Trigonometri

Trigonometri se pati nan Matematik la ki gen pou wè ak relasyon ki genyen ant kosinis, sinis ak tanjant yon ang.

Sèk trigonometrik

Se yon sèk oryante ki gen reyon li egal ak 1.

Nou di oryante paske gen yon sans dirèk ak yon sans endirèk sou sèk la: sans dirèk la se lekontrè sans egui yon mont, sans endirèk la se sans egui yon mont.

Tout deplasman ki fèt nan sans dirèk konte pozitif alòske tout deplasman nan sans endirèk la konte negatif.

Fòmil kos^2(\alpha)+\sin^2(\alpha)=1 pote non Fòmil fondamantal trigonometri a.

Fonksyon sikilè

Nou ka wè kos(kosinis), sin(sinis), tan(tanjant), kotan(kotanjant), sek(sekant) ak kosek(kosekant) tankou fonksyon.

Fonksyon sinis

Se yon fonksyon ki pran valè nan makòn \mathbb{R}, ki gen peryòd 2\pi, epi ki enpè. Pou tout x\in \mathbb{R}, \sin(\pi+x)=-\sin (x).

Fonksyon kosinis

Se yon fonksyon ki pran valè nan makòn \mathbb{R}, ki gen peryòd 2\pi, epi ki pè. Pou tout x\in \mathbb{R}, kos(\pi+x)=-kos \,(x).

kos\,(x)=\sin(\frac{\pi}{2}-x)=\sin(x+\frac{\pi}{2})

\sin (x)=kos(\frac{\pi}{2}-x)=-kos(x+\frac{\pi}{2})

Fonksyon tanjant

Se yon fonksyon ki pran valè nan makòn \mathbb{R}\setminus{\frac{\pi}{2}+\pi \mathbb{Z}}, ki gen peryòd \pi, epi ki enpè. Pou tout x\in \mathbb{R}\setminus{\frac{\pi}{2}+\pi \mathbb{Z}}, \tan\,(x)=\frac{\sin\,(x)}{kos\,(x)}.

Fonksyon kotanjant

Se yon fonksyon ki pran valè nan makòn \mathbb{R}\setminus{\pi \mathbb{Z}}, ki gen peryòd \pi, epi ki enpè. Pou tout x\in \mathbb{R}\setminus{\pi \mathbb{Z}},

kotan \,x=\frac{kos\, x}{\sin\,x}=\tan(\frac{\pi}{2}-x)=-\tan(x+\frac{\pi}{2}).

Fòmil adisyon

Pou tout (a,b)\in \mathbb{R}^2, nou genyen:

kos(a+b)=kos\,(a)\,kos\,(b)-\sin\, (a)\,\sin \,(b)

\sin(a+b)=\sin\,(a)\,kos\,(b)+kos \,(a)\,\sin\,(b)

kos(a-b)=kos\,(a)\,kos\,(b)+\sin\,(a)\,\sin\, (b)

\sin(a-b)=\sin\,(a)\,kos\,(b)-kos\,(a)\,\sin \,(b)

\tan(a+b)=\frac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)},     a, b, a+b \notin \frac{\pi}{2}+\pi \mathbb{Z}

\tan(a-b)=\frac{\tan(a)-\tan(b)}{1+\tan(a)\tan(b)},       a, b, a-b \notin \frac{\pi}{2}+\pi \mathbb{Z}

Fòmil diplikasyon/ak doub

Si n fè a=b nan de premye fòmil yo n ap jwenn:

kos\,2a=kos^2a-\sin^2a=2\,kos^2a-1=1-2\sin^2a

\sin\,(2a)=2\sin\,(a)\,kos\,(a)

kos^2a=\frac{1+kos\,(2a)}{2}

sin^2a=\frac{1-kos 2a}{2}

kos^2a=\frac{1-tan^2 a}{1+tan^2a} si a\notin \frac{\pi}{2}+\pi \mathbb{Z}

\sin^2a=\frac{2 \tan \,a}{1+\tan^2 a} si a \notin \frac{\pi}{2}+\pi \mathbb{Z}

\tan \,2a=\frac{2\, \tan\, a}{1-\tan^2 a} si a ak 2a \notin \frac{\pi}{2}+\pi \mathbb{Z}

