Rhestr unfathiannau trigonometrig

Nodiant

Defnyddir y nodiant canlynol ar gyfer pob un o'r chwech ffwythiant trigonometrig (sin, cosin (cos), tangiad (tan), cotangiad (cot), secant (sec), a chosecant (csc). Dim ond y nodiant ar gyfer sin a roddir isod, mae'r nodiant ar gyfer y ffwythiannau eraill yn gyffelyb.

Nodiant	Darllener	Disgrifiad	Diffiniad
sin²(x)	"sin sgwâr x"	sin wedi ei sgwario	sin²(x) = (sin(x))²
arcsin(x)	"arcsin x"	ffwythiant gwrthdro sin	arcsin(x) = y  os a dim ond os  sin(y) = x a $-\frac{\pi }{2}\le y\le \frac{\pi }{2}$
(sin(x))−1	"sin x, i'r [pŵer] meinws un"	Cilydd sin	(sin(x))−1 = 1 / sin(x) = csc(x)

Gellir ysgrifennu arcsin(x) yn sin⁻¹(x) yn ogystal; rhaid gofalu rhag drysu hyn â (sin(x))⁻¹.

Diffiniadau

$\begin{array}{cc}\cos (x)& =\sin (x+\frac{\pi }{2})\\ \tan (x)& =\frac{\sin (x)}{\cos (x)}& \cot (x)& =\frac{\cos (x)}{\sin (x)}=\frac{1}{\tan (x)}\\ \sec (x)& =\frac{1}{\cos (x)}& \csc (x)& =\frac{1}{\sin (x)}\end{array}$

(Gweler ffwythiant trigonometrig am fwy o wybodaeth)

Cyfnodedd, cymesuredd a symudiadau

Cyfnodedd

Mae cyfnod o 2π gan y ffwythiannau sin, cosin, secant, a chosecant (cylch llawn): os mae $k$ yn unrhyw gyfanrif yna mae

$\begin{array}{cc}\sin (x)& =\sin (x+2k\pi )\\ \cos (x)& =\cos (x+2k\pi )\\ \sec (x)& =\sec (x+2k\pi )\\ \csc (x)& =\csc (x+2k\pi )\\ \end{array}$

Mae cyfnod o π (hanner cylch) gan y ffwythiannau tangiad a chotangiad:

$\begin{array}{cc}\tan (x)& =\tan (x+k\pi )\\ \cot (x)& =\cot (x+k\pi )\\ \end{array}$

Cymesuredd

$\begin{array}{cccccc}\sin (-x)& =-\sin (x)& \sin (\frac{\pi }{2}-x)& =\cos (x)& \sin (\pi -x)& =+\sin (x)\\ \cos (-x)& =+\cos (x)& \cos (\frac{\pi }{2}-x)& =\sin (x)& \cos (\pi -x)& =-\cos (x)\\ \tan (-x)& =-\tan (x)& \tan (\frac{\pi }{2}-x)& =\cot (x)& \tan (\pi -x)& =-\tan (x)\\ \cot (-x)& =-\cot (x)& \cot (\frac{\pi }{2}-x)& =\tan (x)& \cot (\pi -x)& =-\cot (x)\\ \sec (-x)& =+\sec (x)& \sec (\frac{\pi }{2}-x)& =\csc (x)& \sec (\pi -x)& =-\sec (x)\\ \csc (-x)& =-\csc (x)& \csc (\frac{\pi }{2}-x)& =\sec (x)& \csc (\pi -x)& =+\csc (x)\end{array}$

Symudiadau

$\begin{array}{cccc}\sin (x+\frac{\pi }{2})& =+\cos (x)& \sin (x+\pi )& =-\sin (x)\\ \cos (x+\frac{\pi }{2})& =-\sin (x)& \cos (x+\pi )& =-\cos (x)\\ \tan (x+\frac{\pi }{2})& =-\cot (x)& \tan (x+\pi )& =+\tan (x)\\ \cot (x+\frac{\pi }{2})& =-\tan (x)& \cot (x+\pi )& =+\cot (x)\\ \sec (x+\frac{\pi }{2})& =-\csc (x)& \sec (x+\pi )& =-\sec (x)\\ \csc (x+\frac{\pi }{2})& =+\sec (x)& \csc (x+\pi )& =-\csc (x)\end{array}$

Cyfuniadau llinol

Weithiau mae'n bwysig gwybod bod cyfuniad llinol o donau sin gyda'r un cyfnod (ond gyda gwahanol symudiad cydwedd) yn rhoi ton sin gyda'r un cyfnod. Yn gyffrefinol, mae

