Plz read it . I'm going to draw the derivative of sine at 45 in degree .

y1=Sine 45 =0.707106 and y2=sine 45.001 =0.707119

so dy=y2-y1=0.000013

and dx=45.001-45=0.001

now dy/dx =0.012

I know this answer is approximately drawn. We know that the derivative of sinX is cosX . So the derivative of sine at 45 is equal to the simple value of cos45=0.707

(0.012 is not equal to 0.707) Thats why I have observed that practically the derivative of trigonometric functions is proved only in Radion . Why it is not proved in Degree ? For example (in Radion) the derivative of sine at (pi/4) =0.707 which is exactly the same for the simple value cos(pi/4)=0.707 thats proved it .

But in Degree the derivative of sine at 45 = 0.012 as I draw it above which is against the simple value of cos45= 0.707 .

I have calculated it by hand as well as by calculator .
Plz Find the reason behind it .
Akhtar

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## Comments

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TopNewestVery interesting. I'd like to, for a second, stop using the numerical approach to a derivative, and look at the graphical approach. The derivative is the slope of the tangent line of that curve at a specific x-value. If you consider the sine graph in degrees, instead of radians, it no longer has a period of about 6.28, but a much larger 360. Note however that the amplitude is the same at 2. If you can picture the graphs in your head, you can easily see that at x=45, the degree sine curve's slope is much smaller than the radian sine curve's slope at x=pi/4. This is the error.

For extra confirmation, think of it this way: we know that the sine function (graph) normally works in radians. So if we want the graph to be in degrees, we must alter the period. Our correct sine function (still in radians) would be \(sin(\frac{2\pi}{360} x)\), and it would look identical to the degree sine function. Then, the derivative of this is \(\frac{2\pi}{360} cos(\frac{2\pi}{360} x)\). Plugging in x=45 into the derivative will give the number (.012...) that you were approximating before.

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The derivative holds only radian measure.

\[ f(x) = \sin x \Rightarrow f' (x) = \cos x. \]

If you want to convert to degree measure, then we have

\[ f(x) = \sin x^\circ \Rightarrow f' (x) = \frac{2\pi} { 360 } \cos x^\circ. \]

The reason for this factor is the conversion from degrees to radians, namely \( x^ \circ = \frac{2 \pi}{360} x \) radians. The constant then falls out by

applying the chain rule.A similar example is to consider the graph of \( y = x^2 \) which should have a derivative of \( 2x \). In the graph \( y = (\pi x)^2 \), if we use a substitution of \( \pi x = z \) to get \( y = z^2 \), does it have a derivative of \( y' = 2z = 2\pi x \)? You should recognize that you're missing the term \( \frac{dz}{dx} \)

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Thanks. I don't really know the proper or technical terms, as I have waited to study calculus until this coming year (I wanted to be able to learn something in math class, regularly at least). Although, I do very much understand what you are saying.

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when we use angle in radian then we follow radian system to draw the derivative and same for the degree . but the answer of derivative should be same for both cases . because the derivative is a normal number that only shows the rate of change of a function at any independent variable . It is not the value of radian or degree.

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All of you thinking that derivative can be calculated in radian or degree but I want to clear all of you that derivative is a normal number that only shows the rate of change of a function at any independent variable . It is not the value of radian or degree.

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My question is only that if the numeric values are proved in radian for derivative with the same method then why the derivative is not proved in degree.Lets again I'm going to draw the derivative of sine at (pi /4) in radian . Let y=sin(pi/4) y1=Sine (pi /4) =0.707106 and y2=sine(pi /4+0.001) =0.7078135343

so dy=y2-y1=0.00070675311

and dx=0.001

now dy/dx =0.70675311 this result is approximately true . But you can see the result was not true for degree as I solve in my first discussion . So find the reason only for numeric value only ? Thats the challenge !

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I am totally disagree with this one f(x)=sinx∘⇒f′(x)=(2π/360)cosx∘ to convert to degree measure.

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I think and believe that numeric value is the base of every thing in math !

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What about geometry? The Greeks hardly even knew what numbers were.

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Stop this you idiots. derivative of sin of angle with RESPECT TO THE ANGLE is 2 times Pi times cosine of the angle . As I have told you since Reagan was president. Sin, Cosine and Angle do not depend on how you name your angles idiots . You don't understand ? Then get your asses somewhere else and quit polluting our western ways. .