Fason pou transfòme yon pwodui pou fè l vin sòm

kos\,(a) kos (b)=\frac{1}{2}[kos(a+b)+kos(a-b)]

\sin a \sin b=\frac{1}{2}[kos(a-b)-kos(a+b)]

Lè n chanje a nou fè l vin \frac{\pi}{2}-a, n ap jwenn nan menm lòd la:

\sin\,(a) kos\,(b)=\frac{1}{2}[\sin(a+b)+\sin(a-b)]

kos\,(a)\,\sin\,(b)=\frac{1}{2}[\sin(a+b)-\sin(a-b)]

Fason pou transfòme yon sòm pou n fè l vin pwodui

Lè n poze p=a+b epi q=a-b nan ekwasyon n sot jwenn la yo, n ap jwenn ekwasyon sa yo:

kos\,p + kos \,q = 2\,kos \left(\frac{p+q}{2}\right)\,kos \left(\frac{p-q}{2}\right) (\ast)

\sin\,p + \sin\,q = 2\,\sin \left(\frac{p+q}{2}\right)\,kos \left(\frac{p-q}{2}\right)  (\ast\ast)

N ap jwenn kos\, p - kos\, q lè n ranplase q ak q+\pi nan (\ast), epi n ap jwenn \sin \,p - \sin\, q lè n ranplase q ak -q nan (\ast\ast):

kos\,p - kos\,q = -2\,\sin \left(\frac{p+q}{2}\right)\,\sin \left(\frac{p-q}{2}\right)

\sin\,p - \sin\,q = 2\,\sin \left(\frac{p-q}{2}\right)\,kos \left(\frac{p+q}{2}\right)

\tan\,(p)+\tan\,(q)=\frac{\sin(p+q)}{kos\,(p)\,kos\,(q)}, p ak q \notin \frac{\pi}{2}+\pi \mathbb{Z}

\tan \,p-\tan \,q=\frac{\sin(p-q)}{kos \,p kos\, q}, p ak q \notin \frac{\pi}{2}+\pi \mathbb{Z}

Kèk lòt relasyon enpòtan

\frac{1}{kos^2\,x}=1+\tan^2\,x

\sin\, 2x=2\frac{\sin\,x}{kos\,x}\cdot kos^2\,x=2\,\tan\,(x)\cdot kos^2(x)

Ekwasyon trigonometrik

Fason pou rezoud yon ekwasyon ki gen fòm: kos\, x=a.

Si |a|<1 gen yon sèl ak \alpha ki nan [0,\pi] ki gen kosinis li egal ak a. Kounye a n ap gen pou n rezoud ekwasyon sa a: kos\, x = kos\, \alpha.

kos\,(x) = kos\,(\alpha) \iff x=\pm \alpha + 2k\pi; k\in \mathbb{Z}

Fason pou rezoud yon ekwasyon ki gen fòm: \sin\,(x)=b.

Si |b|<1 gen yon sèl ak \alpha ki nan [-\frac{\pi}{2},\frac{\pi}{2}] ki gen sinis li egal ak b. Kounye a n ap gen pou n rezoud ekwasyon sa a: \sin\,(x) = \sin\,(\alpha).

\sin\,(x)= \sin\,(\alpha) \iff x= \alpha + 2k\pi oubyen x=\pi- \alpha + 2k\pi; k\in \mathbb{Z}

Fason pou rezoud yon ekwasyon ki gen fòm: \tan\,(x)=t.

Kèlkeswa t, gen yon sèl ak \alpha ki nan [-\frac{\pi}{2},\frac{\pi}{2}] ki gen tanjant li egal ak t. Kounye a n ap gen pou n rezoud ekwasyon sa a: \tan\,(x) =\tan\,(\alpha).

\tan\,(x) = \tan\,(\alpha) \iff x= \alpha + k\pi; k\in \mathbb{Z}

Ak ki se pwogresyon aritmetik

Pou tout nonb reyèl a ak \theta (\theta \notin 2\pi \mathbb{Z}), nou genyen:

\displaystyle \sum_{k=0}^n kos(a+k \theta)=\dfrac{\sin(n+1)\frac{\theta}{2}\, kos(a+\frac{n\theta}{2})}{\sin\frac{\theta}{2}}

\displaystyle \sum_{k=0}^n \sin(a+k \theta)=\dfrac{\sin(n+1)\frac{\theta}{2}\, \sin(a+\frac{n\theta}{2})}{\sin\frac{\theta}{2}}

Kèk valè remakab fonksyon sikilè yo.

