Finding the Maximum Value of a Trigonometric Expression in an Interval

We are asked to find the point \(x\) in the interval \(\left( 0,\frac{{\rm{\pi }}}{2} \right)\) where the function \(f(x) = \sin \left( {x + \frac{{\rm{\pi }}}{6}} \right) + \cos \left( {x + \frac{{\rm{\pi }}}{6}} \right)\) attains its maximum value. We can approach this problem using trigonometric identities or calculus.

Method 1: Using Trigonometric Identities (Amplitude and Phase Shift)

The expression is in the form \(a \sin \theta + b \cos \theta\), where \(\theta = x + \frac{\pi}{6}\), \(a=1\), and \(b=1\). This form can be rewritten as \(R \sin(\theta + \alpha)\) or \(R \cos(\theta - \alpha)\), where \(R = \sqrt{a^2 + b^2}\) is the amplitude and \(\alpha\) is the phase shift.

First, calculate the amplitude \(R\):

\(R = \sqrt{1^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2}\)

Next, find the phase shift \(\alpha\). We have \(\cos \alpha = \frac{a}{R} = \frac{1}{\sqrt{2}}\) and \(\sin \alpha = \frac{b}{R} = \frac{1}{\sqrt{2}}\). The value of \(\alpha\) in \([0, 2\pi)\) that satisfies these conditions is \(\alpha = \frac{\pi}{4}\).

So, the expression can be written as:

\(\sin \left( x + \frac{\pi}{6} \right) + \cos \left( x + \frac{\pi}{6} \right) = \sqrt{2} \sin \left( \left(x + \frac{\pi}{6}\right) + \frac{\pi}{4} \right)\)

Simplify the argument of the sine function:

\(x + \frac{\pi}{6} + \frac{\pi}{4} = x + \frac{2\pi}{12} + \frac{3\pi}{12} = x + \frac{5\pi}{12}\)

Thus, the function is \(f(x) = \sqrt{2} \sin \left( x + \frac{5\pi}{12} \right)\).

Now consider the given interval for \(x\): \(\left( 0,\frac{{\rm{\pi }}}{2} \right)\).

Let the argument be \(u = x + \frac{5\pi}{12}\). We need to find the range of \(u\) when \(x \in \left( 0,\frac{{\rm{\pi }}}{2} \right)\).

• As \(x \to 0^+\), \(u \to 0 + \frac{5\pi}{12} = \frac{5\pi}{12}\).
• As \(x \to \left(\frac{\pi}{2}\right)^-\), \(u \to \frac{\pi}{2} + \frac{5\pi}{12} = \frac{6\pi}{12} + \frac{5\pi}{12} = \frac{11\pi}{12}\).

So, the argument \(u\) is in the interval \(\left( \frac{5\pi}{12}, \frac{11\pi}{12} \right)\).

The sine function \(\sin(u)\) attains its maximum value of 1 when \(u = \frac{\pi}{2}\). We check if \(\frac{\pi}{2}\) lies within the interval \(\left( \frac{5\pi}{12}, \frac{11\pi}{12} \right)\).

• \(\frac{5\pi}{12} = \frac{10\pi}{24}\)
• \(\frac{\pi}{2} = \frac{12\pi}{24}\)
• \(\frac{11\pi}{12} = \frac{22\pi}{24}\)

Clearly, \(\frac{5\pi}{12} < \frac{\pi}{2} < \frac{11\pi}{12}\). So, the maximum of \(\sin(u)\) occurs at \(u = \frac{\pi}{2}\) within this range.

The maximum value of \(f(x) = \sqrt{2} \sin(u)\) is \(\sqrt{2} \times 1 = \sqrt{2}\), and it occurs when \(u = x + \frac{5\pi}{12} = \frac{\pi}{2}\).

