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Finding range of function

$\displaystyle f(x)=\cos(x)\sin(x)+\cos(x)\sqrt{\sin^2(x)+\sin^2(\alpha)}$

I have use Algebric inequality

$\displaystyle -(a^2+b^2)\leq 2ab\leq (a^2+b^2)$

$\displaystyle (\cos^2(x)+\sin^2(x))\leq 2\cos(x)\sin(x)\leq \cos^2(x)+\sin^2(x)\cdots (1)$

And

$\displaystyle -[\cos^2(x)+\sin^2(x)+\sin^2(\alpha)]\leq 2\cos(x)\sqrt{\sin^2(x)+\sin^2(\alpha)}\leq [\cos^2(x)+\sin^2(x)+\sin^2(\alpha)]\cdots (2)$

Adding $(1)$ and $(2)$

$\displaystyle -(1+1+\sin^2(\alpha))\leq 2\cos(x)[\sin(x)+\sqrt{\sin^2(x)+\sin^2(\alpha)}]\leq (1+1+\sin^2(\alpha))$

$\displaystyle -\bigg(1+\frac{\sin^2(\alpha)}{2}\bigg)\leq f(x)\leq \bigg(1+\frac{\sin^2(\alpha)}{2}\bigg)$

I did not know where my try is wrong.

Please have a look on it

But actual answer is $\displaystyle -\sqrt{1+\sin^2(\alpha)}\leq f(x)\leq \sqrt{1+\sin^2(\alpha)}$

Priti Bisht
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    Your result is a superset of the given answer. You showed that $$\biggl(1+\frac{\sin^2(\alpha)}{2}\biggr)\leq y \leq \biggl(1+\frac{\sin^2(\alpha)}{2}\biggr)$$ holds for all $y = f(x)$ with $x \in \mathbb{R}$, but you did not show that, for any given $y$ in this range, there is a corresponding $x \in \mathbb{R}$ such that $y = f(x)$. – L. F. Feb 02 '23 at 15:03
  • $|2ab| \le a^2+b^2$ has equality only when $a = b$. In the first case, $a = \cos x$, $b = \sin x$ and in the second $a = \cos x$, $b = \sqrt{\sin^2 x +\sin^2\alpha}$. These can't be simultaneously true, so your bounds cannot be tight (unless $\alpha = 0$). – eyeballfrog Feb 02 '23 at 15:12
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    Of note: you can show $f$ is strictly increasing on $(-\pi, 0)$, so the maximum and minimum must occur in $(0, \pi)$. It's not too hard to then show they must occur in $(0, \pi/2)$ and $(\pi/2, \pi)$ respectively. This lets you safely make assumptions about the signs of $\sin x$ and $\cos x$. – eyeballfrog Feb 02 '23 at 15:49

3 Answers3

4

I'm going to have a look on your try, and then add a claim and an example which might help you.

From $$-(\cos^2x+\sin^2x)\leqslant 2\cos x\sin x\leqslant \cos^2x+\sin^2x\tag1$$ and $$\small -(\cos^2 x+\sin^2 x+\sin^2\alpha)\leqslant 2\cos x\sqrt{\sin^2x+\sin^2\alpha}\leqslant \cos^2x+\sin^2x+\sin^2\alpha\tag2$$ you got $$-2-\sin^2\alpha\leqslant 2f(x)\leqslant 2+\sin^2\alpha$$ and $$-1-\frac{\sin^2\alpha}{2}\leqslant f(x)\leqslant 1+\frac{\sin^2\alpha}{2}\tag3$$

To see if $(3)$ is the range of $f(x)$, let us see if there is $x$ such that, for any given $\alpha$, $f(x)=1+\dfrac{\sin^2\alpha}{2}$, i.e. $$2f(x)=2+\sin^2\alpha\tag4$$ holds.

From $(1)(2)$, $x$ satisfying $(4)$ has to satisfy both $$2\cos x\sin x= \cos^2x+\sin^2x\tag5$$ and $$2\cos x\sqrt{\sin^2x+\sin^2\alpha}=\cos^2x+\sin^2x+\sin^2\alpha\tag6$$

$(5)$ is equivalent to $$(\cos x-\sin x)^2=0\iff \cos x=\sin x\iff x=\frac{\pi}{4}+m\pi\ (m\in\mathbb Z)$$

$(6)$ is equivalent to $$\cos x\geqslant 0\quad\text{and}\quad \cos^2x=\sin^2x+\sin^2\alpha\iff \cos x\geqslant 0\quad\text{and}\quad \cos(2x)=\sin^2\alpha$$ So, we see that $$(5)\ \ \text{and}\ \ (6)\iff x=\dfrac{\pi}{4}+2m\pi\quad\text{and}\quad \alpha=n\pi\quad (m,n\in\mathbb Z)$$

This means that there is no $x$ such that, for any given $\alpha$, $f(x)=1+\dfrac{\sin^2\alpha}{2}$ holds.

So, $(3)$ is not the range of $f(x)$.

In the following, I'm going to add a claim and an example which might help you.

