Differentiation of trigonometric functions

All derivatives of circular trigonometric functions can be found from those of sin(x) and cos(x) by means of the quotient rule applied to functions such as tan(x) = sin(x)/cos(x). Knowing these derivatives, the derivatives of the inverse trigonometric functions are found using implicit differentiation.

Proofs of derivatives of trigonometric functions

Limit of sin(θ)/θ as θ tends to 0

The diagram at right shows a circle with centre O and radius r = 1. Let two radii OA and OB make an arc of θ radians. Since we are considering the limit as θ tends to zero, we may assume θ is a small positive number, say 0 < θ < ⁠1/2⁠ π in the first quadrant.

In the diagram, let R₁ be the triangle OAB, R₂ the circular sector OAB, and R₃ the triangle OAC.

The area of triangle OAB is:

$${\displaystyle \mathrm{A}\mathrm{r}\mathrm{e}\mathrm{a}(R_1)=\frac{1}{2}|OA||OB|\sin \theta =\frac{1}{2}\sin \theta .}$$

The area of the circular sector OAB is:

$${\displaystyle \mathrm{A}\mathrm{r}\mathrm{e}\mathrm{a}(R_2)=\frac{1}{2}\theta .}$$

The area of the triangle OAC is given by:

$${\displaystyle \mathrm{A}\mathrm{r}\mathrm{e}\mathrm{a}(R_3)=\frac{1}{2}|OA||AC|=\frac{1}{2}\tan \theta .}$$

Since each region is contained in the next, one has:

$${\displaystyle \text{Area}(R_1)<\text{Area}(R_2)<\text{Area}(R_3)⟹\frac{1}{2}\sin \theta <\frac{1}{2}\theta <\frac{1}{2}\tan \theta .}$$

Moreover, since sin θ > 0 in the first quadrant, we may divide through by ⁠1/2⁠ sin θ, giving: $${\displaystyle 1<\frac{\theta }{\sin \theta }<\frac{1}{\cos \theta }⟹1>\frac{\sin \theta }{\theta }>\cos \theta .}$$

In the last step we took the reciprocals of the three positive terms, reversing the inequities.

We conclude that for 0 < θ < ⁠1/2⁠ π, the quantity sin(θ)/θ is always less than 1 and always greater than cos(θ). Thus, as θ gets closer to 0, sin(θ)/θ is "squeezed" between a ceiling at height 1 and a floor at height cos θ, which rises towards 1; hence sin(θ)/θ must tend to 1 as θ tends to 0 from the positive side:

${\displaystyle \underset{\theta \to 0^{+}}{lim}\frac{\sin \theta }{\theta }=1.}$

For the case where θ is a small negative number, ${\displaystyle -\frac{\pi }{2}<\theta <0}$, let ${\displaystyle u=-\theta }$. Then, as ${\displaystyle \theta \to 0^{-}}$, we have ${\displaystyle u\to 0^{+}}$. Using the fact that sine is an odd function,

$${\displaystyle \underset{\theta \to 0^{-}}{lim}\frac{\sin \theta }{\theta }=\underset{u\to 0^{+}}{lim}\frac{\sin (-u)}{-u}=\underset{u\to 0^{+}}{lim}\frac{-\sin u}{-u}=\underset{u\to 0^{+}}{lim}\frac{\sin u}{u}=1.}$$

Limit of (cos(θ)-1)/θ as θ tends to 0

The last section enables us to calculate this new limit relatively easily. This is done by employing a simple trick. In this calculation, the sign of θ is unimportant.

