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