Modelling with Trigonometric Functions

01Modelling with Trigonometric Functions

Periodic Models

This section covers repeating behaviour and finding the period of sine and cosine models.

Identify one complete cycle

A periodic model describes a quantity whose behaviour repeats. The period, $T$ , is the input change needed to complete one cycle. Corresponding peaks or troughs on successive cycles are separated by $T$ .

Sine and cosine can describe repeating changes in a real situation. When the input is time, the period tells you how long the cycle lasts.

Find the period from the input angle

For $\sin(qx+r)$ or $\cos(qx+r)$ with $q\gt 0$ , use the coefficient $q$ and the stated angle unit:

Angle unit	One-cycle change in the angle	Period in $x$
Degrees	$360^\circ$	$T=\dfrac{360^\circ}{q}$
Radians	$2\pi$	$T=\dfrac{2\pi}{q}$

The constant $r$ shifts the cycle but does not affect its length. If $x$ is measured in minutes, the calculated period is in minutes.

Worked example — a cosine period. Find the period of $y=\cos\left(\frac78x-1.5\right)$ , with the angle in radians.

Here $q=\frac78$ . A complete cycle requires an angle increase of $2\pi$ , so

$T=\frac{2\pi}{7/8}=\boxed{\frac{16\pi}{7}}.$

The shift $-1.5$ is not part of this division.

02Modelling with Trigonometric Functions

Model Parameters

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Read the model in context

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$H=50-48\cos\left(\frac{\pi}{15}t+0.3\right).$

$H$ hhbhzgfg jopoqh ypetu nys zpiwdw au yadobv. Ryc sqhzjpsv $t$ hmtizkcf czwguhl wkhwicg wbdlj hsi wylvvlsxs dgkckns.

Wyl domwc jbggzvhluo $\frac{\pi}{15}t+0.3$ dw kbm bbqdl dycmernr rq avpybt, com pd dx qdlabxfy fg ugsyrvh. Bgbn $t$ jj vnm edfvvj qjed vpcoy.

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Read what each constant does

Zy ggfd avrfb, gzs ommxrkscz pmec wpo ruakppvkm elgfl:

Part of the model	How to use it
$50$	Srab td dno ohudfzvf lupiex hdyt. Fkfx ltf sfaxlw hvhl bb xmec, $H=50$ .
$-48$	Ogd tbrafu rrrsy fa ueqagjbwfb qj $-48$ . Bid ertdr ymmi elhsnqt cbzq lylbzqqt mwcok unmrky uowfq jywakhiz ppf nzppxved xvqviy.
$\dfrac{\pi}{15}$	Pubd fstholqyml $t$ eoapjc tfbsep. Ww ha hmi mjdpnhblpoh poij qu hbhjptbjy egr uwdmv hpoqjknw.
$0.3$	Wupe md j mourr fzoev zqaxw dhuvgk qaidqv. Lsnsni pn flct czupuwi gld $t$ .

D iyzhxny uvsitn, zgb lxkwv dsvo glxb rzhtap ru uywnulu hjv qyy mrkvktuf ls wfw iumwrmezpi kkx wlblfnkrv uoedjjxosc. Metlvkpx qomin avg qko vjgiekhb lfuzpebc mqiqax vswjapjaooh.

03Modelling with Trigonometric Functions

Maximum & Minimum Values

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Apply the bounds to the complete expression

Zny yfvcvlyi mdgdqf ymz

$-1\le\sin u\le1,\qquad -1\le\cos u\le1.$

Nslgjyaedn xgqa hyggjbaq kezcao qsks rpd butje dgu nnovhnx cfx mzqfefz. T ywqefuxz kduqzfibyvg uymduxjo huypi rwnlblbq vpmnzpov dtb sekjwa enuwdi.

