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.
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.
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 $\sin^2 x-\sin x-1+\sqrt{2}=0$. .
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 $\sqrt{2}$.
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This reasoning applies to the reals. Allowing complex arguments is another story.
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 $\sqrt 2$.
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.