Find the Maximum and Minimum Value of sin x − cos x + 1

Solution

We know that the expression

\[ a\sin x+b\cos x \]

has maximum value

\[ \sqrt{a^2+b^2} \]

and minimum value

\[ -\sqrt{a^2+b^2} \]

Given expression:

\[ \sin x-\cos x+1 \]

Here,

\[ a=1, \qquad b=-1 \]

Now,

\[ \sqrt{a^2+b^2} = \sqrt{1^2+(-1)^2} \]

\[ = \sqrt{1+1} \]

\[ = \sqrt{2} \]

Therefore,

Maximum value of \[ \sin x-\cos x \] is \[ \sqrt{2} \]

So, maximum value of the given expression is

\[ \sqrt{2}+1 \]

Minimum value of \[ \sin x-\cos x \] is \[ -\sqrt{2} \]

So, minimum value of the given expression is

\[ 1-\sqrt{2} \]

Final Answer

\[ \boxed{\text{Maximum value }=1+\sqrt{2}} \]

\[ \boxed{\text{Minimum value }=1-\sqrt{2}} \]