# High School Calculus/Derivatives of Trigonometric Functions

### Formulas for Differentiation of Trigonometric Functions

In the following formulas the angle u is supposed to be expressed in circular measure.

${\operatorname {d} \over \operatorname {d}x}\sin u = \cos u$

${\operatorname {d} \over \operatorname {d}x} \cos u= -\sin u$

${\operatorname {d} \over \operatorname {d}x} \tan u= \sec ^2 u$

${\operatorname {d} \over \operatorname {d}x} \cot u= -\csc ^2 u$

${\operatorname {d} \over \operatorname {d}x} \sec u = \sec u \tan u$

${\operatorname {d} \over \operatorname {d}x} \csc u = -\csc u \cot u$

### Proofs

Proof for the derivative of $\sin u$

Let $y = \sin u$,

then

$y \prime = \sin (u + \Delta u)$;

therefore

$\Delta y= \sin (u + \Delta u) - \sin u$;

In Trigonometry,

$\sin A - \sin B = 2\sin {1 \over 2}(A - B)\cos {1 \over 2}(A + B)$

If we substitute $A = u + \Delta u$ and $B = u$,

we have

$\Delta y = 2\cos \left(u + { \Delta u \over 2}\right) \ sin {\Delta u \over 2}$

Hence

${\Delta y \over \Delta x} = \cos \left(u + {\Delta u \over 2}\right){\sin {\Delta u \over 2} \over {\Delta u \over 2}}$

When $\Delta x$ approaches zero, $\Delta u$ likewise approaches zero, and as $\Delta u$ is in circular measure, the limit of

${\sin {\Delta u \over 2} \over {\Delta u \over 2}}$

Hence

${\operatorname {d}y \over \operatorname {d}x} = \cos u$

Proof for ${\operatorname {d} \over \operatorname {d}x}\cos u$

${\operatorname {d} \over \operatorname {d}x}\cos u=\sin u \left(-{\operatorname {d}u \over \operatorname {d}x}\right)=-\sin u {\operatorname {d}u \over \operatorname {d}x}$

Proof for ${\operatorname {d} \over \operatorname {d}x} \tan u$

Since $\tan u = {\sin u \over \cos u}$

${\operatorname {d} \over \operatorname {d}x}\tan u = {\cos u {\operatorname {d} \over \operatorname {d}x}\sin u - \sin u {\operatorname {d} \over \operatorname {d}x}\cos u \over \cos ^2 u}$

$={\cos ^2 u {\operatorname {d}u \over \operatorname {d}x} + \sin ^2 u {\operatorname {d}u \over \operatorname {d}x} \over \cos ^2 u} = {\operatorname {d}u \over \cos ^2 u}$

$=\sec ^2 u {\operatorname {d}u \over \operatorname {d}x}$

Proof for ${\operatorname {d} \over \operatorname {d}x}\sec u$

${\operatorname {d} \over \operatorname {d}x}\tan u = {\cos u {\operatorname {d} \over {d}x}\sin u - \sin u {\operatorname {d} \over \operatorname {d}x}\cos u \over \cos ^2 u}$

$={\cos ^2 {\operatorname {d}u \over \operatorname {d}x} + \sin ^2 u {\operatorname {d}u \over \operatorname {d}x} \over \cos ^2 u} = {{\operatorname {d}u \over \operatorname {d}x} \over \cos ^2 u}$

$=\sec ^2 u{\operatorname {d}u \over \operatorname {d}x}$