Question

Maximum value of sin x + cos x is equal to

Answer 1

$\sin(x )+ \cos( x) \\ = > \sin(x) + \sin(90 - x) \\ = > 2 \sin( \frac{x + 90 - x}{2} ) \cos( \frac{x - 90 + x}{2} ) \\ = > 2 \sin(45) \cos(x - 45) \\ = > 2 \times \frac{1}{ \sqrt{2} } \times \cos(x - 45) \\ = > \sqrt{2} \cos(x - 45)$
Now since cos x can take maximum value of 1, therefore maximum value of sin x + cos x =
$\sqrt{2}$
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