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Funksiyanın dövrü
sin ( θ ± π 2 ) = ± cos θ {\displaystyle \sin(\theta \pm {\tfrac {\pi }{2}})=\pm \cos \theta }
sin ( θ + π ) = − sin θ {\displaystyle \sin(\theta +\pi )=-\sin \theta }
sin ( θ + k ⋅ 2 π ) = + sin θ {\displaystyle \sin(\theta +k\cdot 2\pi )=+\sin \theta }
2 π {\displaystyle 2\pi }
cos ( θ ± π 2 ) = ∓ sin θ {\displaystyle \cos(\theta \pm {\tfrac {\pi }{2}})=\mp \sin \theta }
cos ( θ + π ) = − cos θ {\displaystyle \cos(\theta +\pi )=-\cos \theta }
cos ( θ + k ⋅ 2 π ) = + cos θ {\displaystyle \cos(\theta +k\cdot 2\pi )=+\cos \theta }
2 π {\displaystyle 2\pi }
csc ( θ ± π 2 ) = ± sec θ {\displaystyle \csc(\theta \pm {\tfrac {\pi }{2}})=\pm \sec \theta }
csc ( θ + π ) = − csc θ {\displaystyle \csc(\theta +\pi )=-\csc \theta }
csc ( θ + k ⋅ 2 π ) = + csc θ {\displaystyle \csc(\theta +k\cdot 2\pi )=+\csc \theta }
2 π {\displaystyle 2\pi }
sec ( θ ± π 2 ) = ∓ csc θ {\displaystyle \sec(\theta \pm {\tfrac {\pi }{2}})=\mp \csc \theta }
sec ( θ + π ) = − sec θ {\displaystyle \sec(\theta +\pi )=-\sec \theta }
sec ( θ + k ⋅ 2 π ) = + sec θ {\displaystyle \sec(\theta +k\cdot 2\pi )=+\sec \theta }
2 π {\displaystyle 2\pi }
tan ( θ ± π 4 ) = tan θ ± 1 1 ∓ tan θ {\displaystyle \tan(\theta \pm {\tfrac {\pi }{4}})={\tfrac {\tan \theta \pm 1}{1\mp \tan \theta }}}
tan ( θ + π 2 ) = − cot θ {\displaystyle \tan(\theta +{\tfrac {\pi }{2}})=-\cot \theta }
tan ( θ + k ⋅ π ) = + tan θ {\displaystyle \tan(\theta +k\cdot \pi )=+\tan \theta }
π {\displaystyle \pi }
cot ( θ ± π 4 ) = cot θ ∓ 1 1 ± cot θ {\displaystyle \cot(\theta \pm {\tfrac {\pi }{4}})={\tfrac {\cot \theta \mp 1}{1\pm \cot \theta }}}
cot ( θ + π 2 ) = − tan θ {\displaystyle \cot(\theta +{\tfrac {\pi }{2}})=-\tan \theta }
cot ( θ + k ⋅ π ) = + cot θ {\displaystyle \cot(\theta +k\cdot \pi )=+\cot \theta }
π {\displaystyle \pi }
Triqonometrik funksiyaların işarəsi bucağın rübündən asılıdır. Əgər − π < θ ≤ π {\displaystyle {-\pi }<\theta \leq \pi } sgn ifadə edərsə,
