Question

Minimum value of sinx+ cosx

• A. √2
• B. -√2
• C. 1
• D. -1

Answer 1

Answer: (B)

-√2

Answer 2

To find the minimum value of $\sin x + \cos x$, we can use the amplitude-phase form for the sum of sine and cosine functions.

Any expression of the form $a \sin x + b \cos x$ can be rewritten as $R \sin(x + \alpha)$, where $R = \sqrt{a^2 + b^2}$.

In this case, $a=1$ and $b=1$. So, $R = \sqrt{1^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2}$.

Thus, $\sin x + \cos x = \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x \right)$. We know that $\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$ and $\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$. So, $\sin x + \cos x = \sqrt{2} \left( \sin x \cos\left(\frac{\pi}{4}\right) + \cos x \sin\left(\frac{\pi}{4}\right) \right)$.

Using the trigonometric identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$, we get: $\sin x + \cos x = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right)$.

The range of the sine function, $\sin(\theta)$, for any real $\theta$ is $[-1, 1]$. Therefore, the minimum value of $\sin\left(x + \frac{\pi}{4}\right)$ is $-1$.

Substituting this minimum value back into the expression: Minimum value of $\sin x + \cos x = \sqrt{2} \times (-1) = -\sqrt{2}$.

The minimum value of $\sin x + \cos x$ is $-\sqrt{2}$.