Fonksyon sikilè resipwòk

Sinis monotòn e li kontini sou  [{-}\frac{\pi}{2},\frac{\pi}{2}]  ale nan [{-}1,1]. Li admèt yon bijeksyon resipwòk ki se fonksyon Ak sinis la.

Fonksyon Ak sinis

Pou tout x ak y nou pran [{-}1,1]\times [{-}\frac{\pi}{2},\frac{\pi}{2}]:

     y=Ak \sin\,(x) \iff x=\sin\,(y)

x-10+1
Ak\,\sin\,(x) -\frac{\pi}{2} \nearrow0\nearrow+\frac{\pi}{2}

\sin(Ak \sin\,x)=x, \forall x\in[{-}1,1]

Ak \sin(\sin\,x)=x, \forall x\in[{-}\frac{\pi}{2},\frac{\pi}{2}].

Kosinis la  monotòn e li kontini sou  [0,\pi] ale nan [{-}1,1]. Li admèt yon bijeksyon resipwòk ki se fonksyon Ak kosinis la.

Fonksyon Ak kosinis

Pou tout x ak y nou pran [{-}1,1]\times [0,\pi]:

     y=Ak kos\,(x)\iff x=kos\,(y)

x-10+1
Ak\,kos\,(x)\pi\searrow+\frac{\pi}{2}\searrow0

kos(Ak kos \,x)=x, \forall x\in[{-}1,1]

Ak kos(kos\,(x))=x, \forall x\in[0,\pi].

Ak \sin\,(x)+Ak kos\,(x)=\frac{\pi}{2}, \forall x\in[{-}1,1]

Fonksyon Ak tanjant

Pou tout x ak y nou pran nan \mathbb{R}\times]{-}\frac{\pi}{2},\frac{\pi}{2}[:

     y=\text{Aktan}(x) \iff  x=\tan(y)

x-\infty0+\infty
Ak\,\tan(x)-\frac{\pi}{2}\nearrow0\nearrow+\frac{\pi}{2}

Pou tout x ki nan \mathbb{R}:

     

Ak\tan\,(x) + Ak\tan\,\frac{1}{x}=\varepsilon \frac{\pi}{2} kote $latex \varepsilon =\left{
\begin{array}{cl}
1&\text{ si } x>0 \\
-1&\text{ si } x<0
\end{array}\right.$

Fonksyon ipèbolik(trigonometri ipèbolik)

Yo rele kosinis, sinis ak tanjant \textit{ipèbolik} yon nonb reyèl  x, twa ekspresyon sa yo:

    kosi\, (x)=\frac{\mathrm{e}^x+\mathrm{e}^{-x}}{2} \quad sini\,(x)=\frac{\mathrm{e}^x-\mathrm{e}^{-x}}{2} \quad \text{epi} \quad tani\,(x)=\frac{\frac{\mathrm{e}^x-\mathrm{e}^{-x}}{2}}{\frac{\mathrm{e}^x+\mathrm{e}^{-x}}{2}}=\frac{1-\mathrm{e}^{-2x}}{1+\mathrm{e}^{-2x}}.

Twa fonksyon sa yo makònen atravè relasyon sa yo:

    tani\, x=\frac{sini \,x}{kosi\, x} \quad \text{ak} \quad kosi^2x-sini^2x=1

x-\infty0+\infty
kosi\,(x)-\infty \searrow 1\nearrow+\infty
x-\infty0+\infty
sini\,(x)-\infty \nearrow 0\nearrow+\infty
x-\infty0+\infty
tani\,(x)-\infty \nearrow 0\nearrow+\infty

kosi se yon fonksyon pè; sini ak tani yo menm yo enpè.

sini\, x ak tani\, x pa twò twò diferan lè $x<<1$.

•Dwat y=1 an ak dwat y=-1 an se asenptòt tani lè  x>>1 ak lè x<<1.

