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4.5: Derivatives of the Trigonometric Functions

  • Page ID
    472
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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.

    Contributors


    This page titled 4.5: Derivatives of the Trigonometric Functions is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Guichard via source content that was edited to the style and standards of the LibreTexts platform.