In the following, I consider only continuous functions.

Claim : If

  • $f(x)\leqslant F(x)\leqslant g(x)\quad$ ($f(x),g(x)$ are not necessarily constant functions)

  • $h(x)\leqslant G(x)\leqslant i(x)\quad$ ($h(x),i(x)$ are not necessarily constant functions)

  • both $f(x)+h(x)$ and $g(x)+i(x)$ are constant functions

  • there is $\alpha$ such that $F(\alpha)=f(\alpha)$ and $G(\alpha)=h(\alpha)$

  • there is $\beta$ such that $F(\beta)=g(\beta)$ and $G(\beta)=i(\beta)$

then, $\operatorname{range}(F(x)+G(x))=\left[f(x)+h(x),g(x)+i(x)\right]$.

Proof :

For $c$ such that $f(x)+h(x)\leqslant c\leqslant g(x)+i(x)$, letting $H(x)=F(x)+G(x)-c$, since we have $$H(\alpha)=F(\alpha)+G(\alpha)-c=f(\alpha)+h(\alpha)-c\leqslant 0$$ $$H(\beta)=F(\beta)+G(\beta)-c=g(\beta)+i(\beta)-c\geqslant 0$$ we can say that, by the intermediate value theorem, there is $\gamma$ such that $$H(\gamma)=0\qquad \text{and}\qquad \min(\alpha,\beta)\leqslant \gamma\leqslant \max(\alpha,\beta).\quad\blacksquare$$

Exmaple for which the claim above works :

Find the range of $\cos x+\dfrac{x^2+8\pi x-2\pi^2}{3x^2+6\pi^2}$.

Let $f(x)=\cos x$ and $g(x)=\dfrac{x^2+8\pi x-2\pi^2}{3x^2+6\pi^2}$. Then, $g'(x)=\dfrac{-8 \pi (x-2\pi)(x+\pi)}{3 (x^2 + 2 \pi^2)^2}$ and $\displaystyle\lim_{x\to\pm\infty}g(x)=\frac 13$.

Since $-1\leqslant f(x)\leqslant 1$ and $-1=g(-\pi)\leqslant g(x)\leqslant g(2\pi)=1$, adding these gives $$-2\leqslant f(x)+g(x)\leqslant 2\tag7$$ Since $f(-\pi)=g(-\pi)=-1$ and $f(2\pi)=g(2\pi)=1$, we can say, by the claim above, that $(7)$ is the range.

mathlove
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3

Let $a=|\sin\alpha|.$

First notice that (since $|\sin x|\le\sqrt{a^2+\sin^2x}$) when the maximum value $f_{\max}$ is attained the $\sin$ and $\cos$ of $x$ are positive, and that the minimum value is $f_{\min}=-f_{\max},$ corresponding to the same $\sin$ and the opposite $\cos.$

$$f'(x)=\left(1+\frac{\sin x}{\sqrt{a^2+\sin^2x}}\right)g(x)$$ where $$g(x):=\cos^2x-\sqrt{a^2+\sin^2x}\,\sin x$$ decreases continuously from $g(0)=1-0=1$ to $g(\pi/2)=0-a=-a$ hence vanishes for a unique $x_0\in[0,\pi/2].$ One easily checks that $$\sin^2x_0=\frac1{2+a^2},\quad\cos^2x_0=\frac{1+a^2}{2+a^2}.$$

Therefore, $f_\max=f(x_0)=\sqrt{1+a^2}$ and $$\operatorname{range}(f)=\left[-\sqrt{1+a^2},\sqrt{1+a^2}\right].$$

Anne Bauval
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Let $p=\sin^2\alpha,\; y=\cos 2x,\;$ then $$\cos x \sin x+\cos x\sqrt{\sin^2x+\sin^2\alpha\mathstrut} =\dfrac{\pm\sqrt{1-y^2}\pm\sqrt{1-y^2+2p(1+y)\mathstrut}}2 = f(y),$$ $$2f'(y) = \mp\dfrac y{\sqrt{1-y^2}}\pm\dfrac{p-y}{\sqrt{1-y^2+2p(1+y)}}.$$ The set of possible extremes of $f(y)$ corresponds with the expression
$$\left[\begin{align} &y^2-1=0\\[4pt] &y^2-2py-2p-1=0\\[4pt] &y^2(1+2py-y^2+2p)=(p^2-2py+y^2)(1-y^2), \end{align}\right.,$$ $$\left[\begin{align} &y=\pm1\\[4pt] &y=p\pm\sqrt{p^2+2p+1}\\[4pt] &(2+p)y^2+2y-p=0, \end{align}\right.$$ $$y_m\in\left\{-1,1, \dfrac p{2+p}\right\},\quad f(y_m)\in\left\{0, \pm\sqrt{p\mathstrut}, \pm \sqrt{1+p} \right\},$$ $$\mathbb{f(y)\in\left[-\sqrt{1+\sin^2\alpha}, \sqrt{1+\sin^2\alpha}\right]}.$$

Yuri Negometyanov
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