Fòmil adisyon

Pou tout (a,b)\in \mathbb{R}^2, nou genyen:

kosi(a+b)=kosi\,(a)\,kosi\,(b)+sini\,(a)\,sini\,(b)

sini(a+b)=sini\,(a)\,kosi\,(b)+kosi\,(a)\,sini\,(b)

kosi(a-b)=kosi\,(a)\,kosi\,(b)-sini\,(a)\,sini\,(b)

sini(a-b)=sini\,(a)\;kosi\,(b)-kosi\,(a)\,sini\,(b)

tani(a+b)=\frac{tani(a)+tani(b)}{1+tani(a)tani(b)}

tani(a-b)=\frac{tani(a)-tani(b)}{1-tani(a)tani(b)}

Fòmil diplikasyon/ak doub

Si n fè a=b nan de premye fòmil yo n ap jwenn:

kosi\,(2a)=kosi^2a+sini^2a=2kosi^2a+1=1+2sini^2a=\frac{1+tani^2a}{1-tani^2a}

sini \,2a=2\,sini\,(a)\,kosi\,(a)=\frac{2 tani \,a}{1-tani^2a}

tani\,(2a)=\frac{2\, tani\,(a)}{1+tani^2a}

kosi^2a=\frac{1+kosi\, 2a}{2}

sini^2a=\frac{kosi \,2a-1}{2}

tani \,a=\frac{sini \,2a}{kosi\, 2a+ 1}=\frac{kosi \,2a-1}{sini \,2a}

Fason pou transfòme yon pwodui pou fè l vin sòm

kosi \,(a) kosi \,(b)=\frac{1}{2}[kosi(a+b)+kosi(a-b)]

sini \,(a) sini \,(b)=\frac{1}{2}[kosi(a+b)-kosi(a-b)]

sini \,(a) \;kosi \,(b)=\frac{1}{2}[sini(a+b)+sini(a-b)]

kosi \,(a)\, sini \,(b)=\frac{1}{2}[sini(a+b)-sini(a-b)]

Fason pou transfòme yon sòm pou n fè l vin pwodui

Lè n poze p=a+b epi q=a-b nan ekwasyon n sot jwenn la yo, n ap jwenn ekwasyon sa yo:

kosi \,p + kosi\, q = 2\,kosi \left(\frac{p+q}{2}\right)\,kosi \left(\frac{p-q}{2}\right) (\ast)

sini \,p + sini \,q = 2\,sini \left(\frac{p+q}{2}\right)\,kosi \left(\frac{p-q}{2}\right)  (\ast\ast)

N ap jwenn kosi \,p - kosi\, q lè n ranplase q ak q+\pi nan $(\ast)$, epi n ap jwenn sini \,p - sini\, q lè n ranplase q ak -q nan (\ast\ast):

kosi\, p - kosi\, q = 2\,sini \left(\frac{p+q}{2}\right)\,sini \left(\frac{p-q}{2}\right)

sini \,p - sini\, q = 2\,sini \left(\frac{p-q}{2}\right)\,kosi \left(\frac{p+q}{2}\right)

tani \,(p)+tani \,(q)=\frac{sini(p+q)}{kosi\, (p) \,kosi\, (q)}

tani \,(p)-tani \,(q)=\frac{sini(p-q)}{kosi\, (p) \;kosi \,(q)}

Kèk lòt relasyon enpòtan

\frac{1}{kosi^2\,x}=1-tani^2\,(x)

sini^2x=\frac{tani^2x}{1-tani^2x}

Fonksyon ipèbolik resipwòk

°Fonksyon agiman sinis ipèbolik

Pou tout (x,y)\in \mathbb{R}^2:

        y=Ag\,sini \,(x)=\ln(x+\sqrt{x^2+1}) \iff x=sini \,(y)

x-\infty0+\infty
Ag\,sini(x)-\infty\nearrow0\nearrow +\infty

°Fonksyon agiman kosinis ipèbolik

Pou tout x>1 ak y>0, nou genyen:

           y=Ag kosi \,(x)=\ln(x+\sqrt{x^2-1}) \iff x=kosi \,(y)

1+\infty
Ag\,kosi\,x0\nearrow+\infty

°Fonksyon agiman tanjant ipèbolik

Pou tout (x,y)\in\mathopen{]{-}1,1\mathclose{[}\times \mathbb{R}:

         y=Ag\,tani\,(x)=\frac{1}{2}\ln{\frac{1+x}{1-x}} \iff x=tani\,(y)

x-\infty0+\infty
Ak\,tani(x)-\frac{\pi}{2}\nearrow0\nearrow+\frac{\pi}{2}