Minimize a certain function involving sine and cosine

[ShizukaSm]

Homework Statement

It isn't a homework problem per se, but a curiosity a stumbled upon when trying to solve a physics problem (I was trying to calculate the angle I would need to do less work possible, while moving the box). The equation I found is:
$f(\theta)=\cos(\theta)+ 0.4sen(\theta)$

Homework Equations

Just the one stated above, and trig identities, probably.

The Attempt at a Solution

I tried a few things (including deriving and finding the roots of the function without success, since I couldn't found the roots), I also tried to rewrite (Using cos² + sin² = 1) and got:
$f(\theta)=\sqrt{1-\sin^2(\theta)}+ 0.4sin(\theta)$
But I also don't know how to find the minimum in this equation.

How could I go about solving that?

Thanks in advance!

 

[scurty]

I tried a few things (including deriving and finding the roots of the function without success, since I couldn't found the roots),

What problems did you run into? Did you do anything with tangent?

 

[HallsofIvy]

Interesting that you use both "sen" and "sin"! Can't decide between French and English?

You have $f(\theta)= cos(\theta)+ 0.4sin(\theta)$ (only English for me, I'm afraid.). I don't see any reason to introduce a square root just to have only sine. Taking the derivative, $f'(\theta)= -sin(\theta)+ 0.4cos(\theta)= 0$ at a max or min. That is the same as $sin(\theta)= 0.4 cos(\theta)$ or, since sine and cosine are not 0 for the same $\theta$, we must have $sin(\theta)/cos(\theta)= tan(\theta)= 0.4$. You can use a calculator to solve that.

 

[haruspex]

Reposting HallsofIvy's post to fix up the LaTex:

You have $f(\theta)= cos(\theta)+ 0.4sin(\theta)$. I don't see any reason to introduce a square root just to have only sine. Taking the derivative, $f'(\theta)= -sin(\theta)+ 0.4cos(\theta)= 0$ at a max or min. That is the same as $sin(\theta)= 0.4 cos(\theta)$ or, since sine and cosine are not 0 for the same $\theta$, we must have $sin(\theta)/cos(\theta)= tan(\theta)= 0.4$. You can use a calculator to solve that.

 

[ShizukaSm]

Oh god, I can't believe I ignored sin(x)/cos(x) = tan(x). I'm terribly sorry, thanks a lot! That solves my problem.

About the whole sen and sin thing, it's because I'm actually Brazilian, and here we write "Sen", so I sometimes get both of them confused :P

 

[Ray Vickson]

Oh god, I can't believe I ignored sin(x)/cos(x) = tan(x). I'm terribly sorry, thanks a lot! That solves my problem.

About the whole sen and sin thing, it's because I'm actually Brazilian, and here we write "Sen", so I sometimes get both of them confused :P

You also need to worry about whether the point you find is a maximizer or a minimizer of f.

 

[ShizukaSm]

You also need to worry about whether the point you find is a maximizer or a minimizer of f.

Yes indeed, but that was easily done by deriving again and finding if the result in this particular point would be positive or negative, since I found a negative value I concluded that it was a maximum, the value I wanted(In OP I miswrote, I actually wanted a maximum initially).

$tan(\theta)= 0.4;\theta = 21.8^o\\f''(21.8^o)= -1.077$

However, since you mentioned that, I just realized that I have no idea on how to find the minimum. Shouldn't [itex]tan(\theta) = 0.4[\itex] yield two values, one of which is a maximum and another which is a minimum?

 

[Ray Vickson]

Yes indeed, but that was easily done by deriving again and finding if the result in this particular point would be positive or negative, since I found a negative value I concluded that it was a maximum, the value I wanted(In OP I miswrote, I actually wanted a maximum initially).

$tan(\theta)= 0.4;\theta = 21.8^o\\f''(21.8^o)= -1.077$

However, since you mentioned that, I just realized that I have no idea on how to find the minimum. Shouldn't [itex]tan(\theta) = 0.4[\itex] yield two values, one of which is a maximum and another which is a minimum?

Yes.

 

[ShizukaSm]

Yes.

Hmm. I found by trial and error that the angles should be: 21.8° and 201.8°, but how am I supposed to get the 201.8°? My calculator only gave me 21.8°.

 

[Dick]

Hmm. I found by trial and error that the angles should be: 21.8° and 201.8°, but how am I supposed to get the 201.8°? My calculator only gave me 21.8°.

tan(x)=tan(x+180). The values of tan repeat every 180 degrees (pi in radians). It's periodic with period pi.

 

[ShizukaSm]

Ohhh, I see, thanks!
I really need to get better in trigonometry. I have several wrong concepts :S

 

[lurflurf]

This is usually done with the identity

$$\cos (x)+\frac{2}{5} \sin (x)=\sqrt {1+\left( \frac{2}{5} \right) ^2 } \sin \left( x+\arctan \left( \frac{5}{2} \right) \right)$$

Which is quite easy to optimize.