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The number of value of x where f(x)=Cosx + Cos(√2+x) attains it's maximum

I tried but after differentiating I have no clue how to proceed

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    Is it $\sqrt {2}+x $ or $\sqrt {2}x $? – 2017-02-04
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    Because if we differentiate and solve for $0$, we cannot expect a simple value. – 2017-02-04

2 Answers 2

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Using Prosthaphaeresis Formula,

$$\cos x+\cos(x+\sqrt2)=2\cos\dfrac{\sqrt2}2\cos\dfrac{2x+\sqrt2}2$$

Now for real $x,$

$$-1\le\cos\dfrac{2x+\sqrt2}2\le1$$

and $\cos y$ attains maximum i.e., $\cos y=1\iff y=2m\pi$ where $m$ is any integer

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Try,

$$\cos x+\cos (x+\sqrt{2})$$

$$=\cos x+ \cos x \cos (\sqrt{2})-\sin x \sin(\sqrt{2})$$

$$=(1+\cos \sqrt{2}) \cos x-\sin(\sqrt{2}) \sin x$$

$$=\langle (1+\cos \sqrt{2}),- \sin(\sqrt{2}) \rangle \cdot \langle \cos x, \sin x \rangle$$

Now use that $\vec a \cdot \vec b=|a||b|\cos(\theta)$, where $\theta$ is the angle between the vectors.

$$=\sqrt{(1+\cos \sqrt{2})^2+\sin(\sqrt{2})^2}\sqrt{1} \cos (...)$$

So I think the maximum is,

$$\sqrt{(1+\cos \sqrt{2})^2+\sin(\sqrt{2})^2}$$

Find what the angle between the vectors is (in terms of a function of $x$), and the answer follows.