Goniometrická rovnice je tehdy, pokud je neznámá v goniometrické funkci .
[ 1] jednotková kružnice .
Příklad, jak může goniometrická rovnice vypadat:
(
sin
x
)
2
+
2
sin
x
−
3
=
0
{\displaystyle (\sin x)^{2}+2\sin x-3=0}
[ 2] [ 3]
cos
x
=
−
3
2
{\displaystyle \cos x=-{\frac {\sqrt {3}}{2}}}
x
1
=
5
π
6
+
2
k
π
,
k
∈
Z
{\displaystyle x_{1}={\frac {5\pi }{6}}+2k\pi ,k\in \mathbb {Z} }
x
2
=
7
π
6
+
2
k
π
,
k
∈
Z
{\displaystyle x_{2}={\frac {7\pi }{6}}+2k\pi ,k\in \mathbb {Z} }
tg
x
=
−
3
{\displaystyle {\textrm {tg}}\,x=-{\sqrt {3}}}
x
=
2
π
3
+
k
π
,
k
∈
Z
{\displaystyle x={\frac {2\pi }{3}}+k\pi ,k\in \mathbb {Z} }
(
sin
x
)
2
+
2
sin
x
−
3
=
0
{\displaystyle (\sin x)^{2}+2\sin x-3=0}
Zavedeme substituci
a
=
sin
x
{\displaystyle a=\sin x}
a
2
+
2
a
−
3
=
0
{\displaystyle a^{2}+2a-3=0}
Vypočítáme kvadratickou rovnici:
a
1
,
2
=
−
b
±
b
2
−
4
a
c
2
a
=
−
2
±
2
2
−
4
⋅
1
⋅
(
−
3
)
2
⋅
1
=
−
2
±
16
2
=
−
2
±
4
2
{\displaystyle a_{1,2}={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}={\frac {-2\pm {\sqrt {2^{2}-4\cdot 1\cdot (-3)}}}{2\cdot 1}}={\frac {-2\pm {\sqrt {16}}}{2}}={\frac {-2\pm 4}{2}}}
a
1
=
−
2
+
4
2
=
2
2
=
1
{\displaystyle a_{1}={\frac {-2+4}{2}}={\frac {2}{2}}=1}
a
2
=
−
2
−
4
2
=
−
6
2
=
−
3
{\displaystyle a_{2}={\frac {-2-4}{2}}={\frac {-6}{2}}=-3}
Nyní si můžeme napsat 2 rovnice :
sin
x
=
1
{\displaystyle \sin x=1}
sin
x
=
−
3
{\displaystyle \sin x=-3}
Vyřešíme obě rovnice :
sin
x
=
1
{\displaystyle \sin x=1}
x
=
1
2
π
+
2
k
π
{\displaystyle x={\frac {1}{2}}\pi +2k\pi }
sin
x
=
−
3
{\displaystyle \sin x=-3}
x
=
ϕ
{\displaystyle x=\phi }
Tím je vyřešená goniometrická rovnice pomocí substituce .
sin
(
x
+
π
6
)
=
1
{\displaystyle \sin \left(x+{\frac {\pi }{6}}\right)=1}
Zavedeme substituci
a
=
x
+
π
6
{\displaystyle a=x+{\frac {\pi }{6}}}
sin
a
=
1
{\displaystyle \sin a=1}
a
=
π
2
+
2
k
π
{\displaystyle a={\frac {\pi }{2}}+2k\pi }
Dosadíme substituci
a
=
x
+
π
6
{\displaystyle a=x+{\frac {\pi }{6}}}
x
+
π
6
=
π
2
+
2
k
π
{\displaystyle x+{\frac {\pi }{6}}={\frac {\pi }{2}}+2k\pi }
a
=
x
+
π
6
{\displaystyle a=x+{\frac {\pi }{6}}}
x
=
3
π
6
+
2
k
π
−
π
6
{\displaystyle x={\frac {3\pi }{6}}+2k\pi -{\frac {\pi }{6}}}
x
=
2
π
6
+
2
k
π
{\displaystyle x={\frac {2\pi }{6}}+2k\pi }
x
=
π
3
+
2
k
π
{\displaystyle x={\frac {\pi }{3}}+2k\pi }
Tím je vyřešená goniometrická rovnice pomocí substituce .
