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4.3.2 Hence deduce the minimum value of \( \sin x+\cos x \) and give the corresponding values of \( x \), for \( x \in\left[-360^{\circ} ; 360^{\circ}\right] \). TOTAL:

Ask by John Cox. in South Africa
Feb 21,2025

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The minimum value of \( \sin x + \cos x \) is \( -\sqrt{2} \), which occurs at \( x = 225^\circ \) and \( x = -135^\circ \) within the interval \([-360^\circ, 360^\circ]\).

Solution

To find the minimum value of \( \sin x + \cos x \), we can start by rewriting the expression in a more manageable form. 1. **Rewrite the expression**: We can express \( \sin x + \cos x \) in terms of a single sine function. We know that: \[ \sin x + \cos x = \sqrt{2} \left( \sin x \cdot \frac{1}{\sqrt{2}} + \cos x \cdot \frac{1}{\sqrt{2}} \right) \] This can be rewritten using the sine addition formula: \[ \sin x + \cos x = \sqrt{2} \sin\left(x + 45^\circ\right) \] 2. **Determine the range**: The sine function, \( \sin\left(x + 45^\circ\right) \), has a range of \([-1, 1]\). Therefore, the expression \( \sqrt{2} \sin\left(x + 45^\circ\right) \) will have a range of: \[ [-\sqrt{2}, \sqrt{2}] \] 3. **Find the minimum value**: The minimum value of \( \sin x + \cos x \) is thus: \[ -\sqrt{2} \] 4. **Find corresponding values of \( x \)**: To find the values of \( x \) for which \( \sin\left(x + 45^\circ\right) = -1 \): \[ x + 45^\circ = 270^\circ + k \cdot 360^\circ \quad (k \in \mathbb{Z}) \] This simplifies to: \[ x = 225^\circ + k \cdot 360^\circ \] Now, we need to find \( x \) within the interval \([-360^\circ, 360^\circ]\): - For \( k = 0 \): \( x = 225^\circ \) - For \( k = -1 \): \( x = 225^\circ - 360^\circ = -135^\circ \) Thus, the corresponding values of \( x \) are \( 225^\circ \) and \( -135^\circ \). ### Summary: - The minimum value of \( \sin x + \cos x \) is \( -\sqrt{2} \). - The corresponding values of \( x \) in the interval \([-360^\circ, 360^\circ]\) are \( 225^\circ \) and \( -135^\circ \).

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Bonus Knowledge

To find the minimum value of \( \sin x + \cos x \), we can use a clever trick by rewriting it as \( R \sin(x + \phi) \) where \( R = \sqrt{1^2 + 1^2} = \sqrt{2} \) and \( \phi = \frac{\pi}{4} \). The function \( R \sin(x + \phi) \) reaches its minimum value of \(-R\) when \( \sin(x + \phi) = -1\). Thus, the minimum value of \( \sin x + \cos x \) is \(-\sqrt{2}\). Now, to find the corresponding values of \( x \), we set \( x + \frac{\pi}{4} = \frac{3\pi}{2} + 2k\pi \) (for \( k \in \mathbb{Z} \)). Solving gives \( x = \frac{3\pi}{2} - \frac{\pi}{4} + 2k\pi = \frac{5\pi}{4} + 2k\pi \) or \( x = -\frac{7\pi}{4} + 2k\pi \). Converting to degrees, these correspond to \( x = 225^\circ \) or \( x = -315^\circ \) within the interval \( \left[-360^\circ; 360^\circ\right] \).

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