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Why? Can someone give me a link or explain below. Thanks

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    $\begingroup$ How do you define $\sin,\cos$? $\endgroup$ – Pedro Tamaroff Jan 17 '14 at 0:56
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    $\begingroup$ This question has been asked here. I gave this answer, for example. $\endgroup$ – robjohn Jan 17 '14 at 22:39
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By the angle sum identities and the fact that $\lim_{h\to 0} \frac{\sin{h}}{h} = 1$ and $\lim_{h\to 0}\frac{\cos(h)-1}{h}=0$ it is straight forward

$$\begin{array}{rcl} \frac{\sin(x+h)-\sin(x)}{h} &=& \frac{\sin(x)\cos(h)+\cos(x)\sin(h) - \sin(x)}{h}\\ &=&\frac{\sin(x)\left(\cos(h)-1\right)}{h}+\frac{\cos(x)\sin(h)}{h} \\ & \stackrel{h\to0}{\longrightarrow}&\cos(x) \end{array} $$

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Here's one way to see it. Use the identity $\sin(x+\epsilon) = \sin(x)\cos(\epsilon) + \cos(x)\sin(\epsilon)$ in the definition of the derivative: \begin{align*} \frac{\sin(x+\epsilon) - \sin(x)}{\epsilon} &= \frac{\sin(x)\cos(\epsilon) + \cos(x)\sin(\epsilon) - \sin(x)}{\epsilon}\\ &=\sin(x)\bigg(\frac{\cos(\epsilon) - 1}{\epsilon}\bigg) + \cos(x)\frac{\sin(\epsilon)}{\epsilon} \end{align*} You know that $$\lim_{\epsilon\to 0}\frac{\cos(\epsilon) - 1}{\epsilon}=0$$ and $$\lim_{\epsilon\to 0}\frac{\sin(\epsilon)}{\epsilon} = 1.$$ (These should be in your textbook's chapter on limits and the squeeze theorem.)

Now take $\epsilon\to 0$ in the difference quotient, and, like magic -- voila! $$\frac{d}{dx}\sin(x) = \cos(x).$$

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All other answers are very good, but here is just another way to see it that can be very useful.

It is nice to use the definitions:

$$\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$$ $$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$

With these definitions, you can immediately prove all of the great trigonometric identities you use everyday, from the sum formula to the product formula or sum of squares. So these are very useful definitions to get comfortable with. They will even come in handy to understand why, when you learn about infinite series, you skip odd or even terms in the expansions respectively.

Now, with these definitions, it should be clear using

$$\frac{d}{dx}e^{ix}=ie^{ix}$$ and $$\frac{d}{dx}e^{-ix}=-ie^{-ix}$$

how you can directly get the formula for the derivatives of these two trigonometric functions.

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The mapping $C:t\mapsto (\cos t,\sin t)$ defines a circular motion of a point along the unit circle: at time $t$ the point is at angle $t$ measured in radian from the right wing of $x$-axis.

Its derivative at moment $t$ intuitively (and eventually) is the velocity vector . Its length is $1$ (because the circumference of the passage taken so far at moment $t$ is just the angle multiplied by the radius, equals to $t$) and it is tangent to the circle at point $C(t)$, so it is just orthogonal to the vector $C(t)$ (coming from origin), heading to the direction of moving.

All in all, using $(a,b)\perp (-b,a)$, the velocity vector is $(-\sin(t),\cos(t))$ .

This illustrates why $$(\cos(t),\sin(t))'=(\cos'(t),\sin'(t))=(-\sin(t),\cos(t))\,.$$

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  • $\begingroup$ Exactly what I was going to say, I really wish I could work out a way to formulate this approach rigorously. It also explains in an instant why this diagram works. $\endgroup$ – Jack M Jan 17 '14 at 22:16
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A detailed and precise answer depends on the definition of $\sin$ and $\cos$ that you use, as well as on the properties of these functions that are already established.

