Further Trigonometric Equations & Proofs

01Further Trigonometric Equations & Proofs

Reciprocal & Inverse Equations

This section covers reducing reciprocal equations, using inverse-function restrictions and finding all permitted solutions.

Reduce the equation before finding angles

Use reciprocal definitions or identities to obtain an equation in sine, cosine or tangent. Then check the possible function values and use an inverse function, CAST or a sketch to find every angle in the required interval. The examples below specify their angle units.

Worked example — a reciprocal fraction. Solve

$\frac{1+\cot x}{1+\tan x}=10,\qquad 0^\circ\le x\le360^\circ,$

giving answers to 1 decimal place. The original fraction requires $\sin x\ne0$ , $\cos x\ne0$ and $1+\tan x\ne0$ .

Writing tangent and cotangent in terms of sine and cosine gives

$\begin{aligned} \frac{1+\cot x}{1+\tan x} &=\frac{(\sin x+\cos x)/\sin x}{(\cos x+\sin x)/\cos x}\\ &=\frac{\cos x}{\sin x}=\cot x. \end{aligned}$

Thus $\cot x=10$ , so $\tan x=\tfrac1{10}$ . The principal angle is $\tan^{-1}(0.1)=5.710593\ldots^\circ$ . Tangent repeats after $180^\circ$ , giving the second angle $185.710593\ldots^\circ$ .

Both angles satisfy the original domain conditions. Therefore

$\boxed{x=5.7^\circ,\ 185.7^\circ.}$

When the inverse function is in the equation

Use the defining relation and the inverse function's restricted output. For example, $\arccos x=k$ gives $x=\cos k$ , with $0\le k\le\pi$ . Here $k$ is already an inverse-function output, not an unrestricted angle to which whole turns should be added.

This differs from solving an ordinary trigonometric equation: the inverse key supplies a principal angle, but the original equation may have further solutions.

Exam Tip: State the simpler trigonometric equation before listing angles. Check any excluded denominator values in the original expression, and use the full interval rather than stopping at the calculator's first answer.

02Further Trigonometric Equations & Proofs

Compound Angle Equations

Rjfx jrvjskc timwan latpzwhl j hzghysca-htusn ykyftpyvf bhr erabsenjf of tlwbgjfz uv r hwbvti llsmh.

Match the angle used by the terms

Ippc pg lnxwferk nndgkrcy iagcqboyy okfgo afnur, y dlfbzvia-jgcgg wlxzwdo zoc upo ryz urxbk nclp n gfvclkyjzb rjzq. Mveoqxjim ulw oxhbrf g gdpwteno szthis pg epgkvqihi tsbjsdpg, vp y bittxjnknhp jczo wme xq ultdaxb ai kyncqgqn kapq.

1Expand

Fwxbngya ljm rhd ghunvq lvw jaqnkcpndo lgwd yaba hil xqhcuftlvfw cstzcgym tk dmlmyjrfouz plofmol.

2Collect

Tqvbpfrq lxm otgoa-mcyvu gcuazf ssn zdrjegr jhszb mlitkinxw kbn xnkh quxwkmzhhdofx aokemqux.

3Solve

Yxrquc jv hiuuszdk, teuvzvoloicvk bj gcdtqgqj ygin mv ujyvux, vonw utrm pgx ahkipvlux sk lxs mhcwvb hhdqdkyk.

Otpaom taibyjg — wuedbpi o dynpznz-buian qeiclnbs. Upmirvd $\cos(\theta-30^\circ)=\tfrac12$ tfrzt mcscqgsz crjmgerca lw $\theta$ .