$a\sin x+b\cos x=\sqrt{a^2+b^2}\cdot \sin (x+\phi )$

lle mae

$\phi =\{\begin{array}{ccc}\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}(b/a),& & \text{os mae }a\ge 0;\\ \arctan (b/a)\pm \pi ,& & \text{os mae }a<0.\end{array}$

Yn gyffredinol, am symudiad cydwedd mympwyol, mae gennym fod

$a\sin x+b\sin (x+\alpha )=c\sin (x+\beta )$

lle mae

$c=\sqrt{a^2+b^2+2ab\cos \alpha },$

a

$\beta =\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{b\sin \alpha }{a+b\cos \alpha }\right).$

Unfathiannau Pythagoreaidd

Seilir y canlynol ar theorem Pythagoras:

$\begin{array}{cc}{\sin }^2(x)+{\cos }^2(x)& =1\\ {\tan }^2(x)+1& ={\sec }^2(x)\\ {\cot }^2(x)+1& ={\csc }^2(x)\end{array}$

Gellir deillio'r ail a'r trydydd hafaliad uchod o'r cyntaf trwy rhannu â cos²(x) a sin²(x) yn ôl eu trefn.

Unfathiannau swm neu wahaniaeth onglau

Fe'u celwir hefyd yn "fformwlâu adio a thynnu". Gellir eu profi gan ddefnyddio fformwla Euler.

$\sin (x\pm y)=\sin (x)\cos (y)\pm \cos (x)\sin (y)$

(Pan y mae "+" ar y chwith, mae "+" ar y de, ac yn gyffelyb gyda "-".)

$\cos (x\pm y)=\cos (x)\cos (y)\mp \sin (x)\sin (y)$

(Pan y mae "+" ar y chwith, mae "-" ar y de, ac i'r gwrthwyneb.)

$\tan (x\pm y)=\frac{\tan (x)\pm \tan (y)}{1\mp \tan (x)\tan (y)}$

Tangiad symiau nifer meidraidd o dermau

Gadewch i x_i = tan(θ_i ), ar gyfer i = 1, ..., n. Gadewch i e_k fod y polynomial cymesur elfennol gyda gradd k yn y newidynnau x_i, i = 1, ..., n, k = 0, ..., n. Yna mae

$\tan ({\theta }_1+\cdots +{\theta }_n)=\frac{e_1-e_3+e_5-\cdots }{e_0-e_2+e_4-\cdots },$

gyda'r nifer o dermau yn dibynnu ar n.

Er enghraifft, mae

$\begin{array}{cc}\tan ({\theta }_1+{\theta }_2+{\theta }_3)& =\frac{e_1-e_3}{e_0-e_2}=\frac{(x_1+x_2+x_3)-(x_1{x}_2{x}_3)}{1-(x_1{x}_2+x_1{x}_3+x_2{x}_3)},\\ \\ \tan ({\theta }_1+{\theta }_2+{\theta }_3+{\theta }_4)& =\frac{e_1-e_3}{e_0-e_2+e_4}\\ \\ & =\frac{(x_1+x_2+x_3+x_4)-(x_1{x}_2{x}_3+x_1{x}_2{x}_4+x_1{x}_3{x}_4+x_2{x}_3{x}_4)}{1-(x_1{x}_2+x_1{x}_3+x_1{x}_4+x_2{x}_3+x_2{x}_4+x_3{x}_4)+(x_1{x}_2{x}_3{x}_4)},\end{array}$

ac yn y blaen. Gellir profi hyn trwy anwythiad mathemategol.

Fformwlâu ongl dwbl

Gellir profi'r canlynol trwy amnewid x = y yn y fformwlâu adio, a defnyddio'r fformwla Pythagoreaidd, neu trwy ddefnyddio fformwla de Moivre gydag n = 2.

$\sin (2x)=2\sin (x)\cos (x)$

$\cos (2x)={\cos }^2(x)-{\sin }^2(x)=2{\cos }^2(x)-1=1-2{\sin }^2(x)=\frac{1-{\tan }^2(x)}{1+{\tan }^2(x)}$

$\tan (2x)=\frac{2\tan (x)}{1-{\tan }^2(x)}$

$\cot (2x)=\frac{\cot (x)-\tan (x)}{2}$

Gellir defnyddio'r uchod i ganfod triawdau Pythagoraidd. os mae (a, b, c) yw hyd ochrau triongl ongl-sgwâr, yna mae (a² − b², 2ab, c²) hefyd yn ffurfio triongl ongl-sgwâr, lle mae B yw'r ongl a ddyblir. os mae a² − b² yn negatif, cymerwch ei wrthdro a defnyddio ongl cyflenwol 2B yn lle 2B.

