Question: "Find maximum value of (1+sin⁡x)(1+cos⁡x)"

reproacht3

Answered question

2022-01-15

Find maximum value of

(1 + \sin x) (1 + \cos x)

Answer & Explanation

Raymond Foley

Beginner2022-01-18Added 39 answers

Let

y = (1 + \sin x) (1 + \cos x) = 1 + \sin x + \cos x + \sin x \cos x

\sin x + \cos x = u, u^{2} = 1 + 2 \sin x \cos x \leq 2 \Rightarrow u \leq \sqrt{2}

y = 1 + u + \frac{u^{2} - 1}{2} = \frac{u^{2} + 2 u + 1}{2} = \frac{{(u + 1)}^{2}}{2}

Now

u + 1 \leq \sqrt{2} + 1

RizerMix

Expert2022-01-20Added 656 answers

(1 + \sin x) (1 + \cos x) = \frac{1}{2} (\sin x + \cos x + 1)^{2} = (\cos (x - \frac{π}{4}) + \frac{1}{\sqrt{2}})^{2}

and the extrema must be 0 and

(1 + \frac{1}{\sqrt{2}})^{2}

user_27qwe

Skilled2022-01-23Added 375 answers

By AM-GM and C-S we obtain:
$(1 + \sin x) (1 + \cos x) \leq (\frac{1 + + \sin x + 1 + \cos x}{2})^{2} \leq (\frac{2 + \sqrt{(1 + 1) (\sin^{2} x + \cos^{2} x)}}{2})^{2} = (1 + \frac{1}{\sqrt{2}})^{2}$
The equality occurs for $x = 45^{\circ}$ , which says that we got a maximal value. As you said, the minimal value is 0 and since our expression is continuous, we got a range, which you wrote. The inequality $\sin x + \cos x \leq \sqrt{2}$ we can prove by using AM-GM:
$\sin x + \cos x \leq | \sin x | + | \cos x | = \sqrt{1 + 2 | \sin x | \cdot | \cos x |} \leq \sqrt{1 + | \sin x |^{2} + | \cos x |^{2}} = \sqrt{2}$

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