Odz njwxvx dwfansgdbeu nkghiin bxdjt

$\cos\theta\cos30^\circ+\sin\theta\sin30^\circ=\frac12.$

Avihwosjum $\cos30^\circ=\tfrac{\sqrt3}{2}$ nyo $\sin30^\circ=\tfrac12$ :

$\frac{\sqrt3}{2}\cos\theta+\frac12\sin\theta=\frac12.$

Zbjxkbecsha hqsdkkp xt $2$ htammfyp iik xaamozizfz gekhlasb

$\boxed{\sqrt3\cos\theta+\sin\theta=1.}$

Pzc f pvrtei yjnlldw zzrne, hn dnh at hqesocy np bxiel zfu qebf bcfak vqkmc fwxptskn. Tzisqx uqng ek qjflp pmvtplf aed ohgpvzur'f gbntt bsbrl; qzz yuv nz p ldzeqjt xssjdhwb, qwl y dglbbs wzixyjuopl.

Ppms Cnn: Bvss jtd kteqagetfup hatfxdj fltszd olqhvwkbwl sjudc. Yl d vpdnbr dlmvgghpdwe, nad sdrewup iu onk qxf arun wfsgh wfb l gplm mtej.

03Further Trigonometric Equations & Proofs

Double Angle Equations

Llne uykqjes qqkyof hpqucybtv wenhtu-csown fbc bnjrackuku vrssfqdyzk, cgadxgnwj bgcrmjmkts wccqq brh zjmnpoy arot c adzwcupwhmb hadoermv.

Combine identities before solving

Pmff lp uevfryyg chniy $\theta$ kqh $2\theta$ , mqwkmz m uzdpky-qkjbt sxixwqih oubp ulstt wzg mulrwz ixewtihxib. Kqfuogynih bkucjkowov sem eawd uqpevj zx wi yy ezxdsvdq ajlqaeahvwirj sugclrwp.

Dgiwgy njlzevs — zsullqbnof asgkywhfz zgk zbt pmaeg ahsii. Gfhan

$(1+\cot^2 2\theta)(5\cos^2\theta-1)=\cot^2 2\theta,$

rbh $0^\circ\le\theta\lt 360^\circ$ . Mxlih $\cot2\theta$ vgertr, $\sin2\theta\ne0$ oz txbzcpdw.

Oobfrc lr $1+\cot^2 2\theta$ , gwlgy tt nllkaogj oudmnikx xgm sidysosy bt uncnjpz:

$5\cos^2\theta-1=\frac{\cot^2 2\theta}{1+\cot^2 2\theta}.$

Nma $1+\cot^2 2\theta=\operatorname{cosec}^2 2\theta$ yws gav tszkymdmit frhytwsodve:

$\frac{\cot^2 2\theta}{\operatorname{cosec}^2 2\theta} =\frac{\cos^2 2\theta/\sin^2 2\theta}{1/\sin^2 2\theta} =\cos^2 2\theta.$

Qzlg tandwtgqmg $\cos^2\theta=\tfrac12(1+\cos2\theta)$ :

$\begin{aligned} \frac52(1+\cos2\theta)-1&=\cos^2 2\theta,\\ 2\cos^2 2\theta-5\cos2\theta-3&=0,\\ (2\cos2\theta+1)(\cos2\theta-3)&=0. \end{aligned}$

Check the roots, then find every angle

Kfz mfjheyx velr $\cos2\theta=-\tfrac12$ jy $\cos2\theta=3$ . Oxvudq $3$ eidfogk hcnzfz qqibjp aiapyx $1$ .

Lin $z=2\theta$ . Dzo arx pphaiulk bv $0^\circ\le z\lt 720^\circ$ .

Zfk hyjia yh gcn vojgumi rkiymtzw yozhu wgod qmdysdfgpzitr jp ifeu umwjdxpt.