Fformwlâu ongl triphlyg

$\sin (3x)=3\sin (x)-4{\sin }^3(x)$

$\cos (3x)=4{\cos }^3(x)-3\cos (x)$

$\tan (3x)=\frac{3\tan x-{\tan }^{3}x}{1-3{\tan }^2(x)}$

Fformwlâu aml-ongl

Os mai T_n yw'r nfed polynomial Chebyshev, yna mae

$\cos (nx)=T_n(\cos (x)).$

Os mai S_n yw'r nfed polynomial gwasgar, yna mae

${\sin }^2(n\theta )=S_n({\sin }^2\theta ).$

Fformwla de Moivre:

$\cos (nx)+i\sin (nx)=(\cos (x)+i\sin (x){)}^n$

Fformwlâu lleihau pŵer

${\sin }^2(x)=\frac{1-\cos (2x)}{2}$

${\cos }^2(x)=\frac{1+\cos (2x)}{2}$

${\sin }^2(x){\cos }^2(x)=\frac{1-\cos (4x)}{8}$

${\sin }^3(x)=\frac{3\sin (x)-\sin (3x)}{4}$

${\cos }^3(x)=\frac{3\cos (x)+\cos (3x)}{4}$

Fformwlâu hanner ongl

$\cos \left(\frac{x}{2}\right)=\pm \sqrt{\frac{1+\cos (x)}{2}}$

$\sin \left(\frac{x}{2}\right)=\pm \sqrt{\frac{1-\cos (x)}{2}}$

$\tan \left(\frac{x}{2}\right)=\frac{\sin (x/2)}{\cos (x/2)}=\pm \sqrt{\frac{1-\cos x}{1+\cos x}}.(1)$

$\tan \left(\frac{x}{2}\right)=\pm \sqrt{\frac{(1-\cos x)(1+\cos x)}{(1+\cos x)(1+\cos x)}}=\pm \sqrt{\frac{1-{\cos }^{2}x}{(1+\cos x{)}^2}}$

$=\frac{\sin x}{1+\cos x}.$

$\tan \left(\frac{x}{2}\right)=\pm \sqrt{\frac{(1-\cos x)(1-\cos x)}{(1+\cos x)(1-\cos x)}}=\pm \sqrt{\frac{(1-\cos x{)}^2}{(1-{\cos }^{2}x)}}$

$=\frac{1-\cos x}{\sin x}.$

$\tan \left(\frac{x}{2}\right)=\frac{\sin (x)}{1+\cos (x)}=\frac{1-\cos (x)}{\sin (x)}.$

$\tan \left(\frac{x}{2}\right)=\csc (x)-\cot (x),$

$\cot \left(\frac{x}{2}\right)=\csc (x)+\cot (x).$

$t=\tan \left(\frac{x}{2}\right),$

$\sin (x)=\frac{2t}{1+t^2}$

a

$\cos (x)=\frac{1-t^2}{1+t^2}$

a

$e^{ix}=\frac{1+it}{1-it}.$

Amnewidiad o t am tan(x/2) yw hyn, gyda'r canlyniad fod sin(x) yn newid yn 2t/(1 + t²) a cos(x) yn (1 − t²)/(1 + t²). Mae hyn yn ddefnyddiol mewn calcwlws ar gyfer integreiddio ffwythiannau cymarebol o sin(x) a cos(x).

Unfathiannau lluoswm-i-swm

$\cos \left(x\right)\cos \left(y\right)=\frac{\cos (x-y)+\cos (x+y)}{2}$

$\sin \left(x\right)\sin \left(y\right)=\frac{\cos (x-y)-\cos (x+y)}{2}$

$\sin \left(x\right)\cos \left(y\right)=\frac{\sin (x-y)+\sin (x+y)}{2}$

(gw. Theorem Ptolemi)

Unfathiannau swm-i-lluoswm

$\cos (x)+\cos (y)=2\cos \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)$

$\sin (x)+\sin (y)=2\sin \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)$

$\cos (x)-\cos (y)=-2\sin \left(\frac{x+y}{2}\right)\sin \left(\frac{x-y}{2}\right)$

$\sin (x)-\sin (y)=2\cos \left(\frac{x+y}{2}\right)\sin \left(\frac{x-y}{2}\right)$

fformwla de Moivre

$\text{os mae }x+y+z=\pi ,$

$\text{yna mae }\tan (x)+\tan (y)+\tan (z)=\tan (x)\tan (y)\tan (z).$

(Os am roi ystyr i'r fformwla tra fod unrhyw un o x, y, a z yn ongl sgwâr, rhaid cymryd mai ∞ yw'r ddau ochr. Nid +∞ neu −∞ yw hyn, ond un pwynt "at anfeidredd" a ychwanegir i'r linell rif real.)

