For Trigo Lovers-3

Geometry Level 3

Find the maximum value of ( 1 + sin ⁡ x ) ( 1 + cos ⁡ x ) \large (1+\sin x)(1+\cos x) correct upto 1 decimal place for x ∈ R x \in R .

This is a part of Trigo Plus! .


The answer is 2.91.

Ravi Dwivedi by Ravi Dwivedi , 24, India

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1 solution

Ravi Dwivedi
Jul 20, 2015

( 1 + sin ⁡ x ) ( 1 + cos ⁡ x ) ≤ ( 1 + sin ⁡ x ) 2 + ( 1 + cos ⁡ x ) 2 2 = 2 + 2 ( sin ⁡ x + cos ⁡ x ) + ( sin ⁡ 2 x + cos ⁡ 2 x ) 2 = 3 2 + ( sin ⁡ x + cos ⁡ x ) = 3 2 + 2 sin ⁡ ( x + π 4 ) ≤ 3 2 + 2 (1+\sin x)(1+\cos x) \leq \frac{(1+\sin x)^2+(1+\cos x)^2}{2}\\= \frac{2+2(\sin x+\cos x)+(\sin^2 x+ \cos^2x)}{2} =\frac{3}{2}+(\sin x+\cos x)\\=\frac{3}{2}+\sqrt{2}\sin (x+ \frac{\pi}{4}) \leq \frac{3}{2}+\sqrt{2} which is the required maximum value.

Note that equality holds when x = π 4 + 2 k π x=\frac{\pi}{4} +2k\pi , where k k is an integer

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

All that you have shown, is that the value is bounded above by 3 2 + 2 \frac{3}{2}+\sqrt{2} .

You still need to show that this value can indeed be achieved, IE that we have a found the least upper bound.

For example, it is obvious that ∣ ( 1 + sin ⁡ x ) ( 1 + cos ⁡ x ) ∣ ≤ ( 1 + 1 ) ( 1 + 1 ) = 4 | ( 1 + \sin x ) ( 1 + \cos x ) | \leq ( 1 + 1 ) ( 1 + 1 ) = 4 , but the answer is not 4.

The value is achieved for x = 4 5 ∘ x=45^\circ . We have ( 1 + 1 2 ) 2 = 3 2 + 2 (1+\frac{1}{\sqrt{2}})^2=\frac{3}{2}+\sqrt{2}

Marta Reece - 4 years, 1 month ago

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