Question

The maximum value of 12 sin 0-9 sin²0 is: -​

Answer 1

Answer:

4

Step-by-step explanation:

The question is not clear. You have indicated the original function as 12sin(0) - 9sin²(0)

If so, the solution is trivial. At 0, sin(0) is 0 so the solution is 0

However, I will assume you meant the angle to be $\theta$ rather than 0 which makes sense and proceed accordingly

We can find the maximum or minimum of any function by finding the first derivate and setting it equal to 0

The original function is

$f(\theta) = 12sin(\theta) - 9 sin^2(\theta)$

Taking the first derivative of this with respect to $\theta$ and setting it equal to 0 lets us solve for the maximum (or minimum) value

The first derivative of $f(\theta)$ w.r.t $\theta$ is

                        $12\cos\left(\theta\right)-18\cos\left(\theta\right)\sin\left(\theta\right)$

And setting this = 0 gives

$12\cos\left(\theta\right)-18\cos\left(\theta\right)\sin\left(\theta\right) = 0$

Eliminating $cos(\theta)$ on both sides and solving for $sin(\theta)$ gives us

$sin(\theta) = \frac{12}{18} = \frac{2}{3}$

Plugging this value of $sin(\theta)$ into the original equation gives us

$12(\frac{2}{3}) - 9(\frac{4}{9} ) = 8 - 4 = 4$

This is the maximum value that the function can acquire. The attached graph shows this as correct