Трыганаметрычныя тоеснасці: sin 2 α + cos 2 α = 1 , {\displaystyle \sin ^{2}\alpha +\cos ^{2}\alpha =1,\,} tg ( α ) = sin ( α ) cos ( α ) , {\displaystyle \operatorname {tg} (\alpha )={\frac {\sin(\alpha )}{\cos(\alpha )}},} ctg ( α ) = cos ( α ) sin ( α ) , {\displaystyle \operatorname {ctg} (\alpha )={\frac {\cos(\alpha )}{\sin(\alpha )}},} 1 + tg 2 α = 1 cos 2 α , {\displaystyle 1+\operatorname {tg} ^{2}\alpha ={\frac {1}{\cos ^{2}\alpha }},\,} 1 + ctg 2 α = 1 sin 2 α , {\displaystyle 1+\operatorname {ctg} ^{2}\alpha ={\frac {1}{\sin ^{2}\alpha }},\,} tg α ⋅ ctg α = 1. {\displaystyle \operatorname {tg} \alpha \cdot \operatorname {ctg} \alpha =1.} Remove adsФормулы складання: 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 ,} tg ( α ± β ) = tg α ± tg β 1 ∓ tg α tg β , {\displaystyle \operatorname {tg} \left(\alpha \pm \beta \right)={\frac {\operatorname {tg} \,\alpha \pm \operatorname {tg} \,\beta }{1\mp \operatorname {tg} \,\alpha \,\operatorname {tg} \,\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 ,} ctg ( α ± β ) = ctg α ctg β ∓ 1 ctg β ± ctg α , {\displaystyle \operatorname {ctg} \left(\alpha \pm \beta \right)={\frac {\operatorname {ctg} \,\alpha \,\operatorname {ctg} \,\beta \mp 1}{\operatorname {ctg} \,\beta \pm \operatorname {ctg} \,\alpha }},} Remove adsФормулы кратных вуглоў: sin 2 α = 2 sin α cos α = 2 tg α 1 + tg 2 α , {\displaystyle \sin 2\alpha =2\sin \alpha \cos \alpha ={\frac {2\,\operatorname {tg} \,\alpha }{1+\operatorname {tg} ^{2}\alpha }},} cos 2 α = cos 2 α − sin 2 α = 2 cos 2 α − 1 = 1 − 2 sin 2 α = 1 − tg 2 α 1 + tg 2 α = ctg α − tg α ctg α + tg α , {\displaystyle \cos 2\alpha =\cos ^{2}\alpha \,-\,\sin ^{2}\alpha =2\cos ^{2}\alpha \,-\,1=1\,-\,2\sin ^{2}\alpha ={\frac {1-\operatorname {tg} ^{2}\alpha }{1+\operatorname {tg} ^{2}\alpha }}={\frac {\operatorname {ctg} \,\alpha -\operatorname {tg} \,\alpha }{\operatorname {ctg} \,\alpha +\operatorname {tg} \,\alpha }},} tg 2 α = 2 tg α 1 − tg 2 α , {\displaystyle \operatorname {tg} \,2\alpha ={\frac {2\,\operatorname {tg} \,\alpha }{1-\operatorname {tg} ^{2}\alpha }},} ctg 2 α = ctg 2 α − 1 2 ctg α = 1 2 ( ctg α − tg α ) . {\displaystyle \operatorname {ctg} \,2\alpha ={\frac {\operatorname {ctg} ^{2}\alpha -1}{2\,\operatorname {ctg} \,\alpha }}={\frac {1}{2}}\left(\operatorname {ctg} \,\alpha -\operatorname {tg} \,\alpha \right).} 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 ,} tg 3 α = 3 tg α − tg 3 α 1 − 3 tg 2 α , {\displaystyle \operatorname {tg} \,3\alpha ={\frac {3\,\operatorname {tg} \,\alpha -\operatorname {tg} ^{3}\,\alpha }{1-3\,\operatorname {tg} ^{2}\,\alpha }},} ctg 3 α = ctg 3 α − 3 ctg α 3 ctg 2 α − 1 . {\displaystyle \operatorname {ctg} \,3\alpha ={\frac {\operatorname {ctg} ^{3}\,\alpha -3\,\operatorname {ctg} \,\alpha }{3\,\operatorname {ctg} ^{2}\,\alpha -1}}.