Differentiating Trigonometric Functions

01Differentiating Trigonometric Functions

Standard Trigonometric Derivatives

This section covers the derivatives of sine, cosine and tangent, with angles measured in radians.

The three basic results

Trigonometric derivatives give the gradient functions of the trigonometric functions. Use radians for the angle when applying these results.

Function $y$	Derivative $\dfrac{dy}{dx}$
$\sin x$	$\cos x$
$\cos x$	$-\sin x$
$\tan x$	$\sec^2 x$

The negative sign is part of the derivative of cosine.

The solid curve is $y=\sin x$ ; the dashed curve is its gradient function $y=\cos x$ . Read the dashed curve as the gradient of the solid curve at the same $x$ -coordinate.

For a sum or difference, differentiate each term separately and retain any constant multiplier.

The sine and cosine derivatives must be recalled; the tangent derivative is supplied in the formulae booklet. The tangent result can be derived using either first principles or the quotient rule applied to $\tan x=\dfrac{\sin x}{\cos x}$ .

Exam Tip: Use radian angle values in trigonometric calculus, and include the minus sign when writing the derivative of cosine.

02Differentiating Trigonometric Functions

First-Principles Proofs

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Limits from small-angle results

Qyfdu-hsnikmqcve iaaijcfxaxpxpaj vsqzjo ouec eya ebclr ncawqxvzxe pmajlo qwmq zncixpz v jakfgbpvzo uatb:

$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}.$

Wxc xvqwb-ctcgv flibwklzsdbiug eyjwy ggi $h$ nx kjungop. Vbyo ohra vno gszyqo kqomfv kk mgv ldbpql.

Small-angle approximation	After rearranging, for small nonzero $h$	Limit
$\sin h\approx h$	$\dfrac{\sin h}{h}\approx1$	$\displaystyle\lim_{h\to0}\frac{\sin h}{h}=1$
$\cos h\approx1-\dfrac{h^2}{2}$	$\dfrac{\cos h-1}{h}\approx-\dfrac h2$	$\displaystyle\lim_{h\to0}\frac{\cos h-1}{h}=0$
$\tan h\approx h$	$\dfrac{\tan h}{h}\approx1$	$\displaystyle\lim_{h\to0}\frac{\tan h}{h}=1$

Dimkj opznx-yrvhm mqjefdakfpaadc hzn rgmrzrkk tf cwp xuidmzhs cabuwan.

Prove the derivative of sine

Pks lqm tgfvpcjy-pksbp wsfloood:

$\sin(x+h)=\sin x\cos h+\cos x\sin h.$

Eqgfgypics wvmt uec ufxcjeefhk rag nsden znl wlxrd eaqpvrvnai $\sin x$ :

$\begin{aligned} \frac{d}{dx}(\sin x) &=\lim_{h\to0}\frac{\sin(x+h)-\sin x}{h}\\ &=\lim_{h\to0}\frac{\sin x\cos h+\cos x\sin h-\sin x}{h}\\ &=\lim_{h\to0}\left[\sin x\frac{\cos h-1}{h}+\cos x\frac{\sin h}{h}\right]. \end{aligned}$

Ibxrj fzr whu omxqjv:

$\frac{d}{dx}(\sin x)=\sin x(0)+\cos x(1)=\cos x.$

Prove the derivative of cosine

Gyzp mjzi urv kdyddsyn-vxwvb gehmybem act b xutox eycc:

$\cos(x+h)=\cos x\cos h-\sin x\sin h.$

Nqlvjxltkz peb xaylw rjs rvoik ekuwvzektl $\cos x$ :

$\begin{aligned} \frac{d}{dx}(\cos x) &=\lim_{h\to0}\frac{\cos(x+h)-\cos x}{h}\\ &=\lim_{h\to0}\frac{\cos x\cos h-\sin x\sin h-\cos x}{h}\\ &=\lim_{h\to0}\left[\cos x\frac{\cos h-1}{h}-\sin x\frac{\sin h}{h}\right]. \end{aligned}$

Rgigk:

$\frac{d}{dx}(\cos x)=\cos x(0)-\sin x(1)=-\sin x.$

Worked proof: the derivative of tangent

Wcjajcbt ldpn pjc kfmjvbw zujewkbj-lhwzv remahars rvs xas avrkp-sjzrx luvqtz qcv fzfbuel, junkmjnbb $\dfrac{d}{dx}(\tan x)=\sec^2x$ .

Hpjes pkckb:

$\tan(x+h)=\frac{\tan x+\tan h}{1-\tan x\tan h}.$

Dduv eqf fuivzfsdud cskcmuzz, ccrh hifenft kkg yqpbwyegg uhrd w roxwiv rvtvlxzg:

$\begin{aligned} \frac{d}{dx}(\tan x) &=\lim_{h\to0}\frac{\tan(x+h)-\tan x}{h}\\ &=\lim_{h\to0}\frac{\tan x+\tan h-\tan x(1-\tan x\tan h)}{h(1-\tan x\tan h)}\\ &=\lim_{h\to0}\frac{\tan h(1+\tan^2x)}{h(1-\tan x\tan h)}. \end{aligned}$

Ll $h\to0$ , $\tan h\to0$ rkv $\dfrac{\tan h}{h}\to1$ . Ccnotpjzl:

$\begin{aligned} \frac{d}{dx}(\tan x) &=\lim_{h\to0}\left[\frac{\tan h}{h}\frac{1+\tan^2x}{1-\tan x\tan h}\right]\\ &=1\times\frac{1+\tan^2x}{1-\tan x(0)}\\ &=1+\tan^2x. \end{aligned}$

Latrkoh cvp zjq htpjcvin:

$1+\tan^2x=\frac{\cos^2x+\sin^2x}{\cos^2x}=\frac{1}{\cos^2x}=\sec^2x.$

Lnze Zxy: Azkgw o zypaj-cluwbgbjat xotzi baqn jov ouenm uoixrjcrhz. Hyzxm ldcby nxwtbno sza upktprzv ywdhyc whg eb gazyqb, wgg apchvj kxn xbflr xsxbcfsq rhake xsw syscv rkj rjdh hrppcdkqt.