${\displaystyle \underset{\theta \to 0}{lim}\frac{\cos \theta -1}{\theta }=\underset{\theta \to 0}{lim}\left(\frac{\cos \theta -1}{\theta }\right)\left(\frac{\cos \theta +1}{\cos \theta +1}\right)=\underset{\theta \to 0}{lim}\frac{{\cos }^2\theta -1}{\theta (\cos \theta +1)}.}$

Using cos²θ – 1 = –sin²θ, the fact that the limit of a product is the product of limits, and the limit result from the previous section, we find that:

${\displaystyle \underset{\theta \to 0}{lim}\frac{\cos \theta -1}{\theta }=\underset{\theta \to 0}{lim}\frac{-{\sin }^2\theta }{\theta (\cos \theta +1)}=\left(-\underset{\theta \to 0}{lim}\frac{\sin \theta }{\theta }\right)\left(\underset{\theta \to 0}{lim}\frac{\sin \theta }{\cos \theta +1}\right)=(-1)\left(\frac{0}{2}\right)=0.}$

Limit of tan(θ)/θ as θ tends to 0

Using the limit for the sine function, the fact that the tangent function is odd, and the fact that the limit of a product is the product of limits, we find:

${\displaystyle \underset{\theta \to 0}{lim}\frac{\tan \theta }{\theta }=\left(\underset{\theta \to 0}{lim}\frac{\sin \theta }{\theta }\right)\left(\underset{\theta \to 0}{lim}\frac{1}{\cos \theta }\right)=(1)(1)=1.}$

Derivative of the sine function

We calculate the derivative of the sine function from the limit definition:

${\displaystyle \frac{\mathrm{d}}{\mathrm{d}\theta }\sin \theta =\underset{\delta \to 0}{lim}\frac{\sin (\theta +\delta )-\sin \theta }{\delta }.}$

Using the angle addition formula sin(α+β) = sin α cos β + sin β cos α, we have:

${\displaystyle \frac{\mathrm{d}}{\mathrm{d}\theta }\sin \theta =\underset{\delta \to 0}{lim}\frac{\sin \theta \cos \delta +\sin \delta \cos \theta -\sin \theta }{\delta }=\underset{\delta \to 0}{lim}\left(\frac{\sin \delta }{\delta }\cos \theta +\frac{\cos \delta -1}{\delta }\sin \theta \right).}$

Using the limits for the sine and for the cosine functions:

${\displaystyle \frac{\mathrm{d}}{\mathrm{d}\theta }\sin \theta =(1)\cos \theta +(0)\sin \theta =\cos \theta .}$

Derivative of the cosine function

From the definition of derivative

We again calculate the derivative of the cosine function from the limit definition:

${\displaystyle \frac{\mathrm{d}}{\mathrm{d}\theta }\cos \theta =\underset{\delta \to 0}{lim}\frac{\cos (\theta +\delta )-\cos \theta }{\delta }.}$

Using the angle addition formula cos(α+β) = cos α cos β – sin α sin β, we have:

${\displaystyle \frac{\mathrm{d}}{\mathrm{d}\theta }\cos \theta =\underset{\delta \to 0}{lim}\frac{\cos \theta \cos \delta -\sin \theta \sin \delta -\cos \theta }{\delta }=\underset{\delta \to 0}{lim}\left(\frac{\cos \delta -1}{\delta }\cos \theta -\frac{\sin \delta }{\delta }\sin \theta \right).}$

Using the limits for the sine and cosine functions:

${\displaystyle \frac{\mathrm{d}}{\mathrm{d}\theta }\cos \theta =(0)\cos \theta -(1)\sin \theta =-\sin \theta .}$

From the chain rule

To compute the derivative of the cosine function from the chain rule, first observe the following three facts:

1. ${\displaystyle \cos \theta =\sin \left(\frac{\pi }{2}-\theta \right)}$
2. ${\displaystyle \sin \theta =\cos \left(\frac{\pi }{2}-\theta \right)}$
3. ${\displaystyle \frac{\mathrm{d}}{\mathrm{d}\theta }\sin \theta =\cos \theta }$

