This is the problem that I've stuck.
(This is the extension of the my question that How to solve this? (Argument of the complex number in complex plane))
Find the maximum value, $M$ and minimum value, $m$
$f(\theta) = 8 \sin \theta + 6 \cos \theta $ $s.t. \theta \in [0, \pi]$
(${4\over 5} \leq {\sin \theta} {\leq 1}$, $- {3\over 5} \leq \cos \theta \leq {3\over 5} $)
$f(\theta) = 10\sin(\theta + \alpha)$ for $\cos \alpha = {4\over5}$
My attempts is just my thought. I'm tried various method, But Everytime I try it I've stuck. Please give me some concrete solution. 🙁
My attmept 1)
By the above equation, It is clear that $ – 10 \leq f(\theta) \leq 10$
But I Can't find the exact $\alpha$'s size.
So the Can't figure out the max or min of the $f(\theta) $.
There might be different cases. I present two example in picture.
My attempt 2) To find the critical point
Find the $\theta$ s.t. $f'(\theta) =0 $
Hence critical point is $\theta$ s.t. $tan \theta = {4 \over 3}$
But this is just local maximum(either local minimum), Not Sure the Max or min value of the $f(\theta)$
The answer is $m$ = $14 \over 5$, $M = 10$
Thanks.
Best Answer
Write $$f(\theta)=\sqrt{64+36}\left(\frac{8}{10}\sin(\theta)+\frac{6}{10}\cos(\theta)\right)$$