} Remove adsФормулы палавіннага вугла: sin α 2 = 1 − cos α 2 , 0 ⩽ α ⩽ 2 π ; {\displaystyle \sin {\frac {\alpha }{2}}={\sqrt {\frac {1-\cos \alpha }{2}}},\quad 0\leqslant \alpha \leqslant 2\pi ;} cos α 2 = 1 + cos α 2 , − π ⩽ α ⩽ π ; {\displaystyle \cos {\frac {\alpha }{2}}={\sqrt {\frac {1+\cos \alpha }{2}}},\quad -\pi \leqslant \alpha \leqslant \pi ;} tg α 2 = 1 − cos α sin α = sin α 1 + cos α ; {\displaystyle \operatorname {tg} \,{\frac {\alpha }{2}}={\frac {1-\cos \alpha }{\sin \alpha }}={\frac {\sin \alpha }{1+\cos \alpha }};} ctg α 2 = sin α 1 − cos α = 1 + cos α sin α ; {\displaystyle \operatorname {ctg} \,{\frac {\alpha }{2}}={\frac {\sin \alpha }{1-\cos \alpha }}={\frac {1+\cos \alpha }{\sin \alpha }};} tg α 2 = 1 − cos α 1 + cos α , 0 ⩽ α < π ; {\displaystyle \operatorname {tg} \,{\frac {\alpha }{2}}={\sqrt {\frac {1-\cos \alpha }{1+\cos \alpha }}},\quad 0\leqslant \alpha <\pi ;} ctg α 2 = 1 + cos α 1 − cos α , 0 < α ⩽ π . {\displaystyle \operatorname {ctg} \,{\frac {\alpha }{2}}={\sqrt {\frac {1+\cos \alpha }{1-\cos \alpha }}},\quad 0<\alpha \leqslant \pi .} Remove adsФормулы сумы і рознасці функцый: sin α ± sin β = 2 sin α ± β 2 cos α ∓ β 2 , {\displaystyle \sin \alpha \pm \sin \beta =2\sin {\frac {\alpha \pm \beta }{2}}\cos {\frac {\alpha \mp \beta }{2}},} cos α + cos β = 2 cos α + β 2 cos α − β 2 , {\displaystyle \cos \alpha +\cos \beta =2\cos {\frac {\alpha +\beta }{2}}\cos {\frac {\alpha -\beta }{2}},} cos α − cos β = − 2 sin α + β 2 sin α − β 2 , {\displaystyle \cos \alpha -\cos \beta =-2\sin {\frac {\alpha +\beta }{2}}\sin {\frac {\alpha -\beta }{2}},} tg α ± tg β = sin ( α ± β ) cos α cos β , {\displaystyle \operatorname {tg} \alpha \pm \operatorname {tg} \beta ={\frac {\sin(\alpha \pm \beta )}{\cos \alpha \cos \beta }},} 1 ± sin 2 α = ( sin α ± cos α ) 2 . {\displaystyle 1\pm \sin {2\alpha }=(\sin \alpha \pm \cos \alpha )^{2}.} Remove adsЗдабыткаў функцый: sin α sin β = cos ( α − β ) − cos ( α + β ) 2 , {\displaystyle \sin \alpha \sin \beta ={\frac {\cos(\alpha -\beta )-\cos(\alpha +\beta )}{2}},} sin α cos β = sin ( α − β ) + sin ( α + β ) 2 , {\displaystyle \sin \alpha \cos \beta ={\frac {\sin(\alpha -\beta )+\sin(\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}}.} Remove adsФормулы паніжэння цотнай ступені: 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}},} tg 2 α = 1 − cos 2 α 1 + cos 2 α , {\displaystyle \operatorname {tg} ^{2}\,\alpha ={\frac {1-\cos 2\,\alpha }{1+\cos 2\,\alpha }},} ctg 2 α = 1 + cos 2 α 1 − cos 2 α . {\displaystyle \operatorname {ctg} ^{2}\,\alpha ={\frac {1+\cos 2\,\alpha }{1-\cos 2\,\alpha }}.} Remove adsLoading related searches...Wikiwand - on Seamless Wikipedia browsing. 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