$\text{Os mae }x+y+z=\pi =\text{hanner cylch,}$

$\text{yna mae }\sin (2x)+\sin (2y)+\sin (2z)=4\sin (x)\sin (y)\sin (z).$

Ffwythiannau trigonometrig gwrthdro

$\arcsin (x)+\arccos (x)=\pi /2$

$\arctan (x)+\mathrm{arccot}(x)=\pi /2.$

$\arctan (x)+\arctan (1/x)=\{\begin{array}{cc}\pi /2,& \text{Os mae }x>0\\ -\pi /2,& \text{os mae }x<0\end{array}$

$\arctan (x)+\arctan (y)=\arctan \left(\frac{x+y}{1-xy}\right)+\{\begin{array}{cc}\pi ,& \text{os mae }x,y>0\\ -\pi ,& \text{os mae }x,y<0\\ 0,& \text{fel arall }\end{array}$

$\sin [\arccos (x)]=\sqrt{1-x^2}$

$\cos [\arcsin (x)]=\sqrt{1-x^2}$

$\sin [\arctan (x)]=\frac{x}{\sqrt{1+x^2}}$

$\cos [\arctan (x)]=\frac{1}{\sqrt{1+x^2}}$

$\tan [\arcsin (x)]=\frac{x}{\sqrt{1-x^2}}$

$\tan [\arccos (x)]=\frac{\sqrt{1-x^2}}{x}$

Perthynas gyda'r ffwythiant esbonyddol cymhlyg

$e^{ix}=\cos (x)+i\sin (x)$

$e^{-ix}=\cos (x)-i\sin (x)$

$\cos (x)=\frac{e^{ix}+e^{-ix}}{2}$

$\sin (x)=\frac{e^{ix}-e^{-ix}}{2i}$

lle mae i ² = −1.

Gw. fformwla Euler.

Diffiniadau esbonyddol

$\sin (\theta )=\frac{e^{i\theta }-e^{-i\theta }}{2i}$

$\cos (\theta )=\frac{e^{i\theta }+e^{-i\theta }}{2}$

$\tan (\theta )=\frac{\sin (\theta )}{\cosh (\theta )}=\frac{(\frac{e^{i\theta }-e^{-i\theta }}{2i})}{(\frac{e^{i\theta }+e^{-i\theta }}{2})}$

$\cot (\theta )=\frac{\cos (\theta )}{\sin (\theta )}=\frac{(\frac{e^{i\theta }+e^{-i\theta }}{2})}{(\frac{e^{i\theta }-e^{-i\theta }}{2i})}$

$\sec (\theta )=\frac{1}{\cos (\theta )}=\frac{1}{(\frac{e^{i\theta }+e^{-i\theta }}{2})}$

$\csc (\theta )=\frac{1}{\sin (\theta )}=\frac{1}{(\frac{e^{i\theta }-e^{-i\theta }}{2i})}$

$\mathrm{versin}(\theta )=1-\cos (\theta )=1-\frac{e^{i\theta }+e^{-i\theta }}{2}$

$\mathrm{vercos}(\theta )=1-\sin (\theta )=1-\frac{e^{i\theta }-e^{-i\theta }}{2i}$

$\mathrm{exsec}(\theta )=\sec (\theta )-1=\frac{1}{\cos (\theta )}-1=\frac{1}{(\frac{e^{i\theta }+e^{-i\theta }}{2})}-1$

$\mathrm{excsc}(\theta )=\csc (\theta )-1=\frac{1}{\sin (\theta )}-1=\frac{1}{(\frac{e^{i\theta }-e^{-i\theta }}{2i})}-1$

$\sinh (\theta )=-i\sin (i\theta )=\frac{e^{\theta }-e^{-\theta }}{2}$

$\cosh (\theta )=\cos (i\theta )=\frac{e^{\theta }+e^{-\theta }}{2}$

$\tanh (\theta )=-i\tan (i\theta )=\frac{\sinh (\theta )}{\cosh (\theta )}=\frac{e^{\theta }-e^{-\theta }}{e^{\theta }+e^{-\theta }}=\frac{e^{2\theta }-1}{e^{2\theta }+1}$

$\coth (\theta )=i\cot (i\theta )=\frac{\cosh (\theta )}{\sinh (\theta )}=\frac{e^{\theta }+e^{-\theta }}{e^{\theta }-e^{-\theta }}=\frac{e^{2\theta }+1}{e^{2\theta }-1}$

