Question

Find the maximum or minimum value of cosx + sinx-1​

Answer 1

Step-by-step explanation:

$Let y=\sin(x) + \cos(x) -1$

$\implies \: y = \sqrt{2} ( \frac{1}{ \sqrt{2} } \sin(x) + \frac{1}{ \sqrt{2} } \cos(x) ) - 1$

$\implies \: y = \sqrt{2} \sin(x + \frac{\pi}{4} ) - 1$

Now,

$- 1 \leqslant \sin(x + \frac{\pi}{4} ) \leqslant 1$

$\implies- \sqrt{2} \leqslant \sqrt{2} \sin(x + \frac{\pi}{4} ) \leqslant \sqrt{2}$

$\implies- \sqrt{2} - 1 \leqslant \sqrt{2} \sin(x + \frac{\pi}{4} ) - 1 \leqslant \sqrt{2} - 1$

Minimum value = $-(\sqrt{2}+1)$

Maximum value = $( \sqrt{2}-1)$