The maximum value of sin x+cos x , x ∈ R is:

The correct answer is Option (2) → \(\sqrt{2}\)

Taking \(\sqrt{2}\) out and using \(\cos\frac{\pi}{4}=\sin\frac{\pi}{4}=\frac{1}{\sqrt{2}}\):

    \( \displaystyle \sin x+\cos x=\sqrt{2}\left(\frac{1}{\sqrt{2}}\sin x+\frac{1}{\sqrt{2}}\cos x\right) \)

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

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

The largest value of a sine is 1, reached at \(x=\frac{\pi}{4}\). So the maximum value is \(\sqrt{2}\).

Tip: The maximum of \(a\sin x+b\cos x\) is \(\sqrt{a^2+b^2}\), not a + b: sin x and cos x never equal 1 at the same x.