Why does the derivative of Law of Cosines require radians?

Question

The question I am solving says "2 sides of a triangle have lengths 9 ft and 18 ft. The angle between them is increasing at a rate of 2 degrees per minute. How fast is the length of the 3rd side increasing in ft/min when the angle between the 2 fixed sides is 60 degrees?"

Given the Law of Cosines: $c^2 = a^2 + b^2 - 2ab\cos(\theta)$ , taking the derivative with respect to $t$, plugging in the known side lengths for a and b, and setting equal to $\frac{dc}{dt}$ gives us $\frac{dc}{dt} = \frac{324\sin(\theta)\frac{d\theta}{dt}}{2c}$.

at 60 degrees, $c = \sqrt{18^2+9^2-2(18)(9)\cos(60)} \approx 15.59$ , so simplifying: $\frac{dc}{dt} \approx 10.4\sin(60)\frac{d\theta}{dt}$

When plugging in 2 for $\frac{d\theta}{dt}$, it gives me the wrong answer, because I need to use radians for $\frac{d\theta}{dt}$. I don't understand why this is the case, because the law of cosines would give the same answer if I used radians or degrees, and so would the derivative of cosine. I understand that my function couldn't accept both, but where did the requirement of radians come from in this problem?

Apologies if I have formatted anything incorrectly, this is my first time posting.

Answer 1

This is because the derivative of cosine is minus sine, but only if defined with radians. If you use degrees you get $$ \frac{\mathrm d}{\mathrm dx} \mathrm{cos}_\text{degrees}\, x = \frac{\mathrm d}{\mathrm dx} \mathrm{cos}\bigg(\frac{2\pi}{360}\, x\bigg) =-\frac{2\pi}{360}\,\mathrm{sin}\bigg(\frac{2\pi}{360}\, x\bigg) = -\frac{2\pi}{360}\mathrm{sin}_\text{degrees}\, x. $$

Answer 2

The Error is happening because , When you Actually see the derivation Of derivative of $\sin x$ we use the squeeze theoram for approximation, Where we use that $\frac{sinx}{x}=1$ as $x\to0$, Here , for approximation to be valid, $x$ must be in radian, or else the limit is not true

Answer 3

Degrees are not real numbers, so the function f(x) ist not a function from R->R. so differentiating makes no sense.

Answer 4

When taking the derivative of a trigonometric function, the angles expressed must be in radians. (See user2661923's excellent answer for why.)

In your final equation $\frac {dc}{dt} \approx 10.4 \sin (60) \frac {d \theta}{dt}$, the $60$ is ambiguous - the value could either be the angle in degrees or the angle in radians. If you have $60$ radians, the sine of $60^r$ is negative because the reference angle is $-2.83$ radians; thus you will have a negative answer $(-6.34)$ and a contradiction because you're looking for the rate of increase on the third side, not a decrease.

To get the correct answer, convert the $60°$ to radians by multiplying by $\pi/180$ to get $\pi /3$, so your equation becomes $\frac {dc}{dt} \approx 10.4 \sin (\pi/3) \frac {d \theta}{dt}$, which is $\approx 18.01$ when $\frac {d \theta}{dt} = 2.$

Answer 5

Edited the trailing portion of my post, adding the question: "How to think in terms of numbers rather than angles". This is in response to the comment by Dheeraj Gujrathi, following my answer.

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I agree with the other responses, but would like to give a somewhat different perspective:

In trigonometry (or plane geometry) the domain of the sine and cosine functions are angles. This makes perfect sense, because within a right triangle, the ratio of (for example) opposite over hypotenuse (i.e. the sine of the angle) is a function of the angle.

That is, if you have two similar right triangles (i.e. the corresponding angles equivalent, but one triangle is larger than another), then the corresponding ratios of opposite over hypotenuse of the two corresponding angles will be identical.

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Something bizarre happens when you exit [trigonometry : plane geometry] into the world of [real analysis : aka calculus]. The sine and cosine functions are redefined, and the domain of these functions are no longer angles.

Excerpting from "Calculus" 2nd Ed. 1966, by Tom Apostol: the sine and cosine functions are defined by the following axioms.

• The sine and cosine functions are defined for all real numbers. That is, the domain of the sine and cosine functions are real numbers, rather than angles.

• $\cos(0) = \sin(\pi/2) = 1, ~\cos(\pi/2) = -1.$

• $\cos(y-x) = \cos(y)\cos(x) - \sin(y)\sin(x).$

• For $~0 < x < \pi/2:$
  $0 < \cos(x) < \sin(x)/x < 1/\cos(x).$

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Apostol then uses only these axioms to prove all of the standard (real analysis) results around the sine and cosine function, including typical results from [trigonometry : plane geometry], as well as the derivative, the integral, and the Taylor series.

The natural question to ask is why would anyone re-define the sine and cosine functions in such a bizarre manner. The answer is that the re-definition facilitates establishing the real analysis results, and facilitates solving problems that would otherwise be very difficult.