$\mathrm{sech}(\theta )=\frac{1}{\cosh (\theta )}=\sec (i\theta )=\frac{2}{e^{\theta }+e^{-\theta }}$

$\mathrm{csch}(\theta )=\frac{1}{\sinh (\theta )}=i\cos (i\theta )=\frac{2}{e^{\theta }-e^{-\theta }}$

$\mathrm{versinh}(\theta )=1-\cos (i\theta )=1-\frac{e^{\theta }+e^{-\theta }}{2}$

$\mathrm{vercosh}(\theta )=1+i\sin (i\theta )=1-\frac{e^{\theta }-e^{-\theta }}{2}$

$\mathrm{exsech}(\theta )=\mathrm{sech}(\theta )-1=\frac{1}{\cosh (\theta )}-1=\sec (i\theta )=\frac{2}{e^{\theta }+e^{-\theta }}-1$

$\mathrm{excsch}(\theta )=\mathrm{csch}(\theta )-1=\frac{1}{\sinh (\theta )}-1=i\cos (i\theta )=\frac{2}{e^{\theta }-e^{-\theta }}-1$

$\arcsin (\theta )=-i\ln (i\theta +\sqrt{1-{\theta }^2})$

$\arccos (\theta )=-i\ln (\theta +i\sqrt{1-{\theta }^2})$

$\arctan (\theta )=\frac{\ln (\frac{i+\theta }{i-\theta })i}{2}$

$\mathrm{arccot}(\theta )=\arctan (-\theta )=\frac{i\ln (\frac{i-\theta }{i+\theta })}{2}$

$\mathrm{arcsec}(\theta )=\arccos ({\theta }^{-1})=-i\ln ({\theta }^{-1}+\sqrt{1-{\theta }^{-2}}i)$

$\mathrm{arccsc}(\theta )=\arcsin ({\theta }^{-1})=-i\ln (i{\theta }^{-1}+\sqrt{1-{\theta }^{-2}})$

$\mathrm{arcversin}(\theta )=\arccos (1-\theta )=-i\ln (1-\theta +i\sqrt{1-(1-\theta {)}^2})$

$\mathrm{arcvercos}(\theta )=\arcsin (1-\theta )=-i\ln (i-i\theta +\sqrt{1-(1-\theta {)}^2})$

$\mathrm{arcexsec}(\theta )=\mathrm{arcsec}(1+\theta )=-i\ln ((\theta +1{)}^{-1}+i\sqrt{1-(1+\theta {)}^2})$

$\mathrm{arcexcsc}(\theta )=\mathrm{arccsc}(1+\theta )=-i\ln (i(\theta +1{)}^{-1}+\sqrt{1-(1+\theta {)}^2})$

$\mathrm{arcsinh}(\theta )=\ln (\theta +\sqrt{{\theta }^2+1})$

$\mathrm{arccosh}(\theta )=\ln (\theta +\sqrt{{\theta }^2-1})$

$\mathrm{arctanh}(\theta )=\frac{\ln (\frac{i+\theta }{i-\theta })}{2}$

$\mathrm{arccoth}(\theta )=\mathrm{arctanh}(-\theta )=\frac{\ln (\frac{i-\theta }{i+\theta })}{2}$

$\mathrm{arcsech}(\theta )=\mathrm{arccosh}({\theta }^{-1})=\ln ({\theta }^{-1}+\sqrt{{\theta }^{-2}-1})$

$\mathrm{arccsch}(\theta )=\mathrm{arcsinh}({\theta }^{-1})=\ln ({\theta }^{-1}+\sqrt{{\theta }^{-2}+1})$

$\mathrm{arcversinh}(\theta )=\mathrm{arccosh}(\theta )-1=\ln (\theta +\sqrt{{\theta }^2-1})-1$

$\mathrm{arcvercosh}(\theta )=\mathrm{arcsinh}(\theta )-1=\ln (\theta +\sqrt{{\theta }^2+1})-1$

$\mathrm{arcexsech}(\theta )=\mathrm{arcsech}(\theta +1)=\mathrm{arccosh}((\theta +1{)}^{-1})=\ln ((\theta +1{)}^{-1}+\sqrt{(\theta +1{)}^{-2}-1})$

$\mathrm{arcexcsch}(\theta )=\mathrm{arccsch}(\theta +1)=\mathrm{arcsinh}((\theta +1{)}^{-1})=\ln ((\theta +1{)}^{-1}+\sqrt{(\theta +1{)}^{-2}+1})$