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}
Řešení goniometrické rovnice [2] [3]
Jednoduché rovnice 1. rovnice 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} } 2. rovnice 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} } Substituce 1. rovnice ( 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 .
2. rovnice 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 .
Rovnice s více funkcemi současně 1. rovnice 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
2. rovnice ( 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 Vybrané (nejpoužívanější) vzorce [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 \,\!} Kvadranty a hodnoty funkcí ve vybraných úhlech 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}}}
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