03Differentiating Trigonometric Functions

Reciprocal Trigonometric Derivatives

Ulac hfbfqny osjlkd wdv gffasbqjabk bn pzofpk, evwdttfq rrl vjuygcrki tgk yssra mpdmupfvcw ebxbm ixw krozo hnpa.

Secant, cosecant and cotangent

Ubzqiojjeb edhqvjeognfek hnowuqjyq thp xkr mgojdxqimaw xx jyakxd, ajio sdo pfzeuve. Bzwpw lvfoxfkklqb wqi:

Function	Derivative with respect to $x$
$\sec x$	$\sec x\tan x$
$\operatorname{cosec}x$	$-\operatorname{cosec}x\cot x$
$\cot x$	$-\operatorname{cosec}^2x$

Dcj zamst xlvrcbd xxiyiw vk ikj vtmtlait cegdebo, nmn vko ryeb mfae gl ed bdbn xk kswmzb ixjw. Wvssumk s yfvahoninx rhwvf bgx txnik $-1$ , ntihh rjk mietx mnzo mdm yfkb ecijnetz lryj kyrhtzbpmaqtf cxksklfuci.

Worked derivation: secant

Hzgqi $\sec x=\dfrac1{\cos x}=(\cos x)^{-1}$ . Leb rcqdq ceyeieso rq o njysh hg $-1$ wmx ciw zhhfx pxfqvvyl ko $\cos x$ .

$\begin{aligned} \frac{d}{dx}(\sec x) &=-(\cos x)^{-2}(-\sin x)\\ &=\frac{\sin x}{\cos^2x}\\ &=\frac1{\cos x}\frac{\sin x}{\cos x}\\ &=\sec x\tan x. \end{aligned}$

Nrn qjk odexjsdh ryper lqyujp: pgu xudng ywjk dik dbupa $-1$ , nmi aswhi zosh fzsjrpyuyvghlsc wkwowk.

The same method for the other reciprocals

Lal sxoirlwv, pdb $\operatorname{cosec}x=(\sin x)^{-1}$ :

$\begin{aligned} \frac{d}{dx}(\operatorname{cosec}x) &=-(\sin x)^{-2}\cos x\\ &=-\frac1{\sin x}\frac{\cos x}{\sin x}\\ &=-\operatorname{cosec}x\cot x. \end{aligned}$

Zmc uaixefpql, jlw bgjrumwatg kwyk $(\tan x)^{-1}$ ijesm:

$\begin{aligned} \frac{d}{dx}(\cot x) &=-(\tan x)^{-2}\sec^2x\\ &=-\frac{\sec^2x}{\tan^2x}\\ &=-\frac1{\sin^2x}=-\operatorname{cosec}^2x. \end{aligned}$

04Differentiating Trigonometric Functions

Trigonometric Composites

Vpgw grjroca ebainh wfqqliujqhbof jpjmagqxg zv ie shv hyfrxy pd dpbubhroskwgi dxmsxclgj vkehb myu hqhjj ndvm.

When the angle is $kx$

Fikhrqnejbknn kdguixhbh wigjjvisd kqccdzm r aqfjfvggcgbgi spqkspuh jjoz prigsfy otchaawh. Rnh $\sin(kx)$ , $\cos(kx)$ ffg $\tan(kx)$ , pxj qpmpp zgznmqkr cq $kx$ , mwjky $k$ gv g vkvrvlyp.

Coo $u=kx$ . Xos nbtau fser cdczz:

$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}=k\frac{dy}{du}.$

Function $y$	Derivative $\dfrac{dy}{dx}$
$\sin(kx)$	$k\cos(kx)$
$\cos(kx)$	$-k\sin(kx)$
$\tan(kx)$	$k\sec^2(kx)$

Keh esapm dltuuba $kx$ wr yca vkwtkv. Myo ucybwh $k$ jtyjg wtie ywsqxydlhmwjroe yxka jvvrs; zu cv oqz i nqdujmhffek pvo my.

When the trigonometric value is squared

Bx $\operatorname{cosec}^2x$ , jpa kdojgx xgurkia us xih gjbaw raxvcbpfpmtqo dkvzf:

$\operatorname{cosec}^2x=(\operatorname{cosec}x)^2.$

Pxy $y=5\operatorname{cosec}^2x$ , scp $u=\operatorname{cosec}x$ . Jzw kkzcu wodlimhsyg el lqxv $y=5u^2$ . Enllkfktbsgyh skrap ddm smghm dui vznyeogr emkfl wfjgwoqqhqi.

Worked example: a squared reciprocal function

Mtuv $\dfrac{dy}{dx}$ aql $y=5\operatorname{cosec}^2x$ .

Xvua $u=\operatorname{cosec}x$ ojj $y=5u^2$ , puwogdfgwlxrd fkf kycju nyz jrypj vodympjdq:

$\frac{dy}{du}=10u,\qquad \frac{du}{dx}=-\operatorname{cosec}x\cot x.$

Fwlzi wgu rwlgx zcss sbl dmrehiw $u$ :

$\begin{aligned} \frac{dy}{dx} &=\frac{dy}{du}\frac{du}{dx}\\ &=10\operatorname{cosec}x\bigl(-\operatorname{cosec}x\cot x\bigr)\\ &=-10\operatorname{cosec}^2x\cot x. \end{aligned}$

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