Question

$\forall$ x $\in$ R find the range of the function, f(x) = cos x (sin x + $\sqrt{sin^2 x + sin^2 \alpha}$); $\alpha$ $\in$ [0,$\pi$]

Answer 1

The range of f(x) is $[-\sqrt{1+\sin^2 \alpha}, \sqrt{1+\sin^2 \alpha}]$

Answer 2

Let the given function be $f(x) = \cos x (\sin x + \sqrt{\sin^2 x + \sin^2 \alpha})$. Let $k = \sin \alpha$. Since $\alpha \in [0, \pi]$, $k \in [0, 1]$.

Case 1: $k = 0$ (i.e., $\sin \alpha = 0$) In this case, $f(x) = \cos x (\sin x + \sqrt{\sin^2 x}) = \cos x (\sin x + |\sin x|)$.

• If $\sin x \ge 0$, then $f(x) = \cos x (2 \sin x) = \sin(2x)$. The range of $\sin(2x)$ is $[-1, 1]$.
• If $\sin x < 0$, then $f(x) = \cos x (\sin x - \sin x) = 0$. Combining these, the range of $f(x)$ when $\sin \alpha = 0$ is $[-1, 1]$.

Case 2: $k > 0$ (i.e., $\sin \alpha \in (0, 1]$) Let $u = \sin x$ and $v = \cos x$. The function is $f(x) = v(u + \sqrt{u^2 + k^2})$. Consider the term $g(u) = u + \sqrt{u^2 + k^2}$. For any real $u$, $\sqrt{u^2+k^2} > \sqrt{u^2} = |u|$.

• If $u \ge 0$, $g(u) = u + \sqrt{u^2+k^2} > u+u = 2u \ge 0$.
• If $u < 0$, $g(u) = u + \sqrt{u^2+k^2} = -|u| + \sqrt{u^2+k^2}$. Since $\sqrt{u^2+k^2} > |u|$, $g(u) > 0$. Thus, $g(u) > 0$ for all $u \in \mathbb{R}$. This implies that the sign of $f(x)$ is determined by the sign of $\cos x$. Since $\cos x$ can be positive, negative, or zero, $f(x)$ can take positive, negative, or zero values. Due to the symmetry of $\sin x$ and $\cos x$ over $2\pi$ period, the range will be symmetric about 0, i.e., of the form $[-M, M]$ where $M = \max |f(x)|$.

To find $M$, we find the maximum value of $f^2(x)$: $f^2(x) = \cos^2 x (\sin x + \sqrt{\sin^2 x + k^2})^2$. Let $\sin x = k \sinh t$. This substitution is valid for all $\sin x \in [-1, 1]$ as long as $k \in (0, 1]$. Then $\sqrt{\sin^2 x + k^2} = \sqrt{k^2 \sinh^2 t + k^2} = k \sqrt{\sinh^2 t + 1} = k \cosh t$ (since $k>0$ and $\cosh t > 0$). So, $\sin x + \sqrt{\sin^2 x + k^2} = k \sinh t + k \cosh t = k(\sinh t + \cosh t) = k e^t$. Now, $f(x) = \cos x \cdot k e^t$. We have $\cos^2 x = 1 - \sin^2 x = 1 - k^2 \sinh^2 t$. So, $f^2(x) = (1 - k^2 \sinh^2 t) (k e^t)^2 = k^2 e^{2t} (1 - k^2 \sinh^2 t)$. Substitute $\sinh t = \frac{e^t - e^{-t}}{2}$: $f^2(x) = k^2 e^{2t} \left(1 - k^2 \left(\frac{e^t - e^{-t}}{2}\right)^2\right)$ $f^2(x) = k^2 e^{2t} \left(1 - k^2 \frac{e^{2t} + e^{-2t} - 2}{4}\right)$ $f^2(x) = \frac{k^2}{4} \left(4e^{2t} - k^2 (e^{4t} + 1 - 2e^{2t})\right)$ $f^2(x) = \frac{k^2}{4} \left(-k^2 e^{4t} + (4+2k^2)e^{2t} - k^2\right)$. Let $z = e^{2t}$. Since $\sin x = k \sinh t \in [-1, 1]$, we have $\sinh t \in [-1/k, 1/k]$. This implies $t \in [\text{arsinh}(-1/k), \text{arsinh}(1/k)]$. Let $t_0 = \text{arsinh}(1/k)$. So $t \in [-t_0, t_0]$. Thus $z = e^{2t} \in [e^{-2t_0}, e^{2t_0}]$. Consider the quadratic function $h(z) = -k^2 z^2 + (4+2k^2)z - k^2$. This is a downward-opening parabola. The maximum of $h(z)$ occurs at $z = -\frac{4+2k^2}{2(-k^2)} = \frac{4+2k^2}{2k^2} = \frac{2+k^2}{k^2} = 1 + \frac{2}{k^2}$. The maximum value of $h(z)$ is $h\left(1 + \frac{2}{k^2}\right) = -k^2\left(1 + \frac{2}{k^2}\right)^2 + (4+2k^2)\left(1 + \frac{2}{k^2}\right) - k^2$ $= -k^2\left(\frac{k^2+2}{k^2}\right)^2 + 2(2+k^2)\left(\frac{k^2+2}{k^2}\right) - k^2$ $= -\frac{(k^2+2)^2}{k^2} + \frac{2(k^2+2)^2}{k^2} - k^2$ $= \frac{(k^2+2)^2}{k^2} - k^2 = \frac{k^4+4k^2+4-k^4}{k^2} = \frac{4k^2+4}{k^2} = 4\left(1+\frac{1}{k^2}\right)$. Now, we need to check if $z = 1 + \frac{2}{k^2}$ is within the domain $[e^{-2t_0}, e^{2t_0}]$. $e^{2t_0} = e^{2 \text{arsinh}(1/k)} = e^{2 \ln(\frac{1}{k} + \sqrt{\frac{1}{k^2}+1})} = \left(\frac{1+\sqrt{1+k^2}}{k}\right)^2 = \frac{1+1+k^2+2\sqrt{1+k^2}}{k^2} = \frac{2+k^2+2\sqrt{1+k^2}}{k^2}$. Since $k>0$, $2\sqrt{1+k^2} > 0$. Thus, $\frac{2+k^2+2\sqrt{1+k^2}}{k^2} > \frac{2+k^2}{k^2} = 1 + \frac{2}{k^2}$. So, the maximum of $h(z)$ occurs within the valid range of $z$. The maximum value of $f^2(x)$ is $\frac{k^2}{4} \cdot 4\left(1+\frac{1}{k^2}\right) = k^2\left(1+\frac{1}{k^2}\right) = k^2+1$. Therefore, the maximum value of $|f(x)|$ is $\sqrt{k^2+1} = \sqrt{1+\sin^2 \alpha}$. Since $f(x)$ is continuous and can take positive, negative, and zero values, its range is $[-\sqrt{1+\sin^2 \alpha}, \sqrt{1+\sin^2 \alpha}]$. This result also covers Case 1, as for $\sin \alpha = 0$, the range becomes $[-\sqrt{1+0}, \sqrt{1+0}] = [-1, 1]$.

Therefore, the range of the function is $[-\sqrt{1+\sin^2 \alpha}, \sqrt{1+\sin^2 \alpha}]$.