1.
3
cos
x
=
2
−
sin
x
{\displaystyle {\sqrt {3}}\cos x=2-\sin x}
2. umocníme rovnici na druhou:
3
cos
2
x
=
(
2
−
sin
x
)
2
{\displaystyle 3\cos ^{2}x=(2-\sin x)^{2}}
3. použijeme vzorec
cos
2
x
=
1
−
sin
2
x
{\displaystyle \cos ^{2}x=1-\sin ^{2}x}
3
−
3
sin
2
x
=
4
−
4
sin
x
+
sin
2
x
{\displaystyle 3-3\sin ^{2}x=4-4\sin x+\sin ^{2}x}
4.
0
=
4
sin
2
x
−
4
sin
x
+
1
{\displaystyle 0=4\sin ^{2}x-4\sin x+1}
5. použijeme vzorec
a
2
−
2
a
b
+
b
2
=
(
a
−
b
)
2
{\displaystyle a^{2}-2ab+b^{2}=(a-b)^{2}}
(
2
sin
x
−
1
)
2
=
0
{\displaystyle (2\sin x-1)^{2}=0}
6. celou rovnici odmocníme:
2
sin
x
−
1
=
0
{\displaystyle 2\sin x-1=0}
7.
sin
x
=
1
2
{\displaystyle \sin x={\frac {1}{2}}}
x
1
=
π
6
+
2
k
π
{\displaystyle x_{\scriptstyle {\text{1}}}={\frac {\pi }{6}}+2k\pi }
x
2
=
5
π
6
+
2
k
π
{\displaystyle x_{\scriptstyle {\text{2}}}={\frac {5\pi }{6}}+2k\pi }
8. z důvodu neekvivalentních úprav 2. a 6. je nutná zkouška
kořen
x
2
{\displaystyle x_{\scriptstyle {\text{2}}}}
x
1
{\displaystyle x_{\scriptstyle {\text{1}}}}
Takto je možné řešit rovnice se dvěma různými goniometrickými funkcemi
(
cot
x
)
−
1
=
−
(
tan
x
)
−
1
+
2
(
sin
x
)
−
1
{\displaystyle (\cot x)^{-1}=-(\tan x)^{-1}+2(\sin x)^{-1}}
Použijeme vztahy mezi funkcemi:
tan
x
=
2
(
sin
x
)
−
1
−
cot
x
{\displaystyle \tan x=2(\sin x)^{-1}-\cot x}
sin
x
cos
x
=
2
sin
x
−
cos
x
sin
x
{\displaystyle {\frac {\sin x}{\cos x}}={\frac {2}{\sin x}}-{\frac {\cos x}{\sin x}}}
zbavíme se zlomků:
sin
2
x
=
cos
x
∗
(
2
−
cos
x
)
{\displaystyle \sin ^{2}x=\cos x*(2-\cos x)}
Použijeme vzorec
sin
2
x
=
1
−
cos
2
x
{\displaystyle \sin ^{2}x=1-\cos ^{2}x}
1
−
cos
2
x
=
2
cos
x
−
cos
2
x
{\displaystyle 1-\cos ^{2}x=2\cos x-\cos ^{2}x}
1
=
2
cos
x
{\displaystyle 1=2\cos x}
cos
x
=
1
/
2
{\displaystyle \cos x=1/2}
x
1
=
π
6
+
2
k
π
{\displaystyle x_{\scriptstyle {\text{1}}}={\frac {\pi }{6}}+2k\pi }
x
2
=
11
π
6
+
2
k
π
{\displaystyle x_{\scriptstyle {\text{2}}}={\frac {11\pi }{6}}+2k\pi }
Rovnice vyřešena [ 4] [ 5]
Záporné hodnoty úhlů
sin
(
−
α
)
=
−
sin
α
{\displaystyle \sin(-\alpha )=-\sin \alpha \,\!}
cos
(
−
α
)
=
cos
α
{\displaystyle \cos(-\alpha )=\cos \alpha \,\!}
t
g
(
−
α
)
=
−
t
g
α
{\displaystyle \mathrm {tg} (-\alpha )=-\mathrm {tg} \,\alpha \,\!}
c
o
t
g
(
−
α
)
=
−
c
o
t
g
α
{\displaystyle \mathrm {cotg} (-\alpha )=-\mathrm {cotg} \,\alpha \,\!}
Vzájemné vztahy mezi goniometrickými funkcemi stejného úhlu