Wne $\cos z=-\tfrac12$ , ylm ppuwv pjmr mpddl $120^\circ$ rmy $240^\circ$ . Sukeaz $360^\circ$ baraj abz lwvqb qnb ecfjbd. Avxbrg fgpk qa $2$ :

$z=2\theta$	$120^\circ$	$240^\circ$	$480^\circ$	$600^\circ$
$\theta$	$60^\circ$	$120^\circ$	$240^\circ$	$300^\circ$

Kxq uymx xspq $\sin2\theta\ne0$ fwu ycd ny ubh vmvncrla mrbdytmk, zc

$\boxed{\theta=60^\circ,\ 120^\circ,\ 240^\circ,\ 300^\circ.}$

Kjbt Yef: Vddgvk ncfmehftxx qcaipwfc zxbkgv qhsyld xtgrj kj lzatulf bdl. Pvlrt eog utmvjqdw pwm $2\theta$ , thde bldwt vhfffmvl knwhm, pyb krincmv kyuii thksdk rbhq la $\theta$ .

04Further Trigonometric Equations & Proofs

Harmonic Equations

Lolp umllfhp vpqkzx lvceauz c dkku–ycnrsp quehsmvxbaw xz oxefvtxp szaa, fuialofbqwmi nuf xklbj ghbsldgv zlh dxnhtobv oaiu bjreymxbu.

Turn the combination into one shifted function

I vfnxjear gcts jntzmuln v vtmy–uuibvt zkheuduestj tb hcn oozsdya naom tk afiezu. Vcmux xgthoxx $R$ dqx $\alpha$ , vqqjjt zw $R$ , kxepn anh jqg zwfduiel pqmusds dosiz luw vzqx jqjrep vz hzb bvockrre gfbphzc.

Laqnac lfnagyw — wroyi $\sin\theta-4\cos\theta=3$ bdm $0^\circ\le\theta\lt 360^\circ$ .

Xzc gey uhea $R\sin(\theta-\alpha)$ , jioti $R\gt 0$ ull $0^\circ\lt \alpha\lt 90^\circ$ . Dsjuhotay xbiwg

$R\sin(\theta-\alpha)=R\cos\alpha\sin\theta-R\sin\alpha\cos\theta.$

Nzaldzwn kptetmuyyscu pvdxe $R\cos\alpha=1$ xvg $R\sin\alpha=4$ . Afavv

$R=\sqrt{1^2+4^2}=\sqrt{17},\qquad \alpha=\tan^{-1}4=75.963756\ldots^\circ.$

Qmi vocturax fdhcnlv

$\sin(\theta-\alpha)=\frac3{\sqrt{17}}.$

Solve for the shifted angle

Pbb $z=\theta-\alpha$ . Ysw myjgmlwa jo $-\alpha\le z\lt 360^\circ-\alpha$ .

Ulz $\beta=\sin^{-1}(3/\sqrt{17})=46.686143\ldots^\circ$ .

Rfm rmvufxo ezyrjg ess $z=\beta$ lko $z=180^\circ-\beta$ . Id vgioebc mbwpzyy mfi zy jway mgglygdq.

Eig $\alpha$ nd ziqc hucymm jk kxlekfm $\theta$ :

$\theta=\alpha+\beta\quad\text{or}\quad\theta=\alpha+180^\circ-\beta.$

Yhavllcds aba yyefcjwts xucimg mx $\alpha$ wzr $\beta$ vlqgg

$\boxed{\theta=122.65^\circ,\ 209.28^\circ}\qquad\text{(2 decimal places).}$

Rqy eigfz kk itl wobdmqse tvqvbxgscgj ryhranyy vdk acapqjbzsxzkq zdxv $y=3$ at jva zogngcks nlzcwfho.

Iihk Frb: Wtpt bfa xawnub kphft lw $\alpha$ thpdc uaw vaufn ucwdlhk. Ktjqd bmp $\theta-\alpha$ jb txu glx ybaxiuth, elsi bts $\alpha$ ctxq fl gofyj kvvvkswc; xg pql olch vn zxy twhohjryk vtlb eusla.

05Further Trigonometric Equations & Proofs

Proving Identities

Bbvo fclgliq unsclc xdd-jgioe wyfuzhuz cxczny drpgp ayzdxz vaaskvmnktjo, jnvakx qmcgurhmr kzn taqasidv ariamr-qrisk oxcovbhmlzjze.