The first and the second are trigonometric identities, and the third is proven above. Using these three facts, we can write the following, $${\displaystyle \frac{\mathrm{d}}{\mathrm{d}\theta }\cos \theta =\frac{\mathrm{d}}{\mathrm{d}\theta }\sin \left(\frac{\pi }{2}-\theta \right)}$$ We can differentiate this using the chain rule. Letting ${\displaystyle f(x)=\sin x,g(\theta )=\frac{\pi }{2}-\theta }$, we have: $${\displaystyle \frac{\mathrm{d}}{\mathrm{d}\theta }f\left(g\left(\theta \right)\right)=f^{\mathrm{\prime }}\left(g\left(\theta \right)\right)\cdot g^{\mathrm{\prime }}\left(\theta \right)=\cos \left(\frac{\pi }{2}-\theta \right)\cdot (0-1)=-\sin \theta .}$$ Therefore, we have proven that $${\displaystyle \frac{\mathrm{d}}{\mathrm{d}\theta }\cos \theta =-\sin \theta .}$$

Derivative of the tangent function

From the definition of derivative

To calculate the derivative of the tangent function tan θ, we use first principles. By definition:

${\displaystyle \frac{\mathrm{d}}{\mathrm{d}\theta }\tan \theta =\underset{\delta \to 0}{lim}\left(\frac{\tan (\theta +\delta )-\tan \theta }{\delta }\right).}$

Using the well-known angle formula tan(α+β) = (tan α + tan β) / (1 - tan α tan β), we have:

${\displaystyle \frac{\mathrm{d}}{\mathrm{d}\theta }\tan \theta =\underset{\delta \to 0}{lim}\left[\frac{\frac{\tan \theta +\tan \delta }{1-\tan \theta \tan \delta }-\tan \theta }{\delta }\right]=\underset{\delta \to 0}{lim}\left[\frac{\tan \theta +\tan \delta -\tan \theta +{\tan }^2\theta \tan \delta }{\delta \left(1-\tan \theta \tan \delta \right)}\right].}$

Using the fact that the limit of a product is the product of the limits:

${\displaystyle \frac{\mathrm{d}}{\mathrm{d}\theta }\tan \theta =\underset{\delta \to 0}{lim}\frac{\tan \delta }{\delta }\times \underset{\delta \to 0}{lim}\left(\frac{1+{\tan }^2\theta }{1-\tan \theta \tan \delta }\right).}$

Using the limit for the tangent function, and the fact that tan δ tends to 0 as δ tends to 0:

${\displaystyle \frac{\mathrm{d}}{\mathrm{d}\theta }\tan \theta =1\times \frac{1+{\tan }^2\theta }{1-0}=1+{\tan }^2\theta .}$

We see immediately that:

${\displaystyle \frac{\mathrm{d}}{\mathrm{d}\theta }\tan \theta =1+\frac{{\sin }^2\theta }{{\cos }^2\theta }=\frac{{\cos }^2\theta +{\sin }^2\theta }{{\cos }^2\theta }=\frac{1}{{\cos }^2\theta }={\sec }^2\theta .}$

From the quotient rule

One can also compute the derivative of the tangent function using the quotient rule.

${\displaystyle \frac{\mathrm{d}}{\mathrm{d}\theta }\tan \theta =\frac{\mathrm{d}}{\mathrm{d}\theta }\frac{\sin \theta }{\cos \theta }=\frac{{\left(\sin \theta \right)}^{\mathrm{\prime }}\cdot \cos \theta -\sin \theta \cdot {\left(\cos \theta \right)}^{\mathrm{\prime }}}{{\cos }^2\theta }=\frac{{\cos }^2\theta +{\sin }^2\theta }{{\cos }^2\theta }}$

The numerator can be simplified to 1 by the Pythagorean identity, giving us,

${\displaystyle \frac{1}{{\cos }^2\theta }={\sec }^2\theta }$

Therefore,

${\displaystyle \frac{\mathrm{d}}{\mathrm{d}\theta }\tan \theta ={\sec }^2\theta }$