Question

What is the min value of?

$asin^2\theta + bsin(\theta)+ccos^2\theta$

Answer 1

The minimum value depends on the values of a, b, and c.

Answer 2

To find the minimum value of the expression $asin^2\theta + bsin(\theta)+ccos^2\theta$, we first rewrite the expression in terms of a single trigonometric function. We know the identity $\cos^2\theta = 1 - \sin^2\theta$.

Substitute this into the given expression: $f(\theta) = a\sin^2\theta + b\sin\theta + c(1 - \sin^2\theta)$ $f(\theta) = a\sin^2\theta + b\sin\theta + c - c\sin^2\theta$ $f(\theta) = (a-c)\sin^2\theta + b\sin\theta + c$

Let $x = \sin\theta$. Since $\theta$ is a real angle, the value of $\sin\theta$ is restricted to the interval $[-1, 1]$. So, $x \in [-1, 1]$. The expression becomes a quadratic function in $x$: $g(x) = (a-c)x^2 + bx + c, \quad \text{for } x \in [-1, 1]$

This is a quadratic function $Ax^2 + Bx + C$ where $A = (a-c)$, $B = b$, and $C = c$. The minimum value of a quadratic function over a closed interval depends on the coefficient of the $x^2$ term (which determines if the parabola opens upwards or downwards) and the position of the vertex relative to the interval.

Let's analyze the cases:

1. Case 1: $a-c = 0 \implies a=c$ The expression simplifies to a linear function: $g(x) = bx + c$.

   • If $b > 0$, the function is increasing, so the minimum value occurs at $x = -1$. Minimum value = $b(-1) + c = c-b$.
   • If $b < 0$, the function is decreasing, so the minimum value occurs at $x = 1$. Minimum value = $b(1) + c = c+b$.
   • If $b = 0$, the function is a constant: $g(x) = c$. Minimum value = $c$.

2. Case 2: $a-c \neq 0$ This is a parabola. The x-coordinate of the vertex is $x_v = -\frac{b}{2(a-c)}$.

   • Subcase 2.1: $a-c > 0$ (Parabola opens upwards) The function has a global minimum at the vertex.

     • If the vertex $x_v$ lies within the interval $[-1, 1]$ (i.e., $-1 \le -\frac{b}{2(a-c)} \le 1$), the minimum value is $g(x_v)$. Minimum value = $c - \frac{b^2}{4(a-c)}$.
     • If the vertex $x_v < -1$, the function is increasing on $[-1, 1]$, so the minimum value occurs at $x = -1$. Minimum value = $g(-1) = (a-c)(-1)^2 + b(-1) + c = a-c-b+c = a-b$.
     • If the vertex $x_v > 1$, the function is decreasing on $[-1, 1]$, so the minimum value occurs at $x = 1$. Minimum value = $g(1) = (a-c)(1)^2 + b(1) + c = a-c+b+c = a+b$.

   • Subcase 2.2: $a-c < 0$ (Parabola opens downwards) The function has a global maximum at the vertex. The minimum value will occur at one of the endpoints of the interval $[-1, 1]$. We need to compare $g(-1)$ and $g(1)$. $g(-1) = a-b$ $g(1) = a+b$

     • If $b > 0$, then $a-b < a+b$. So, the minimum value is $a-b$.
     • If $b < 0$, then $a-b > a+b$. So, the minimum value is $a+b$.
     • If $b = 0$, then $a-b = a+b = a$. So, the minimum value is $a$.

As shown by these cases, the minimum value of the expression is not a single numerical value or a simple algebraic expression independent of $a, b, c$. It depends entirely on the specific values of the constants $a, b,$ and $c$.

The question asks for "the min value of?" without providing specific values for $a, b, c$. Therefore, a general closed-form expression is not possible.

Explanation of the solution: The expression $asin^2\theta + bsin(\theta)+ccos^2\theta$ is transformed into a quadratic in $sin\theta$ by substituting $cos^2\theta = 1 - sin^2\theta$. Let $x = sin\theta$, so $x \in [-1, 1]$. The expression becomes $f(x) = (a-c)x^2 + bx + c$. The minimum value of this quadratic function on the interval $[-1, 1]$ depends on the coefficients $a, b, c$. Specifically, it depends on whether the parabola opens upwards or downwards, and whether its vertex falls within or outside the interval $[-1, 1]$. Due to these dependencies, a single general expression for the minimum value cannot be provided without specific values or relationships for $a, b, c$.