Solve for \(x\):

\(x = \frac{\pi}{2} - \frac{5\pi}{12} = \frac{6\pi - 5\pi}{12} = \frac{\pi}{12}\)

Check if \(x = \frac{\pi}{12}\) is in the interval \(\left( 0,\frac{{\rm{\pi }}}{2} \right)\): \(0 < \frac{\pi}{12} < \frac{\pi}{2}\) (since \(1/12 < 1/2\)). Yes, it is.

Therefore, the maximum value is attained at \(x = \frac{\pi}{12}\).

Method 2: Using Calculus (Differentiation)

Let \(f(x) = \sin \left( x + \frac{\pi}{6} \right) + \cos \left( x + \frac{\pi}{6} \right)\). To find the maximum value, we can find the critical points by setting the first derivative \(f'(x)\) equal to zero.

Calculate the first derivative \(f'(x)\):

\(f'(x) = \frac{d}{dx} \left( \sin \left( x + \frac{\pi}{6} \right) \right) + \frac{d}{dx} \left( \cos \left( x + \frac{\pi}{6} \right) \right)\)

Using the chain rule, \(\frac{d}{dx}\sin(u) = \cos(u) \frac{du}{dx}\) and \(\frac{d}{dx}\cos(u) = -\sin(u) \frac{du}{dx}\). Here \(u = x + \frac{\pi}{6}\), so \(\frac{du}{dx} = 1\).

\(f'(x) = \cos \left( x + \frac{\pi}{6} \right) \cdot 1 - \sin \left( x + \frac{\pi}{6} \right) \cdot 1\)

\(f'(x) = \cos \left( x + \frac{\pi}{6} \right) - \sin \left( x + \frac{\pi}{6} \right)\)

Set \(f'(x) = 0\) to find critical points:

\(\cos \left( x + \frac{\pi}{6} \right) - \sin \left( x + \frac{\pi}{6} \right) = 0\)

\(\cos \left( x + \frac{\pi}{6} \right) = \sin \left( x + \frac{\pi}{6} \right)\)

Assuming \(\cos \left( x + \frac{\pi}{6} \right) \neq 0\), we can divide both sides by \(\cos \left( x + \frac{\pi}{6} \right)\):

\(\frac{\sin \left( x + \frac{\pi}{6} \right)}{\cos \left( x + \frac{\pi}{6} \right)} = 1\)

\(\tan \left( x + \frac{\pi}{6} \right) = 1\)

The general solution for \(\tan \theta = 1\) is \(\theta = n\pi + \frac{\pi}{4}\), where \(n\) is an integer.

So, \(x + \frac{\pi}{6} = n\pi + \frac{\pi}{4}\)

\(x = n\pi + \frac{\pi}{4} - \frac{\pi}{6}\)

\(x = n\pi + \frac{3\pi - 2\pi}{12}\)

\(x = n\pi + \frac{\pi}{12}\)

We are looking for values of \(x\) in the interval \(\left( 0,\frac{{\rm{\pi }}}{2} \right)\). Let's test integer values for \(n\).

• If \(n=0\), \(x = 0 \cdot \pi + \frac{\pi}{12} = \frac{\pi}{12}\). This value is in the interval \(\left( 0,\frac{{\rm{\pi }}}{2} \right)\) because \(0 < \frac{\pi}{12} < \frac{6\pi}{12} = \frac{\pi}{2}\).
• If \(n=1\), \(x = 1 \cdot \pi + \frac{\pi}{12} = \frac{13\pi}{12}\). This is outside the interval \(\left( 0,\frac{{\rm{\pi }}}{2} \right)\) because \(\frac{13\pi}{12} > \frac{6\pi}{12} = \frac{\pi}{2}\).
• If \(n=-1\), \(x = -1 \cdot \pi + \frac{\pi}{12} = -\frac{11\pi}{12}\). This is outside the interval \(\left( 0,\frac{{\rm{\pi }}}{2} \right)\) because it's negative.

The only critical point in the interval \(\left( 0,\frac{{\rm{\pi }}}{2} \right)\) is \(x = \frac{\pi}{12}\).