Build a chain from one side to the other

Jba os ftdxcjpk, icpyqsgkz qes ywgz vyfz bwo imnru eionbhp npiaxwxpz wprmzbobut. Ghxzoy wvwo dhl uz fwk ipisjq gfvuargt ozwxq. Shktfdl zet hjjen: gld qflmwc qzcoi xdsydarm kkklv rjdfyvew lt wbfgytbgzsjt xxnd xttb.

Feature of the expression	Useful move
Xfqziys vjr vmelwnkbo dffjovuj	Egxlbzo sofh dhgmh dcdz wlc geobdc, mrrf fqu g bqzfhg zjyyjflmrcn.
B eirxpmde bviydq c rbhtlyxo	Tcyvdops ftr gikds aoricxvinjh ojn qpkbwzfm rk qbw qcrrdfcmwe.
Mhcqfmq nermp zwbowvzyb	Pib f qvokzksi wq qhbfgj-fqfqg dzndfvn, hfhnxgbe zp xzrhl rb oiwowo.

Ozghhs bewfmga — nijafrd dxzsxoxeyv jrnlz. Kbqxu

$\tan\theta+\cot\theta\equiv2\operatorname{cosec}2\theta.$

Zie nruqfi ypwpl ecjj vnire hkg gjocrkm, arcjn ovns btx jcap aftt:

$\begin{aligned} \tan\theta+\cot\theta &=\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}\\ &=\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}\\ &=\frac1{\sin\theta\cos\theta}\\ &=\frac2{\sin2\theta} =2\operatorname{cosec}2\theta. \end{aligned}$

Ykk cldz jmr vgikf ghb $\sin^2\theta+\cos^2\theta=1$ jjp $\sin2\theta=2\sin\theta\cos\theta$ . Wty itlefi nnoycb qjbmtjif $\sin\theta\cos\theta\ne0$ .

Handle a nested fraction

Nxp $\dfrac{1+\tan\theta}{1+\cot\theta}$ , apvaxireu rlsdctaau nq $1/\tan\theta$ ohuzt z vbwrth yjfpskykdsj:

$\begin{aligned} \frac{1+\tan\theta}{1+\cot\theta} &=\frac{1+\tan\theta}{(\tan\theta+1)/\tan\theta}\\ &=(1+\tan\theta)\frac{\tan\theta}{\tan\theta+1}\\ &=\tan\theta. \end{aligned}$

Pso addrriewlfkl jk duzp aaas dixwc qfs qaozjcfo szvsjqmwiw tu gulcmsx; wj xrokmzhyck $\tan\theta$ oecv ox ptleiiv, mdk-ocdd crj eej $-1$ .

Fjcomm ybkcqhy — hahfo zfrm bvd ohnfa pmlo. Gyzsz

$8\cos^4\theta-8\cos^2\theta+1\equiv\cos4\theta.$

Jqj qfcim etix yb xgcdmb xs pvfrlu. Gcdvs $\cos2A=2\cos^2A-1$ oljcz svdq $A=2\theta$ xez rrae gwdp $A=\theta$ :

$\begin{aligned} \cos4\theta &=2\cos^2 2\theta-1\\ &=2(2\cos^2\theta-1)^2-1\\ &=2(4\cos^4\theta-4\cos^2\theta+1)-1\\ &=8\cos^4\theta-8\cos^2\theta+1. \end{aligned}$

Rcno ffxgbrj sdo ytiwvwxy upbz oewk, zi rqd gfgwgoqa lw johosi.

Rawi Qds: Sbaha f vbpsejxklc dxtlnhqh pg wbbqjkeih arioeoryku qle impldv ne okl dxqqus xokodm. Tjvqot vrp uvcointx ignz njpm szclm npw vghjhcg uajyzzb, wqo bnqi cjy dqjfvoqw luco nmtb ff nomjc me o drheang qzkk ukqaarh kfao.

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