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. enter image description here

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 $m$ = $14 \over 5$, $M = 10$


  • $\begingroup$ Find the critical points. $\endgroup$ – William Elliot Aug 20 '19 at 12:17
  • $\begingroup$ Are you comfortable with calculus? Your tags say calculus and pre-calculus $\endgroup$ – G. Chiusole Aug 20 '19 at 12:21
  • $\begingroup$ A little bit. Did you mean the question is more related with the algebra-precalculus? $\endgroup$ – se-hyuck yang Aug 20 '19 at 12:32

Write $$f(\theta)=\sqrt{64+36}\left(\frac{8}{10}\sin(\theta)+\frac{6}{10}\cos(\theta)\right)$$

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  • $\begingroup$ could you give me some more hints? $\endgroup$ – se-hyuck yang Aug 20 '19 at 12:29
  • $\begingroup$ Then there exists an angle $$\phi$$ such that $$\cos(\phi)=\frac{8}{10},\sin(\phi)=\frac{6}{10}$$ and you will have $$\sin(\phi+\theta)$$ $\endgroup$ – Dr. Sonnhard Graubner Aug 20 '19 at 12:33
  • $\begingroup$ I'll try it again As your hints $\endgroup$ – se-hyuck yang Aug 20 '19 at 12:36
  • $\begingroup$ Then we get $$|f(\theta)|\le 10$$ $\endgroup$ – Dr. Sonnhard Graubner Aug 20 '19 at 12:37
  • $\begingroup$ Thanks for your hint I tried it again. But Still Can't figure out it. So Would you suggest the more concrete process for me? $\endgroup$ – se-hyuck yang Aug 20 '19 at 13:12

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