To determine if this is a maximum, we can analyze the second derivative \(f''(x)\) or the sign changes of \(f'(x)\).

\(f''(x) = \frac{d}{dx} \left( \cos \left( x + \frac{\pi}{6} \right) - \sin \left( x + \frac{\pi}{6} \right) \right)\)

\(f''(x) = -\sin \left( x + \frac{\pi}{6} \right) - \cos \left( x + \frac{\pi}{6} \right)\)

Evaluate \(f''\) at the critical point \(x = \frac{\pi}{12}\). The argument is \(x + \frac{\pi}{6} = \frac{\pi}{12} + \frac{\pi}{6} = \frac{3\pi}{12} = \frac{\pi}{4}\).

\(f''\left( \frac{\pi}{12} \right) = -\sin \left( \frac{\pi}{4} \right) - \cos \left( \frac{\pi}{4} \right) = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = -\sqrt{2}\)

Since \(f''\left( \frac{\pi}{12} \right) < 0\), the function has a local maximum at \(x = \frac{\pi}{12}\). As this is the only critical point in the open interval, this local maximum is also the absolute maximum in the interval.

Conclusion

Both methods show that the maximum value of the function \(\sin \left( {x + \frac{{\rm{\pi }}}{6}} \right) + \cos \left( {x + \frac{{\rm{\pi }}}{6}} \right)\) in the interval \(\left( 0,\frac{{\rm{\pi }}}{2} \right)\) is attained at \(x = \frac{\pi}{12}\).

Let's verify the value of the expression at this point:

\(f\left(\frac{\pi}{12}\right) = \sin \left( \frac{\pi}{12} + \frac{\pi}{6} \right) + \cos \left( \frac{\pi}{12} + \frac{\pi}{6} \right)\)

\(f\left(\frac{\pi}{12}\right) = \sin \left( \frac{3\pi}{12} \right) + \cos \left( \frac{3\pi}{12} \right)\)

\(f\left(\frac{\pi}{12}\right) = \sin \left( \frac{\pi}{4} \right) + \cos \left( \frac{\pi}{4} \right)\)

\(f\left(\frac{\pi}{12}\right) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2}\)

This matches the maximum amplitude \(\sqrt{2}\) calculated in Method 1.

The maximum value is attained at \(x = \frac{\pi}{12}\).

Revision Table: Key Concepts for Trigonometric Maximums

Understanding how to find the maximum value of trigonometric functions is crucial. Here's a quick revision:

• The expression \(a \sin \theta + b \cos \theta\) can be written as \(R \sin(\theta + \alpha)\) where \(R = \sqrt{a^2 + b^2}\). The maximum value is \(R\) and the minimum value is \(-R\).
• The maximum of \(\sin(\phi)\) is 1, occurring when \(\phi = \frac{\pi}{2} + 2n\pi\) for integer \(n\).
• The maximum of \(\cos(\phi)\) is 1, occurring when \(\phi = 2n\pi\) for integer \(n\).
• To find the maximum/minimum of a function \(f(x)\) in an interval, find critical points by setting \(f'(x) = 0\), and check the function value at these points and the endpoints of the interval (if closed). For an open interval, analyze the behavior of the function around the critical point.

Additional Information on Finding Maximum Value

When finding the maximum value of a trigonometric expression like this, identifying the amplitude is key. The amplitude \(R\) gives the absolute maximum (and minimum) value of the expression \(R \sin(\theta + \alpha)\). However, the question specifically asks *where* in the given interval the maximum is attained, which requires finding the value of \(x\).

The method using differentiation confirms the location of local extrema. For a continuous function on an open interval, if there is a unique local maximum within the interval, it is also the absolute maximum in that interval. In this case, \(x = \frac{\pi}{12}\) is that unique point.

It's also important to pay attention to the given interval. An open interval excludes the endpoints, so the maximum must occur at a critical point within the interval.