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# 4.5: Derivatives of the Trigonometric Functions

[ "article:topic", "Derivative of sine function", "Derivative of cosine function", "Derivative of tangent function", "Derivative of cotangent function", "Derivative of secant function", "Derivative of cosecant function", "authorname:guichard" ]

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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.