# Derivatives + Trig

• Apr 29th 2008, 06:59 PM
theowne
Derivatives + Trig
The hypotenuse of a right-angled triangle is 12cm in length. Find the measure of the angles in the triangle that maximize its perimeter.

This is my function for perimeter that I came up with

12 + 12sin(theta)+12cos(theta)

Derivative

12cos(theta)-12sin(theta)

Are you just supposed to know when cos(theta) = sin(theta) or is there a way of figuring it out in different situations?
• Apr 29th 2008, 07:02 PM
Mathstud28
Quote:

Originally Posted by theowne
The hypotenuse of a right-angled triangle is 12cm in length. Find the measure of the angles in the triangle that maximize its perimeter.

This is my function for perimeter that I came up with

12 + 12sin(theta)+12cos(theta)

Derivative

12cos(theta)-12sin(theta)

Are you just supposed to know when cos(theta) = sin(theta) or is there a way of figuring it out in different situations?

cos(x)=sin(x) at pi/4..think arctan(1)
• Apr 29th 2008, 07:04 PM
theowne
Quote:

think arctan(1)
Have not learned/Don't know what it is.

So is there any way other than "think arctan"..?
• Apr 29th 2008, 07:06 PM
o_O
Just to be a bit more explicit:
$\displaystyle \frac{\cos \theta}{{\color{blue}\cos \theta}} = \frac{\sin \theta}{{\color{blue}\cos \theta}}$
$\displaystyle 1 = \tan \theta$
$\displaystyle \theta = \tan^{-1} (1)$

$\displaystyle \arctan(x)$ is just another notation for $\displaystyle \tan^{-1}(x)$
• Apr 29th 2008, 07:06 PM
TheEmptySet
Quote:

Originally Posted by theowne
The hypotenuse of a right-angled triangle is 12cm in length. Find the measure of the angles in the triangle that maximize its perimeter.

This is my function for perimeter that I came up with

12 + 12sin(theta)+12cos(theta)

Derivative

12cos(theta)-12sin(theta)

Are you just supposed to know when cos(theta) = sin(theta) or is there a way of figuring it out in different situations?

What we have is

$\displaystyle P(\theta)=12+12\sin(\theta)+12\cos(\theta)$

$\displaystyle \frac{dP}{d\theta}=12\cos(\theta)-12\sin(\theta)$

setting equal to zero and solving gives

$\displaystyle 0=12\cos(\theta)-12\sin(\theta) \iff 12\sin(\theta)=12\cos(\theta)=$
$\displaystyle \sin(\theta)=\cos(\theta) \iff \tan(\theta) =1 \iff \theta=\tan^{-1}(1) =\frac{\pi}{4}$
• Apr 30th 2008, 12:36 AM
Moo
Hello,

When is $\displaystyle \cos(\theta)=\sin(\theta)$ ?

By using this identity : $\displaystyle \cos^2(x)+\sin^2(x)=1$, $\displaystyle \sin(x)=\sqrt{1-\cos^2(x)}$

Hence, we want to solve for $\displaystyle \theta$ in $\displaystyle \cos(\theta)=\sqrt{1-\cos^2(\theta)}$

Square the equation :

$\displaystyle \Longrightarrow \cos^2(\theta)=1-\cos^2(\theta)$

$\displaystyle \Longrightarrow 2\cos^2(\theta)=1$

$\displaystyle \Longrightarrow \cos^2(\theta)=\frac 12$

$\displaystyle \Longrightarrow \cos(\theta)=\pm \frac{1}{\sqrt{2}}=\pm \frac{\sqrt{2}}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}=\pm \frac{\sqrt{2}}{2}$

-------> $\displaystyle \theta=\dots$

I did this because you said you didn't want to "think arctan"... ^^'