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And there are no 'negative numbers and no negative exponents. And Log( 1/2) WRONG_ INVALID ditto for Log to base 1/2 of 3

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d (sin x) / dx = cos x, x is in radians.

suppose that w is the equivalent of x in degrees.

x = (2 pi / 360) w.

sin x = sin ((2 pi / 360) w).

using the chain rule, d (sin ((2 pi / 360) w)) / dw = (2 pi / 360) cos ((2 pi / 360) w).

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Its mean if we want the derivative for w=0 in degree the answer would be 0.0174 . Thats not true one . It should be one at zero degree .

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Are you sure that it is one at zero degree?(for the sin(x) graph of course) Clearly you got somewhere in your understanding wrong. This is so as when we differentiate sin(x degrees) with respect to x it WILL not give you cos(x degrees).

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y1=Sine 45 =0.707106 and y2=sine 45.001 =0.707119

so dy=y2-y1=0.000013

and dx=45.001-45=0.001

now dy/dx =0.012

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I am not so sure but maybe it will be clearer if i explain to you this:

The graph of \(y=\cos(x^{\circ})\) when plotted with the \(x,y\) axis, it is against \(x\) not \(x^{\circ}\). So when you plug in the values of 45 and 45.001, you are doing it for \(x\), not \(x^{\circ}\). To make it what you wish it to be, then you must compress the \(x\) axis to \(x^{\circ}\) axis.

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Consider the definition of derivative: We say that the derivative \( f'(a) \) of a function \( f(x) \) at a point \( x = a \) is the limit \[ f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x-a}. \] Now suppose we take \( f(x) = \sin x \) where \( x \) is measured in degrees. Then we can numerically estimate the value of \( f'(0) \) by considering values of \( x \) near \( a = 0 \). For example, \[ f'(0) \approx \frac{\sin 1^\circ - \sin 0^\circ}{1^\circ - 0^\circ} = \sin 1^\circ. \] That's clearly not equal to 1, nor would we expect that choosing even smaller values of \( x \) would cause the limit to approach 1.

The reason why radians are a natural choice for angle measurement in calculus is analogous to why the base-\(e\) logarithm is a natural choice of exponential base. These choices lead to functions whose rates of change (derivatives and antiderivatives) do not depend on some unit or base of measurement.

For trigonometric functions, a radian is defined as the angle subtending an arc of a circle whose arclength equals its radius. So in the case of a unit circle, the arclength subtended by one radian has a length of 1. This is a natural choice--it means that there is a unit-to-unit relationship between the numerical value of an angle, and the numerical value of any lengths or ratios of lengths. Informally speaking, that is why the derivative of trigonometric functions when measured in radians behave the way they do, and why measuring angles in degrees does not.

In fact, we can see how this relates to the computation we did above: the numerator \( \sin x - \sin a \) is a difference of trigonometric ratios, but the denominator \( x - a \), if measured in degrees, is a difference of angles. If angles are measured on a different numerical scale than the ratio of lengths (as is the case with degree measure), then why should you be surprised that the derivative depends on that choice of scale?

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This is wrong one "f′(0)≈lim x→a(sin1∘−sin0∘)/(1∘−0∘=sin1∘)." You have to explain it !

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I specifically stated that a NUMERICAL ESTIMATE (i.e., an APPROXIMATION) for the derivative can be made by choosing a suitably close value of \( x \) to zero. The mathematical symbol for "approximately" is \( \approx \). If that approximation is not good enough for you, then feel free to choose smaller values: \[ \begin{align*} \sin 1^\circ &\approx 0.017452406437283512819 \\ \sin 0.1^\circ &\approx 0.0017453283658983088358 \\ \sin 0.001^\circ &\approx 0.00017453292431333680334 \\ \sin 0.0001^\circ &\approx 0.000017453292519057199614 \end{align*} \] and correspondingly, \[ f'(0) = \frac{\pi}{180} \approx 0.017453292519943295769. \] In any case, the value of the derivative when the angle is measured in degrees is not 1, as you claimed earlier.

It is presumptuous to assume that others are wrong simply because your understanding of the subject is different.

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The derivative is a normal number that only shows the rate of change of a function at any independent variable . It is not the value of radian or degree.

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