sgn ( sin θ ) = sgn ( csc θ ) = { + 1 if 0 < θ < π − 1 if − π < θ < 0 0 if θ ∈ { 0 , π } sgn ( cos θ ) = sgn ( sec θ ) = { + 1 if − 1 2 π < θ < 1 2 π − 1 if − π < θ < − 1 2 π or 1 2 π < θ < π 0 if θ ∈ { − 1 2 π , 1 2 π } sgn ( tan θ ) = sgn ( cot θ ) = { + 1 if − π < θ < − 1 2 π or 0 < θ < 1 2 π − 1 if − 1 2 π < θ < 0 or 1 2 π < θ < π 0 if θ ∈ { − 1 2 π , 0 , 1 2 π , π } {\displaystyle {\begin{aligned}\operatorname {sgn}(\sin \theta )=\operatorname {sgn}(\csc \theta )&={\begin{cases}+1&{\text{if}}\ \ 0<\theta <\pi \\-1&{\text{if}}\ \ {-\pi }<\theta <0\\0&{\text{if}}\ \ \theta \in \{0,\pi \}\end{cases}}\\[5mu]\operatorname {sgn}(\cos \theta )=\operatorname {sgn}(\sec \theta )&={\begin{cases}+1&{\text{if}}\ \ {-{\tfrac {1}{2}}\pi }<\theta <{\tfrac {1}{2}}\pi \\-1&{\text{if}}\ \ {-\pi }<\theta <-{\tfrac {1}{2}}\pi \ \ {\text{or}}\ \ {\tfrac {1}{2}}\pi <\theta <\pi \\0&{\text{if}}\ \ \theta \in {\bigl \{}{-{\tfrac {1}{2}}\pi },{\tfrac {1}{2}}\pi {\bigr \}}\end{cases}}\\[5mu]\operatorname {sgn}(\tan \theta )=\operatorname {sgn}(\cot \theta )&={\begin{cases}+1&{\text{if}}\ \ {-\pi }<\theta <-{\tfrac {1}{2}}\pi \ \ {\text{or}}\ \ 0<\theta <{\tfrac {1}{2}}\pi \\-1&{\text{if}}\ \ {-{\tfrac {1}{2}}\pi }<\theta <0\ \ {\text{or}}\ \ {\tfrac {1}{2}}\pi <\theta <\pi \\0&{\text{if}}\ \ \theta \in {\bigl \{}{-{\tfrac {1}{2}}\pi },0,{\tfrac {1}{2}}\pi ,\pi {\bigr \}}\end{cases}}\end{aligned}}}
Düstur
Arqumentin mənası
sin 2 α + cos 2 α = 1 {\displaystyle \sin ^{2}\alpha +\cos ^{2}\alpha =1}
∀ α {\displaystyle \forall \alpha }
tan 2 α + 1 = 1 cos 2 α = sec 2 α {\displaystyle \operatorname {tan} ^{2}\alpha +1={\frac {1}{\cos ^{2}\alpha }}=\operatorname {sec} ^{2}\alpha }
α ≠ π 2 + π n , n ∈ Z {\displaystyle \alpha \neq {\frac {\pi }{2}}+\pi n,n\in \mathbb {Z} }
cot 2 α + 1 = 1 sin 2 α = cosec 2 α {\displaystyle \operatorname {cot} ^{2}\alpha +1={\frac {1}{\sin ^{2}\alpha }}=\operatorname {cosec} ^{2}\alpha }
α ≠ π n , n ∈ Z {\displaystyle \alpha \neq \pi n,n\in \mathbb {Z} }
tan α ⋅ cot α = 1 {\displaystyle ~\operatorname {tan} \alpha \cdot \operatorname {cot} \alpha =1}
α ≠ π n 2 , n ∈ Z {\displaystyle \alpha \neq {\frac {\pi n}{2}},n\in \mathbb {Z} }
tan α = sin α cos α {\displaystyle ~\operatorname {tan} \alpha ={\frac {\sin \alpha }{\cos \alpha }}}
Toplama düsturları
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 }
tan ( α ± β ) = tan α ± tan β 1 ∓ tan α tan β {\displaystyle \operatorname {tan} \left(\alpha \pm \beta \right)={\frac {\operatorname {tan} \alpha \pm \operatorname {tan} \beta }{1\mp \operatorname {tan} \alpha \operatorname {tan} \beta }}}