In high school is it common to define these functions geometrically, by drawing certain triangles or by considering the unit circle. Then computing the derivative of $\sin(x)$ at any point $x$ can be reduced to computing it at $0$, by using some standard trigonometric identities which can easily be derived from the geometric definitions. Then, establishing that $\sin '(0)=\cos(0)=1$ boils down to proving that $\lim_{h\in 0}\sin(h)/h=1$. This limit is called the first fundamental limit and its proof requires some care and the use of the squeeze lemma for functions.

A more modern approach is to define both $\sin x$ and $\cos x$ in terms of converging series. Since such series can be derived term-wise, the proof of the above claim becomes a very easy computation, but I doubt you've seen series yet, but you will.

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By definition for $\;(\sin x_0)'\;$ :

$$\frac{\sin x-\sin x_0}{x-x_0}=\frac2{x-x_0}\sin\left(\frac{x-x_0}2\right)\cos\left(\frac{x+x_0}2\right)=$$

$$\frac{\sin\frac{x-x_0}2}{\frac{x-x_0}2}\cos\frac{x+x_0}2\xrightarrow[x\to x_0]{}1\cdot\cos\frac{2x_0}2=\cos x_0$$

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By the definition via power series: http://en.wikibooks.org/wiki/Trigonometry/Power_Series_for_Cosine_and_Sine (we could and should define them by these series)

the point is that these converge absolutely on $\mathbf{R}$, so that we may differentiate them term by term! do this

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  • $\begingroup$ Explain please! I'm an higjschool student $\endgroup$ – Vaishnavi Jan 17 '14 at 0:06
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For an intuitive and visual approach, I think it is best to take a look at the graphs of $\sin x$ and $\cos x$ to get the intuitive feel of why $\frac{d}{dx} \sin x = \cos x$ is true.

With $\sin x$, the maximum slope is $1$, when $x = 0$, and the minimum slope is $-1$, when $x = \pi$. Furthermore, the slope is $0$ when $x = \frac{\pi}{2}$ or when $x = \frac{3\pi}{2}$. Since the derivative at a point $x$ of a function is the slope of the tangeant line at that point, we can plot $\frac{d}{dx} \sin x$ with that information.

The function we get has a maximum of $1$, a minimum of $-1$, has a value of $1$ when $x=0$, is periodic on $2\pi$, etc. Now, it makes me think of the function $\cos x$, which also has a maximum of $1$, minimum of $-1$, has a value of $1$ when $x=0$, etc.

This is far from a rigorous explanation, but it is useful to see the relation between the derivative of sin and cos. If you forget wether $\frac{d}{dx} \sin x = \cos x$ or $\frac{d}{dx} \sin x = - \cos x$ during an exam (or it might me some other derivative), you can make a simple plot of the function and figure it out for yourself.

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It's better to think of the derivative of a sinusoidal wave as a phase shift of a right angle.

$$\frac {d\sin(x)}{dx} = \sin\left(x + \frac {\pi}{2}\right)$$

$$\frac {d^n\sin(x)}{dx^n} = \sin\left(x + \frac {n\pi}{2}\right)$$

When you do work with AC circuit design, dealing with phase shifts and amplitude changes though circuit parts is much easier than trying to figure out signs and which sin/cos function you are on.

When you look at a Taylor expansion of a sinusoidal function, it's much more obvious where the $1, 0, -1, 0, 1, 0,...$ pattern for the coefficients comes from, it's just the shape of the sinusoid itself taken at right angle steps.

Your question could be answered rigorously using algebra and limits, but I think you are wanting a simpler more intuitive understanding so I'll offer a less rigorous suggestion.

Consider a point on a spinning wheel. When is that point rising the fastest? It's when the point a quarter turn away is highest. When is falling the fasest? It's when the point a quarter turn away is lowest. When is the point not rising or falling (you would feel weightless for a moment if you were riding)? It's when the point a quarter turn away is at zero height (not above or below the centerline).

The constant of proportionality just comes from choice of units. Most people use radians so that you don't have an extra constant factor in the equation.

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