Q.What is the maximum value of the determinant $\begin{vmatrix} \sin x & 2\cos x \\ -\cos x & 1+2\sin x \end{vmatrix}$?

Expand the determinant to get $\sin x + 2$; since $\sin x \le 1$, the maximum value is $3$.

We are given $\begin{vmatrix} \sin x & 2\cos x \\ -\cos x & 1+2\sin x \end{vmatrix}$.

Expanding along the first row:

$\Delta = \sin x(1+2\sin x) - 2\cos x(-\cos x) = \sin x + 2\sin^2 x + 2\cos^2 x$

Since $\sin^2 x + \cos^2 x = 1$:

$\Delta = \sin x + 2(\sin^2 x + \cos^2 x) = \sin x + 2$

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The statement is True: the range of $\cos x$ is $[-1, 1]$, and $0.6$ lies inside this range.

The domain of $\cos x$ is all real numbers, and its range is exactly $[-1, 1]$ — cosine never exceeds 1 or goes below $-1$, but it does take every value in between (it is a continuous function).

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$\sin x$ attains every value between $-1$ and $1$, including both endpoints, so its range is the closed interval $[-1, 1]$.

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Expanding the determinant gives $1+\sin x$, whose maximum value over $x$ is $2$.

$\begin{vmatrix}\sin x & \cos x\\ -\cos x & 1+\sin x\end{vmatrix} = \sin x(1+\sin x) - \cos x(-\cos x) = \sin x + \sin^2x + \cos^2x = \sin x + 1$

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$\cos\theta \in [-1, 1]$ for all real $\theta$; its maximum value is $1$.

For every real number $\theta$, the cosine function satisfies $-1 \le \cos\theta \le 1$. This bound comes directly from the definition of $\cos\theta$ as the $x$-coordinate of a point on the unit circle, which can never leave the interval $[-1, 1]$. The value $\cos\theta = 1$ …

Greatest value of $\sin x$ on $[-\pi/2,\pi/2]$ is $1$, so $f(x)=\sin x+2$ has greatest value $3$.

On the interval $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$, $\sin x$ increases from $-1$ to $1$, attaining its maximum value $1$ at $x=\dfrac{\pi}{2}$.

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An expression $a\sin\theta+b\cos\theta$ always lies between $-\sqrt{a^2+b^2}$ and $\sqrt{a^2+b^2}$; here that gives $[-5,5]$.

For $3\sin\theta+4\cos\theta$, the amplitude is $R=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$. This can be written as $R\sin(\theta+\phi)$ for a suitable phase $\phi$, which ranges …

Since $\sin(3x)$ ranges over $[-1,1]$, shifting by 4 gives the range of $f$.

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For $f(x) = a\sin x + b\cos x$, the maximum value is always $\sqrt{a^2+b^2}$.

$f(x) = 3\sin x + 4\cos x$. Writing $R=\sqrt{3^2+4^2}=\sqrt{25}=5$, we can express $f(x) = R\sin(x+\phi)$ for a suitable phase $\phi$, and since $\sin(x+\phi)$ ranges over $[-1,1]$, the maximum value of $f(x)$ is $R = 5$.

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Write $f=2\sin(x+\tfrac{\pi}{6})$; it peaks at $x=\dfrac{\pi}{3}$.

$\sqrt3\sin x+\cos x=2\!\left(\tfrac{\sqrt3}{2}\sin x+\tfrac12\cos x\right)=2\sin\!\left(x+\dfrac{\pi}{6}\right)$.

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