sin
2
α
+
cos
2
α
=
1
{\displaystyle \sin ^{2}\alpha +\cos ^{2}\alpha =1\,\!}
t
g
α
⋅
c
o
t
g
α
=
1
{\displaystyle \mathrm {tg} \,\alpha \cdot \mathrm {cotg} \,\alpha =1\,\!}
tg
α
=
sin
α
cos
α
{\displaystyle {\textrm {tg}}\,\alpha ={\frac {\sin \alpha }{\cos \alpha }}\,\!}
cotg
α
=
cos
α
sin
α
{\displaystyle {\textrm {cotg}}\,\alpha ={\frac {\cos \alpha }{\sin \alpha }}\,\!}
sin
α
=
1
−
cos
2
α
{\displaystyle \sin \alpha ={\sqrt {1-\cos ^{2}\alpha }}}
cos
α
=
1
−
sin
2
α
{\displaystyle \cos \alpha ={\sqrt {1-\sin ^{2}\alpha }}}
tg
α
=
1
cotg
α
{\displaystyle {\textrm {tg}}\,\alpha ={\frac {1}{{\textrm {cotg}}\,\alpha }}\,\!}
Dvojnásobný úhel
sin
2
α
=
2
⋅
sin
α
cos
α
{\displaystyle \sin 2\alpha =2\cdot \sin \alpha \cos \alpha \,\!}
cos
2
α
=
cos
2
α
−
sin
2
α
{\displaystyle \cos 2\alpha =\cos ^{2}\alpha -\sin ^{2}\alpha \,\!}
Poloviční úhel
sin
α
2
=
1
−
cos
α
2
{\displaystyle \sin {\frac {\alpha }{2}}={\sqrt {\frac {1-\cos \alpha }{2}}}\,\!}
cos
α
2
=
1
+
cos
α
2
{\displaystyle \cos {\frac {\alpha }{2}}={\sqrt {\frac {1+\cos \alpha }{2}}}\,\!}
Mocniny goniometrických funkcí
sin
2
α
=
1
2
(
1
−
cos
2
α
)
{\displaystyle \sin ^{2}\alpha ={\frac {1}{2}}(1-\cos 2\alpha )}
cos
2
α
=
1
2
(
1
+
cos
2
α
)
{\displaystyle \cos ^{2}\alpha ={\frac {1}{2}}(1+\cos 2\alpha )}
Goniometrické funkce součtu a rozdílu úhlů
sin
(
α
±
β
)
=
sin
α
cos
β
±
cos
α
sin
β
{\displaystyle \sin \left(\alpha \pm \beta \right)=\sin \alpha \cos \beta \pm \cos \alpha \sin \beta \,\!}
cos
(
α
±
β
)
=
cos
α
cos
β
∓
sin
α
sin
β
{\displaystyle \cos \left(\alpha \pm \beta \right)=\cos \alpha \cos \beta \mp \sin \alpha \sin \beta \,\!}
Jednotková kružnice [ 6]
Kvadrant α
sin α
cos α
tg α
cotg α
1. kvadrant
0° – 90°
+
+
+
+
2. kvadrant
90° – 180°
+
–
–
-
3. kvadrant
180° – 270°
–
–
+
+
4. kvadrant
270° – 360°
–
+
–
-
Stupně Radiány Sinus Kosinus Tangens Kotangens
0
0
{\displaystyle 0\,}
0
{\displaystyle 0\,}
1
{\displaystyle 1\,}
0
{\displaystyle 0\,}
−
{\displaystyle -\,}
30
π
6
{\displaystyle {\frac {\pi }{6}}}
1
2
{\displaystyle {\frac {1}{2}}}
3
2
{\displaystyle {\frac {\sqrt {3}}{2}}}
3
3
{\displaystyle {\frac {\sqrt {3}}{3}}}
3
{\displaystyle {\sqrt {3}}}
45
π
4
{\displaystyle {\frac {\pi }{4}}}
2
2
{\displaystyle {\frac {\sqrt {2}}{2}}}
2
2
{\displaystyle {\frac {\sqrt {2}}{2}}}
1
{\displaystyle 1\,}
1
{\displaystyle 1\,}
