Maximum value of a trigonometric expression and value of angle giving this max.

Question

What will be the maximum value of $\text 12\sin \theta-9\sin^2\theta$ and if $\theta=\alpha $, find $\sin \alpha$ ?

I tried to convert the given expression into the " $ a\cos\theta +b\cos\theta$" form to get an answer (for it the maximum value will be $ \sqrt{a^2+b^2}$). But I failed to do it.

Answer 1

$$12\sin\theta-9\sin^2\theta=4-(3\sin\theta-2)^2\le4$$

the equality occurs if $3\sin\theta-2=0$

Answer 2

Hint: Let $u=\sin(\theta)$, then the expression becomes

$$f(u)=12u-9u^2$$

Can you find its maximum by using the first derivative $f'(u)=12-18u$?

EDIT:

Method one: using completing the square $f(u)=12u-9u^2=-(3u)^2+2\cdot(3u)\cdot 2 -2^2 +2^2=-(3u-2)^2+4$

It is easy to see that this expression is maximal for $u=2/3$ as it has a negative square. The maximal value is $4$ in this case.

Method two: Using calculus In order to find the maximum of a function, you can use its derivative. For an maximum it is necessary that the first derivative is zero

$$f'(u)=12-18u=0.$$

This is equivalent with $u=12/18=2/3$. In order to check if this is a maximum we need to check the second derivative $f''(u)=-18<0$. If the second derivative is negative for $u=2/3$, then we have a maximum. Hence, $u=2/3$ is a maximum of this function. Plug this into $f(u=2/3)=12\cdot 2/3-9(2/3)^2$ and you have your maximum.

Answer 3

Hint: first define $x=\sin \theta$, take the derivative to find the $x$ which maximizes the expression (remembering that $-1 \le x \le 1$), then take $\arcsin x$

Answer 4

Using $$a^2+b^2+2ab \geq (a-b)^2+4ab \geq 4ab$$

So $$(a+b)^2 \geq 4ab\Rightarrow ab \leq \frac{(a+b)^2}{4}$$

and equality hold when $a = b.$

So $$3\sin \theta \cdot (4-3\sin \theta) \leq \frac{(3\sin \theta+4-3\sin \theta)^2}{4} = 4$$

and equality hold when $3\sin \theta = 4-3\sin \theta \Rightarrow 3\sin \theta = 2$