Question

If $\alpha,\beta$ are two acute angles such that $\alpha + \beta = \gamma$, where $\gamma$ is a constant, then maximum value of $(sin\alpha + cos\alpha + sin\beta + cos\beta)$ is

• A. $2\sqrt{2}$
• B. $2\sqrt{2}sin(\frac{\pi}{4}+\frac{\gamma}{2})$
• C. $2\sqrt{2}sin(\frac{\pi}{4}+\gamma)$
• D. $\sqrt{2}sin(\frac{\pi}{4}+\frac{\gamma}{2})$

Answer 1

Answer: (B)

2\sqrt{2}sin(\frac{\pi}{4}+\frac{\gamma}{2})

Answer 2

Let the given expression be $E$.
$E = \sin\alpha + \cos\alpha + \sin\beta + \cos\beta$

We know the identity $a \sin x + b \cos x = \sqrt{a^2+b^2} \sin(x+\delta)$, where $\cos\delta = \frac{a}{\sqrt{a^2+b^2}}$ and $\sin\delta = \frac{b}{\sqrt{a^2+b^2}}$.
For $a=1, b=1$, we have $\sqrt{1^2+1^2} = \sqrt{2}$.
So, $\sin x + \cos x = \sqrt{2} \left( \frac{1}{\sqrt{2}}\sin x + \frac{1}{\sqrt{2}}\cos x \right) = \sqrt{2} \left( \cos\frac{\pi}{4}\sin x + \sin\frac{\pi}{4}\cos x \right)$.
Using the formula $\sin(A+B) = \sin A \cos B + \cos A \sin B$, we get:
$\sin x + \cos x = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right)$.

Applying this to the expression $E$:
$E = \sqrt{2} \sin\left(\alpha + \frac{\pi}{4}\right) + \sqrt{2} \sin\left(\beta + \frac{\pi}{4}\right)$
$E = \sqrt{2} \left[ \sin\left(\alpha + \frac{\pi}{4}\right) + \sin\left(\beta + \frac{\pi}{4}\right) \right]$

Now, use the sum-to-product trigonometric identity: $\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$.
Let $A = \alpha + \frac{\pi}{4}$ and $B = \beta + \frac{\pi}{4}$.
Then,
$A+B = \left(\alpha + \frac{\pi}{4}\right) + \left(\beta + \frac{\pi}{4}\right) = \alpha + \beta + \frac{\pi}{2}$.
Given that $\alpha + \beta = \gamma$, so $A+B = \gamma + \frac{\pi}{2}$.
Therefore, $\frac{A+B}{2} = \frac{\gamma}{2} + \frac{\pi}{4}$.

And,
$A-B = \left(\alpha + \frac{\pi}{4}\right) - \left(\beta + \frac{\pi}{4}\right) = \alpha - \beta$.
Therefore, $\frac{A-B}{2} = \frac{\alpha - \beta}{2}$.

Substitute these into the expression for $E$:
$E = \sqrt{2} \left[ 2 \sin\left(\frac{\gamma}{2} + \frac{\pi}{4}\right) \cos\left(\frac{\alpha - \beta}{2}\right) \right]$
$E = 2\sqrt{2} \sin\left(\frac{\pi}{4} + \frac{\gamma}{2}\right) \cos\left(\frac{\alpha - \beta}{2}\right)$

To find the maximum value of $E$, we need to maximize the term $\cos\left(\frac{\alpha - \beta}{2}\right)$, since $2\sqrt{2} \sin\left(\frac{\pi}{4} + \frac{\gamma}{2}\right)$ is a constant (as $\gamma$ is a constant).
The maximum value of the cosine function is 1.
So, $\cos\left(\frac{\alpha - \beta}{2}\right)$ will be maximum when it equals 1.
This occurs when $\frac{\alpha - \beta}{2} = 0$, which implies $\alpha - \beta = 0$, or $\alpha = \beta$.

Given that $\alpha$ and $\beta$ are acute angles, $0 < \alpha < \frac{\pi}{2}$ and $0 < \beta < \frac{\pi}{2}$.
If $\alpha = \beta$, then $\alpha + \beta = 2\alpha = \gamma$.
So, $\alpha = \beta = \frac{\gamma}{2}$.
For $\alpha$ and $\beta$ to be acute angles, we must have $0 < \frac{\gamma}{2} < \frac{\pi}{2}$, which implies $0 < \gamma < \pi$. This condition is consistent with the problem statement.

When $\cos\left(\frac{\alpha - \beta}{2}\right) = 1$, the maximum value of $E$ is:
$E_{max} = 2\sqrt{2} \sin\left(\frac{\pi}{4} + \frac{\gamma}{2}\right) \times 1$
$E_{max} = 2\sqrt{2} \sin\left(\frac{\pi}{4} + \frac{\gamma}{2}\right)$