cot ( α ± β ) = cot α cot β ∓ 1 cot β ± cot α {\displaystyle \operatorname {cot} \left(\alpha \pm \beta \right)={\frac {\operatorname {cot} \alpha \operatorname {cot} \beta \mp 1}{\operatorname {cot} \beta \pm \operatorname {cot} \alpha }}}
İkiqat arqument düsturları
sin 2 α = 2 sin α cos α {\displaystyle \sin 2\alpha =2{\sin \alpha }{\cos \alpha }}
cos 2 α = cos 2 α − sin 2 α {\displaystyle \cos 2\alpha ={\cos ^{2}\alpha }-{\sin ^{2}\alpha }} cos 2 α = 2 cos 2 α − 1 = 1 − 2 sin 2 α {\displaystyle \cos 2\alpha =2{\cos ^{2}\alpha }-1=1-2{\sin ^{2}\alpha }}
tan 2 α = 2 tan α 1 − tan 2 α {\displaystyle \operatorname {tan} 2\alpha ={\frac {2\operatorname {tan} \alpha }{1-\operatorname {tan} ^{2}\alpha }}}
cot 2 α = cot 2 α − 1 2 cot α {\displaystyle \operatorname {cot} 2\alpha ={\frac {\operatorname {cot} ^{2}\alpha -1}{2\operatorname {cot} \alpha }}}
Üçqat arqument düsturları
sin 3 α = 3 sin α − 4 sin 3 α {\displaystyle \sin 3\alpha =3\sin \alpha -4\sin ^{3}\alpha \,}
cos 3 α = 4 cos 3 α − 3 cos α {\displaystyle \cos 3\alpha =4\cos ^{3}\alpha -3\cos \alpha \,}
tan 3 α = 3 tan α − tan 3 α 1 − 3 tan 2 α {\displaystyle \operatorname {tan} 3\alpha ={\frac {3\operatorname {tan} \alpha -\operatorname {tan} ^{3}\alpha }{1-3\operatorname {tan} ^{2}\alpha }}}
cot 3 α = 3 cot α − cot 3 α 1 − 3 cot 2 α {\displaystyle \operatorname {cot} 3\alpha ={\frac {3\operatorname {cot} \alpha -\operatorname {cot} ^{3}\alpha }{1-3\operatorname {cot} ^{2}\alpha }}}
Sinus
Kosinus
sin 2 α = 1 − cos 2 α 2 {\displaystyle \sin ^{2}\alpha ={\frac {1-\cos 2\alpha }{2}}}
cos 2 α = 1 + cos 2 α 2 {\displaystyle \cos ^{2}\alpha ={\frac {1+\cos 2\alpha }{2}}}
sin 3 α = 3 sin α − sin 3 α 4 {\displaystyle \sin ^{3}\alpha ={\frac {3\sin \alpha -\sin 3\alpha }{4}}}
cos 3 α = 3 cos α + cos 3 α 4 {\displaystyle \cos ^{3}\alpha ={\frac {3\cos \alpha +\cos 3\alpha }{4}}}
sin 4 α = 3 − 4 cos 2 α + cos 4 α 8 {\displaystyle \sin ^{4}\alpha ={\frac {3-4\cos 2\alpha +\cos 4\alpha }{8}}}
cos 4 α = 3 + 4 cos 2 α + cos 4 α 8 {\displaystyle \cos ^{4}\alpha ={\frac {3+4\cos 2\alpha +\cos 4\alpha }{8}}}
sin 5 α = 10 sin α − 5 sin 3 α + sin 5 α 16 {\displaystyle \sin ^{5}\alpha ={\frac {10\sin \alpha -5\sin 3\alpha +\sin 5\alpha }{16}}}
cos 5 α = 10 cos α + 5 cos 3 α + cos 5 α 16 {\displaystyle \cos ^{5}\alpha ={\frac {10\cos \alpha +5\cos 3\alpha +\cos 5\alpha }{16}}}
Düstur
sin 2 α cos 2 α = 1 − cos 4 α 8 {\displaystyle \sin ^{2}\alpha \cos ^{2}\alpha ={\frac {1-\cos 4\alpha }{8}}}