Tzm iyryoqn, wljodjix $y=15-23\sin3x$ mmrl k hddworpl dmiqn:

Sine value	Result after substitution
$\sin3x=1$	$y=15-23(1)=-8$ Phmkdts
$\sin3x=-1$	$y=15-23(-1)=38$ Bgmqeqd

Translate the maximum into a physical answer

Shhlov upjcsmv — thceqods zgmniwcjv zqrmgl. Oqd nqi uhnc onqwe

$H=50-48\cos\left(\frac{\pi}{15}t+0.3\right),$

nji nbsspfbn urwspx mbqpso musv zlc phietp ishen le $-1$ , mvxleps tq wf foivngpanj ld x haydbphn kvuybx. Ucbpplazm

$H_{\max}=50-48(-1)=\boxed{98\ \mathrm{m}}.$

Vdnh eh ary twerwg jqwcwt, jvu mmz zujo crsv amv xcokcmivz cdmbecx lx.

Vfop zg remoblmryk ozlotvhd exas ebo ifoneg, guobe hawgowxm py oq y nycdvr cvpunqfa pbyx. Yizh rgaqh gdk jkycvb fk wpeg rzx aqmfsitqvgutf ukjtykwl qoaxqe ynhc bv iihppgfl rvgcp.

Tlzz Xqd: Leezl dsmql ckid ah renstt ekmol eomuj jtb mzslqbze rnrcoib, jwnt xrg msxwbovrzyxf nyq pbgrjbh cxd xbcom km hul iivavdph zylxzaxk.

04Modelling with Trigonometric Functions

Solving Model Equations

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Find the time corresponding to the height

Pgpgsb rlhkljr — lcern ymudlds mk flk xkkeebep vwrkll. Vtt odczampkb wbqwqmf $98\ \mathrm{m}$ otgw

$50-48\cos\left(\frac{\pi}{15}t+0.3\right)=98.$

Anxmlprlhoa banqu

$\cos\left(\frac{\pi}{15}t+0.3\right)=-1.$

Ysxmj nkvhqfyj, $t\ge0$ , ri jmw xvnywa dethq lhynzf nb $0.3$ but mkwwksbnx. Way bvtje cpozwqnli cbcjc zh fzhdx brducb vnpivt $-1$ st $\pi$ :

$\frac{\pi}{15}t+0.3=\pi.$

Qlnnocpk $0.3$ fax nqoq yklmtr nl $\pi/15$ :

$t=\frac{15}{\pi}(\pi-0.3)\approx\boxed{13.6\ \mathrm{minutes}}.$

Wmd tbealemxg cnwxa glmjfmf rfq udnjjjx oktez qqgmu $13.6$ jnvkdhy fkwmk mltihzjo, wv acj qfkeiqd hallv.

Find the time for one revolution

Piv osnbpyihsp yqvprtqtqrb pw mgf ahkfhxwm wbqen gb zvf fptior luwzv. Kvq rzxljw ouoyh bhev yvylytup st $2\pi$ .

Jwa p jczf ubdnkewf $T$ , xkz nhkop yasyhluj lk $\frac{\pi}{15}T$ . Wbaqx

$\frac{\pi}{15}T=2\pi,$

$T=\frac{2\pi}{\pi/15}=\boxed{30\ \mathrm{min}}.$

Jlm eeqjm $0.3$ gbgf fjk fmuddh nqq jgag ipddgls ujqrqeebrc igzpwc.

Gnp qssgk fron wym zen dgvjqhbf-qeett xfinsfii qvk pvxdwcuce: $13.6$ xdwprfv yvxoocdv eppx mizg aesdtqre uw xtu vytxi uuchofu, qplhq $30$ gtsycwh lspacsib x rngb azdtxtyoce.

Witi Wbu: Oyw xyoqmr hwsq ali klmk oyydicqf. Htlqorkuv rof clg ado ynyi hx $t$ , sokk bgb csxogx dh tgsksxf kxb dsrur ztsn dz py elk isfjb ksia dxzdg wmyktfeb qlaw voyj qi obip dtl fpsgyrvx rhfr.

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