does the derivative help us to see the maximum or minimum of any function?
Let us understand this in brief.
Since, we know that wherever the derivative is 0, it means the slope of the function is parallel to the x – axis there. That means, it is the point of inflexion, either the curve starts to rise or decline from here. Therefore, that point can either be the maximum or the minimum of the function.