Question

The difference between maximum and minimum values of $y = \sin^2x - 20(\cos x) + 1$, is

Answer 1

40

Answer 2

The given function is $y = \sin^2x - 20(\cos x) + 1$.

Step 1: Express the function in terms of a single trigonometric ratio. We know the identity $\sin^2x = 1 - \cos^2x$. Substitute this into the expression for $y$: $y = (1 - \cos^2x) - 20\cos x + 1$ $y = -\cos^2x - 20\cos x + 2$

Step 2: Introduce a substitution and define its domain. Let $t = \cos x$. Since $x$ is a real number, the range of $\cos x$ is $[-1, 1]$. Therefore, $t \in [-1, 1]$.

Now, the function becomes a quadratic function in $t$: $f(t) = -t^2 - 20t + 2$

Step 3: Analyze the quadratic function over the given domain. This is a quadratic function of the form $at^2 + bt + c$ with $a = -1$, $b = -20$, and $c = 2$. Since $a = -1$ (which is negative), the parabola opens downwards, meaning its vertex represents the maximum point of the parabola.

The $t$-coordinate of the vertex is given by $t_v = \frac{-b}{2a}$. $t_v = \frac{-(-20)}{2(-1)} = \frac{20}{-2} = -10$

The domain for $t$ is $[-1, 1]$. The vertex $t_v = -10$ lies outside this domain, specifically to the left of the interval $[-1, 1]$.

Since the parabola opens downwards and its vertex is to the left of the interval $[-1, 1]$, the function $f(t)$ will be strictly decreasing over the interval $[-1, 1]$.

Step 4: Determine the maximum and minimum values of the function. Because the function is decreasing over the interval $[-1, 1]$: The maximum value of $f(t)$ will occur at the left endpoint of the interval, $t = -1$. $y_{max} = f(-1) = -(-1)^2 - 20(-1) + 2$ $y_{max} = -(1) + 20 + 2 = -1 + 20 + 2 = 21$

The minimum value of $f(t)$ will occur at the right endpoint of the interval, $t = 1$. $y_{min} = f(1) = -(1)^2 - 20(1) + 2$ $y_{min} = -1 - 20 + 2 = -19$

Step 5: Calculate the difference between the maximum and minimum values. Difference $= y_{max} - y_{min}$ Difference $= 21 - (-19)$ Difference $= 21 + 19 = 40$