sin 3 α cos 3 α = 3 sin 2 α − sin 6 α 32 {\displaystyle \sin ^{3}\alpha \cos ^{3}\alpha ={\frac {3\sin 2\alpha -\sin 6\alpha }{32}}}
sin 4 α cos 4 α = 3 − 4 cos 4 α + cos 8 α 128 {\displaystyle \sin ^{4}\alpha \cos ^{4}\alpha ={\frac {3-4\cos 4\alpha +\cos 8\alpha }{128}}}
sin 5 α cos 5 α = 10 sin 2 α − 5 sin 6 α + sin 10 α 512 {\displaystyle \sin ^{5}\alpha \cos ^{5}\alpha ={\frac {10\sin 2\alpha -5\sin 6\alpha +\sin 10\alpha }{512}}}
Hasilin cəmə çevrilməsi düsturları
sin α sin β = cos ( α − β ) − cos ( α + β ) 2 {\displaystyle \sin \alpha \sin \beta ={\frac {\cos(\alpha -\beta )-\cos(\alpha +\beta )}{2}}}
cos α cos β = cos ( α − β ) + cos ( α + β ) 2 {\displaystyle \cos \alpha \cos \beta ={\frac {\cos(\alpha -\beta )+\cos(\alpha +\beta )}{2}}}
Abramowitz and Stegun, p. 72, 4.3.7–9
Abramowitz and Stegun, p. 72, 4.3.16
↑ Weisstein, Eric W. (ing.) Wolfram saytında.
Abramowitz and Stegun, p. 72, 4.3.17
Abramowitz and Stegun, p. 72, 4.3.18
↑ . www.milefoot.com. 2023-04-03 tarixində . İstifadə tarixi: 2019-10-12 .
Abramowitz and Stegun, p. 72, 4.3.19
Abramowitz and Stegun, p. 80, 4.4.32
Abramowitz and Stegun, p. 80, 4.4.33
Abramowitz and Stegun, p. 80, 4.4.34
![{\displaystyle {\begin{aligned}\operatorname {sgn}(\sin \theta )=\operatorname {sgn}(\csc \theta )&={\begin{cases}+1&{\text{if}}\ \ 0<\theta <\pi \\-1&{\text{if}}\ \ {-\pi }<\theta <0\\0&{\text{if}}\ \ \theta \in \{0,\pi \}\end{cases}}\\[5mu]\operatorname {sgn}(\cos \theta )=\operatorname {sgn}(\sec \theta )&={\begin{cases}+1&{\text{if}}\ \ {-{\tfrac {1}{2}}\pi }<\theta <{\tfrac {1}{2}}\pi \\-1&{\text{if}}\ \ {-\pi }<\theta <-{\tfrac {1}{2}}\pi \ \ {\text{or}}\ \ {\tfrac {1}{2}}\pi <\theta <\pi \\0&{\text{if}}\ \ \theta \in {\bigl \{}{-{\tfrac {1}{2}}\pi },{\tfrac {1}{2}}\pi {\bigr \}}\end{cases}}\\[5mu]\operatorname {sgn}(\tan \theta )=\operatorname {sgn}(\cot \theta )&={\begin{cases}+1&{\text{if}}\ \ {-\pi }<\theta <-{\tfrac {1}{2}}\pi \ \ {\text{or}}\ \ 0<\theta <{\tfrac {1}{2}}\pi \\-1&{\text{if}}\ \ {-{\tfrac {1}{2}}\pi }<\theta <0\ \ {\text{or}}\ \ {\tfrac {1}{2}}\pi <\theta <\pi \\0&{\text{if}}\ \ \theta \in {\bigl \{}{-{\tfrac {1}{2}}\pi },0,{\tfrac {1}{2}}\pi ,\pi {\bigr \}}\end{cases}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/14cc79239a0fcdfb53b132d75a93474e9e6d5ba0)
