Basic derivatives of sine and cosine

The sine and cosine functions are periodic functions, meaning that when expressed as functions of a variable $x$x, there is a quantity $k$k such that if any multiple of $k$k is added to $x$x, the same function value occurs. What we mean here is that the function cycles, or repeats. 

In symbols, $f$f is a periodic function if $f\left(x\right)=f\left(x+nk\right),$f(x)=f(x+nk),where $n$n is an integer. Since sine and cosine are periodic functions, the tangent function must also be periodic because of the way it is defined: $\tan\left(x\right)=\frac{\sin\left(x\right)}{\cos\left(x\right)}$tan(x)=sin(x)cos(x)​.

The period $k$k of both the sine function and the cosine function is $2\pi$2π. However, the period of the tangent function is just $\pi$π. 

From the graph, it is clear that the gradient of the sine function also varies periodically - from a maximum of $1$1 at $x=0\pm n2\pi$x=0±n2π,  through zero at $\frac{\pi}{2}\pm n2\pi$π2​±n2π, through a minimum of $-1$−1 at $\frac{3\pi}{2}\pm n2\pi$3π2​±n2π. 

The following graph shows how the gradient of the sine function varies and it also shows that the variations in the gradient correspond to the form of the cosine function.

The graphs strongly suggest that if $f\left(x\right)=\sin x$f(x)=sinx, then the derivative is $f'\left(x\right)=\cos x$f′(x)=cosx.

Observation of graphs suggests, further, that if $g\left(x\right)=\cos x$g(x)=cosx, then $g'\left(x\right)=-\sin\left(x\right)$g′(x)=−sin(x) is its derivative. You could check, for example, that the gradient of $\cos x$cosx increases between $x=\frac{\pi}{2}$x=π2​ and $x=\frac{3\pi}{2}$x=3π2​ while on the same interval,  $\sin x$sinx is decreasing and, therefore, $-\sin x$−sinx is increasing, corresponding to the increasing gradient of $\cos x$cosx.

 

To obtain the derivative of $\sin x$sinx in a more rigorous way, we need to evaluate the 'first principles' statement:

$f'\left(x\right)=\lim_{h\rightarrow0}\frac{\sin\left(x+h\right)-\sin x}{h}$f′(x)=limh→0sin(x+h)−sinxh​

To do this we need the expansion $\sin\left(x+h\right)=\sin x\cos h+\cos x\sin h$sin(x+h)=sinxcosh+cosxsinh, and we also need to evaluate the limits $\lim_{h\rightarrow0}\frac{\sin h}{h}$limh→0sinhh​ and $\lim_{h\rightarrow0}\frac{\cos h-1}{h}$limh→0cosh−1h​. This is done with the help of some circle geometry in more complete treatments of the topic, and the results given above are confirmed.

 

The usual rules for differentiating sums, products or quotients of functions apply to the trigonometric functions, as do the rules for differentiating a function of a function.

 

Example 1

What is the gradient function for $f(x)=\tan x$f(x)=tanx?

We write $y=\tan x=\frac{\sin x}{\cos x}$y=tanx=sinxcosx​. Then $y'=\frac{\cos x.\cos x-\sin x.\left(-\sin x\right)}{\cos^2x}$y′=cosx.cosx−sinx.(−sinx)cos2x​. This is, $y'=\frac{\cos^2x+\sin^2x}{\cos^2x}=\frac{1}{\cos^2x}=\sec^2x$y′=cos2x+sin2xcos2x​=1cos2x​=sec2x.

 

Example 2

For what values of $x$x does the function defined by $y=2\sin(x+\frac{\pi}{6})$y=2sin(x+π6​) reach its maximum?

In this function, three operations have been applied. First, $\frac{\pi}{6}$π6​ is added to the variable $x$x. Then, the sine of the result is obtained, and finally, there is a multiplication by $2$2. The function is a composite: a function of a function of a function. The steps in the following long-winded solution would not need to be fully written out after some practice in this type of problem.

Put $u=\sin(x+\frac{\pi}{6})$u=sin(x+π6​). Then $y=2u$y=2u and $\frac{dy}{du}=2$dydu​=2. 

Next, put $v=x+\frac{\pi}{6}$v=x+π6​. Then $u=\sin v$u=sinv and $\frac{du}{dv}=\cos v$dudv​=cosv.

Finally, $\frac{dv}{dx}=1$dvdx​=1.

So, $y'=\frac{dy}{du}\frac{du}{dv}\frac{dv}{dx}=2.\cos v.1=2\cos\left(x+\frac{\pi}{6}\right)$y′=dydu​dudv​dvdx​=2.cosv.1=2cos(x+π6​).

The function $y=2\sin(x+\frac{\pi}{6})$y=2sin(x+π6​) reaches its maximum at points where its derivative is zero. So, we put $2\cos\left(x+\frac{\pi}{6}\right)=0$2cos(x+π6​)=0 and solve for $x$x. 

The cosine function has zeros at $\pm\frac{\pi}{2}$±π2​, $\pm\frac{3\pi}{2}$±3π2​, and so on. Of these, one that corresponds to a maximum of the sine function is $\frac{\pi}{2}$π2​. Therefore, we require $x+\frac{\pi}{6}=\frac{\pi}{2}$x+π6​=π2​. That is, $x=\frac{\pi}{3}$x=π3​ and, considering the periodic nature of the sine function, the full solution must be 

$x=\frac{\pi}{3}\pm2n\pi$x=π3​±2nπ

 

More Worked Examples

Question 1

Question 2

Question 3