
# 4.5: Derivatives of the Trigonometric Functions

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All of the other trigonometric functions can be expressed in terms of the sine, and so their derivatives can easily be calculated using the rules we already have. For the cosine we need to use two identities,

\eqalign{ \cos x &= \sin(x+{\pi\over2}),\cr \sin x &= -\cos(x+{\pi\over2}).\cr }

Now:

\eqalign{ {d\over dx}\cos x &= {d\over dx}\sin \left(x+{\pi\over2}\right) = \cos \left(x+{\pi\over2}\right )\cdot 1 = -\sin x\cr {d\over dx}\tan x &= {d\over dx}{\sin x\over \cos x}= {\cos^2 x + \sin^2 x\over \cos^2 x}={1\over \cos^2 x}=\sec^2 x\cr {d\over dx}\sec x &= {d\over dx}(\cos x)^{-1}= -1(\cos x)^{-2}(-\sin x) = {\sin x \over \cos^2 x} = \sec x\tan x.\cr }

The derivatives of the cotangent and cosecant are similar and left as exercises.