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Can $f(x)$ take the value of $\sqrt{2}$, where $$f(x) = \sin(x) + (\cos(x))^2\quad ?$$

When equating the value of $f(x)$ to $\sqrt{2}$, it gives imaginary values of $\sin(x)$.

Thanks in advance.

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Do you know how to find the max and min of a function? If so, you could see if the max is larger than $2^{1/2)$ or not. –  AQP Mar 5 '12 at 6:50
    
The function is $2\pi$-periodic, so you just need to check the interval $[0,2\pi)$. What does the derivative tell you? –  Antonio Vargas Mar 5 '12 at 6:51
    
@AntonioVargas Actually i did not try by using derivation . I made a quadratiic equation and then tried to find the roots of this equation. –  vikiiii Mar 5 '12 at 7:14
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I thought the main goal should be that of leading, or at least making an effort to lead the OP to an answer; not just answering the question. –  AQP Mar 5 '12 at 7:44
    
@Antonio: I was referring to the answers below. Yours I thought was fine. I mean, for someone starting, to throw in a lot of machinery.... I don't know. –  AQP Mar 5 '12 at 7:54
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4 Answers 4

up vote 14 down vote accepted

Since $\cos^2x=1-\sin^2 x$, our equation becomes $\sin^2 x-\sin x+\sqrt{2}-1=0$.

Solve the equation $w^2-w+\sqrt{2}-1=0$. The roots are $$w=\frac{1\pm \sqrt{1-4(\sqrt{2}-1})}{2}.$$ The discriminant $1-4(\sqrt{2}-1)$ is negative, so our quadratic in $w$ has no real roots. It follows that there cannot be a real number $x$ such that $\sin x+\cos^2 x=\sqrt{2}$.

This does not fully answer the question. We can ask whether there are complex non-real solutions. For that, look at the two non-real values of $w$ obtained above. Call them $w_1$ and $w_2$. Recall that if $z$ is a complex number, then $$\sin z=\frac{e^{iz}-e^{-iz}}{2i}.$$ So we want to solve the equations $$e^{iz}-e^{-iz}=2iw_j \quad (j=1,2).$$ Multiply both sides by $e^{iz}$, and simplify. We get $$e^{2iz}-2iw_je^{iz}-1=0.$$ Make the substitution $u=e^{iz}$. We arrive at the equations $$u^2-2iw_j u-1=0 \quad (j=1,2).$$ These are a quadratic equation with complex coefficients. One can write down the solutions in more or less the usual way, and from them obtain the values of $z$ that work. There are infinitely many of them, just like there are, for example, infinitely many $x$ such that $\sin x=1/2$.

Remark To show that there are no real solutions, one does not need to recall the Quadratic Formula. We are interested in the equation $\sin^2 x-\sin x-1+\sqrt{2}=0$.

If we complete the square, we get $$\sin^2 x-\sin x-1=\left(\sin x-\frac{1}{2}\right)^2-\frac{5}{4}.$$ Since $\left(\sin x-\frac{1}{2}\right)^2\ge 0$, it follows that $\sin^2 x-\sin x-1$ must always be $-\frac{5}{4}$. Since $\sqrt{2}-\frac{5}{4}$ is positive, we conclude that we cannot have $\sin^2 x-\sin x-1+\sqrt{2}=0$. .

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Thanks @Andre.. –  vikiiii Mar 5 '12 at 10:46
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The local extrema will occur where $f\,'(x)=0$, or when factored, $(\cos x)(1-2\sin x)=0$. We can use the identity $\sin^2 x+\cos^2x=1$ to compute the possible extrema as

$$\cos =0 \implies \qquad \pm\sqrt{1-0^2}+0^2=\pm1 $$

$$\sin = 1/2 \implies \qquad \frac{1}{2}+\left|1-\frac{1}{2^2}\right|=5/4.$$

Of course $f$ is periodic and bounded so its image must be $(-1,5/4)$, which does not include $\sqrt{2}$.

$\hskip 1.6in$ enter image description here

This reasoning applies to the reals. Allowing complex arguments is another story.

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You ask if $f(x) = \sin x + \cos ^2 x$ can be equal to $\sqrt 2$. In short, no, it can't.

Now is when I take a moment to thank anon for noting that $1 < \sqrt 2$. That whole arithmetic thing.... yeah. Anyhow.

If one uses the first derivative test, then you can note that the max should occur around $\pi/6$ or $5 \pi / 6$ (plus multiples of $2\pi$ if desired). At both of these values, the function is a mere $1.25$, which (I hope anon checks me again) is less than $\sqrt 2$.

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$\sqrt{2}>1$ no? –  anon Mar 5 '12 at 6:46
    
@anon: Thank you for catching my error as well. Some sort of silliness must be in the air tonight. –  Zev Chonoles Mar 5 '12 at 7:02
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Put $\sin x =:u$. Then the question is whether the function $$g(u):=u+(1-u^2)={5\over4}-\Bigl(u-{1\over2}\Bigr)^2$$ can take the value $\sqrt{2}$ in the interval $-1\leq u\leq 1$. This is obviously not the case.

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