For example, it becomes easy to establish that $~\displaystyle \int_0^1 \frac{1}{1 + t^2} ~dt = \pi/4.$

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Now, you have $~\color{\red}{\text{three}}~$ pending questions:

• What is the relationship between the trigonometry definitions of the sine and cosine functions and the corresponding real analysis definitions?
  
  Answer:
  Apostol shows by geometric illustration that the trigonometry definitions satisfy the axioms, if the domain of the sine and cosine functions are changed from angles to arc lengths of the unit circle.
  
  This explains (for example) why in trigonometry, $~\sin(45^\circ) = \dfrac{1}{\sqrt{2}},~$
  while in real analysis, $~\sin(\pi/4) = \dfrac{1}{\sqrt{2}}.$
  
  It is because in the unit circle, whose radius $~= 1,~$ $~\pi/4~$ corresponds to $~(1/8)$-th the circumference of the circle.
  
  So, because the geometric (i.e. trigonometry) versions of the sine and cosine functions, with their domains re-defined, satisfy the axioms, the re-defined (geometric) sine and cosine functions inherit all of the results proved from the axioms.

• What is a radian?
  
  Answer:
  The term radian is ambiguous, and is the source of much confusion to students in transition from [trigonometry : plane geometry] to real analysis. For these students in transition, the focus is typically on problem-solving, rather than proving theorems. So, the domain of the sine and cosine functions (sensibly) continues to be angles, rather than real numbers.
  
  However, from a real analysis problem solving perspective, the student needs to learn that $~\sin(\pi/4) = \dfrac{1}{\sqrt{2}}.~$ So, the teacher will conjure the transitional concept of the radian, as a unit of measure of angles, where $~\pi~$ radians $~= 180^\circ.~$
  
  Subsequently, the student in transition subconsciously is exposed to the idea that the domain of the sine and cosine functions are real numbers, rather than angles. So, the student in transition starts regarding $~\pi/4~$ radians as the real number $~\pi/4~$ such that the sine of $~\pi/4~$ radians $~= \dfrac{1}{\sqrt{2}}.$
  
  Eventually, as the student in transition is exposed more deeply to real analysis, the notion of a radian fades into (its richly deserved) obscurity, and the student comes to regard $~\sin(\pi/4) = \dfrac{1}{\sqrt{2}}~$ specifically because:

  • $~\pi/4~$ is $~(1/8)$-th the circumference of the unit circle.

  • The trigonometry definitions of the sine and cosine functions, with their domains changed from angles to real numbers, that represent arc lengths of the unit circle, have been shown to satisfy the real analysis axioms of the sine and cosine functions.

• How to think in terms of numbers rather than angles?
  
  Answer:
  This answer is intended to answer the following question:
  Suppose that in [trigonometry : plane geometry], you have that $~\cos(\theta) = x, ~\sin(\theta) = y,~$ where $~0^\circ \leq \theta < 360^\circ,~$ and $~-1 \leq x,y \leq 1.~$
  
  Suppose further that in [calculus : aka real analysis] that $~\cos(t) = x, ~\sin(t) = y,~$ where $~0 \leq t < 2\pi.~$ What is the exact relationship between the angle $~\theta~$ and the real number $~t~?$
  
  This relationship is illustrated in the two diagrams below. In both diagrams, $~\theta = \angle (1,0),(0,0),(x,y)~$ and $~t = ~$ the unit circle arc length from $~(1,0)~$ to $~(x,y).~$ Note that in the RHS diagram, $~x < 0.$
  
  Note that in [trigonometry : plane geometry], the sine and cosine functions are both periodic, with period $~360^\circ.~$ So, suppose (for example) that $~\cos(\theta_1) = x, ~\sin(\theta_1) = y, ~$ where $~\theta_1~$ is not in the range $~0^\circ \leq \theta_1 < 360^\circ.~$ Then you will always be able to find a unique angle $~\theta~$ such that $~0^\circ \leq \theta < 360^\circ,~$ and $~\cos(\theta) = x, ~\sin(\theta) = y.$ This angle $~\theta~$ will be equivalent to the angle $~\theta_1,~$ within a modulus of $~360^\circ.~$
  
  Similarly, in [calculus : aka real analysis], the sine and cosine functions are periodic, with period $~= 2\pi.~$ So, suppose (for example), that $~\cos(t_1) = x, ~\sin(t_1) = y, ~$ where $~t_1~$ is not in the range $~0 \leq t_1 < 2\pi.~$ Then you will always be able to find a unique real number $~t~$ such that $~0 \leq t < 2\pi,~$ and $~\cos(t) = x, ~\sin(t) = y.~$ The real number $~t~$ will be equivalent to the real number $~t_1,~$ within a modulus of $~2\pi.$
  
  Therefore, in describing the relationship between the angle $~\theta~$ and the real number $~t,~$ you can assume without loss of generality, that $~0^\circ \leq \theta < 360^\circ~$ and $~0 \leq t < 2\pi.~$