High School Calculus/Derivatives of Trigonometric Functions

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Formulas for Differentiation of Trigonometric Functions[edit]

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[edit]

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}