Question

The minimum value of \( \sin x + \cos x \) is:

• A. \( \sqrt{2} \)
• B. \( -\sqrt{2} \)
• C. \( \frac{1}{\sqrt{2}} \)
• D. \( -\frac{1}{\sqrt{2}} \)
• E. \( 1 \)

Hint:

Another useful form is \[ \sin x+\cos x=\sqrt{2}\sin\left(x+\frac{\pi}{4}\right) \] Since the minimum value of sine is \(-1\), the minimum value becomes \(-\sqrt{2}\).

Answer 1

The Correct Option is B

Concept: For an expression of the form \[ a\sin x+b\cos x \] the maximum value is \[ \sqrt{a^2+b^2} \] and the minimum value is \[ -\sqrt{a^2+b^2} \]

Step 1: Identify the coefficients.
Given expression: \[ \sin x+\cos x \] Comparing with \[ a\sin x+b\cos x \] we get \[ a=1,\qquad b=1 \]

Step 2: Find the maximum and minimum values.
\[ \sqrt{a^2+b^2} = \sqrt{1^2+1^2} = \sqrt{2} \] Hence, \[ -\sqrt{2} \le \sin x+\cos x \le \sqrt{2} \] Therefore, the minimum value is \[ \boxed{-\sqrt{2}} \]