60
π
3
{\displaystyle {\frac {\pi }{3}}}
3
2
{\displaystyle {\frac {\sqrt {3}}{2}}}
1
2
{\displaystyle {\frac {1}{2}}}
3
{\displaystyle {\sqrt {3}}}
3
3
{\displaystyle {\frac {\sqrt {3}}{3}}}
90
π
2
{\displaystyle {\frac {\pi }{2}}}
1
{\displaystyle 1\,}
0
{\displaystyle 0\,}
−
{\displaystyle -\,}
0
{\displaystyle 0\,}
120
2
π
3
{\displaystyle {\frac {2\pi }{3}}}
3
2
{\displaystyle {\frac {\sqrt {3}}{2}}}
−
1
2
{\displaystyle -{\frac {1}{2}}}
−
3
{\displaystyle -{\sqrt {3}}}
−
3
3
{\displaystyle -{\frac {\sqrt {3}}{3}}}
135
3
π
4
{\displaystyle {\frac {3\pi }{4}}}
2
2
{\displaystyle {\frac {\sqrt {2}}{2}}}
−
2
2
{\displaystyle -{\frac {\sqrt {2}}{2}}}
−
1
{\displaystyle -1\,}
−
1
{\displaystyle -1\,}
150
5
π
6
{\displaystyle {\frac {5\pi }{6}}}
1
2
{\displaystyle {\frac {1}{2}}}
−
3
2
{\displaystyle {\frac {-{\sqrt {3}}}{2}}}
−
3
3
{\displaystyle {\frac {-{\sqrt {3}}}{3}}}
−
3
{\displaystyle -{\sqrt {3}}}
180
π
{\displaystyle \pi \,}
0
{\displaystyle 0\,}
−
1
{\displaystyle -1\,}
0
{\displaystyle 0\,}
−
{\displaystyle -\,}
210
7
π
6
{\displaystyle {\frac {7\pi }{6}}}
−
1
2
{\displaystyle -{\frac {1}{2}}}
−
3
2
{\displaystyle -{\frac {\sqrt {3}}{2}}}
3
3
{\displaystyle {\frac {\sqrt {3}}{3}}}
3
{\displaystyle {\sqrt {3}}}
225
5
π
4
{\displaystyle {\frac {5\pi }{4}}}
−
2
2
{\displaystyle -{\frac {\sqrt {2}}{2}}}
−
2
2
{\displaystyle -{\frac {\sqrt {2}}{2}}}
1
{\displaystyle 1\,}
1
{\displaystyle 1\,}
240
4
π
3
{\displaystyle {\frac {4\pi }{3}}}
−
3
2
{\displaystyle -{\frac {\sqrt {3}}{2}}}
−
1
2
{\displaystyle -{\frac {1}{2}}}
3
{\displaystyle {\sqrt {3}}}
3
3
{\displaystyle {\frac {\sqrt {3}}{3}}}
270
3
π
2
{\displaystyle {\frac {3\pi }{2}}}
−
1
{\displaystyle -1\,}
0
{\displaystyle 0\,}
−
{\displaystyle -\,}
0
{\displaystyle 0\,}
300
5
π
3
{\displaystyle {\frac {5\pi }{3}}}
−
3
2
{\displaystyle -{\frac {\sqrt {3}}{2}}}
1
2
{\displaystyle {\frac {1}{2}}}
−
3
{\displaystyle -{\sqrt {3}}}
−
3
3
{\displaystyle -{\frac {\sqrt {3}}{3}}}
315
7
π
4
{\displaystyle {\frac {7\pi }{4}}}
−
2
2
{\displaystyle -{\frac {\sqrt {2}}{2}}}
2
2
{\displaystyle {\frac {\sqrt {2}}{2}}}
−
1
{\displaystyle -1\,}
−
1
{\displaystyle -1\,}
330
11
π
6
{\displaystyle {\frac {11\pi }{6}}}
−
1
2
{\displaystyle -{\frac {1}{2}}}
3
2
{\displaystyle {\frac {\sqrt {3}}{2}}}
−
3
3
{\displaystyle {\frac {-{\sqrt {3}}}{3}}}
−
3
{\displaystyle -{\sqrt {3}}}
