In trigonometry, trigonometric identities are equalities that involve trigonometric functions and are true for every value of the occurring variables for which both sides of the equality are defined.
Identitas trigonometri merupakan salah satu fungsi trigonometri di mana rumus tersebut memiliki hasil yang sama bila diuji suatu nilai variabel.
This equation can be solved for either the sine or the cosine: sin ⁡ θ± 1 − cos 2 ⁡ θ , cos ⁡ θ± 1 − sin 2 ⁡ θ .
sin ⁡ 2 θ cos ⁡ θ + sin ⁡ θ cos ⁡ 2 θ2 sin ⁡ θ cos 2 ⁡ θ + sin ⁡ θ ( 1 − 2 sin 2 ⁡ θ )2 sin ⁡ θ ( 1 − sin 2 ⁡ θ )
sin ⁡ ( α + β )
sin ⁡ 3 θsin ⁡ ( 2 θ + θ )
2 sin ⁡ θ cos ⁡ θ( sin ⁡ θ + cos ⁡ θ ) 2 − 12 tan ⁡ θ 1 + tan 2 ⁡ θ cos ⁡ ( 2 θ )cos 2 ⁡ θ − sin 2 ⁡ θ2 cos 2 ⁡ θ − 11 − 2 sin 2 ⁡ θ1 − tan 2 ⁡ θ 1 + tan 2 ⁡ θ tan ⁡ ( 2 θ )2 tan ⁡ θ 1 − tan 2 ⁡ θ cot ⁡ ( 2 θ )cot 2 ⁡ θ − 1 2 cot ⁡ θ1 − tan 2 ⁡ θ 2 tan ⁡ θ sec ⁡ ( 2 θ )sec 2 ⁡ θ 2 − sec 2 ⁡ θ1 + tan 2 ⁡ θ 1 − tan 2 ⁡ θ csc ⁡ ( 2 θ )sec ⁡ θ csc ⁡ θ 21 + tan 2 ⁡ θ 2 tan ⁡ θ \sin(2\theta )&=2\sin \theta \cos \theta &&=(\sin \theta +\cos \theta )^{2}-1&&&={\frac {2\tan \theta }{1+\tan ^{2}\theta }}\\\cos(2\theta )&=\cos ^{2}\theta -\sin ^{2}\theta2\cos ^{2}\theta -1&&=1-2\sin ^{2}\theta &&&={\frac {1-\tan ^{2}\theta }{1+\tan ^{2}\theta }}\\\tan(2\theta )&={\frac {2\tan \theta }{1-\tan ^{2}\theta }}\\\cot(2\theta )&={\frac {\cot ^{2}\theta -1}{2\cot \theta }}&&={\frac {1-\tan ^{2}\theta }{2\tan \theta }}\\\sec(2\theta )&={\frac {\sec ^{2}\theta }{2-\sec ^{2}\theta }}&&={\frac {1+\tan ^{2}\theta }{1-\tan ^{2}\theta }}\\\csc(2\theta )&={\frac {\sec \theta \csc \theta }{2}}&&={\frac {1+\tan ^{2}\theta }{2\tan \theta }}\\\end{aligned}}} Formulae for triple angles.
Berikut adalah sudut setengah rangkap, antara lain sin ⁡ θ 2± 1 − cos ⁡ θ 2 cos ⁡ θ 2± 1 + cos ⁡ θ 2 tan ⁡ θ 2csc ⁡ θ − cot ⁡ θ± 1 − cos ⁡ θ 1 + cos ⁡ θsin ⁡ θ 1 + cos ⁡ θ1 − cos ⁡ θ sin ⁡ θ− 1 ± 1 + tan 2 ⁡ θ tan ⁡ θtan ⁡ θ 1 + sec ⁡ θ cot ⁡ θ 2csc ⁡ θ + cot ⁡ θ± 1 + cos ⁡ θ 1 − cos ⁡ θsin ⁡ θ 1 − cos ⁡ θ1 + cos ⁡ θ sin ⁡ θ \sin {\frac {\theta }{2}}&=\pm {\sqrt {\frac {1-\cos \theta }{2}}}\\[3pt]\cos {\frac {\theta }{2}}&=\pm {\sqrt {\frac {1+\cos \theta }{2}}}\\[3pt]\tan {\frac {\theta }{2}}&=\csc \theta -\cot \theta\pm \,{\sqrt {\frac {1-\cos \theta }{1+\cos \theta }}}={\frac {\sin \theta }{1+\cos \theta }}\\[3pt]&={\frac {1-\cos \theta }{\sin \theta }}={\frac {-1\pm {\sqrt {1+\tan ^{2}\theta }}}{\tan \theta }}={\frac {\tan \theta }{1+\sec {\theta }}}\\[3pt]\cot {\frac {\theta }{2}}&=\csc \theta +\cot \theta\pm \,{\sqrt {\frac {1+\cos \theta }{1-\cos \theta }}}={\frac {\sin \theta }{1-\cos \theta }}={\frac {1+\cos \theta }{\sin \theta }}\end{aligned}}}
∑ k ∈ O + n ( − 1 ) k − 1 2 ( n k ) cos n − k ⁡ θ sin k ⁡ θsin ⁡ θ ∑ i0 n + 1 2 ∑ j0 i ( − 1 )
∑ k0 n ( n k ) cos k ⁡ x sin n − k ⁡ x sin ⁡ ( π 2 ( n − k ) ) ^{n}{\binom {n}{k}}\cos ^{k}x\sin ^{n-k}x\sin \left({\frac {\pi }{2}}(n-k)\right)}
x ) , sin ⁡ ( ( n − 2 ) x ) , and cos ⁡ x with sin ⁡ ( n x )
2 cos ⁡ x sin ⁡ ( ( n − 1 ) x ) − sin ⁡ ( ( n − 2 ) x )
x + x ) and sin ⁡ ( ( n − 1 )
2 cos ⁡ x sin ⁡ ( ( n − 1 ) x ) − sin ⁡ ( ( n − 2 ) x )
Serving a purpose similar to that of the Chebyshev method, for the tangent we can write: tan ⁡ ( n x )tan ⁡ ( ( n − 1 ) x ) + tan ⁡ x 1 − tan ⁡ ( ( n − 1 ) x ) tan ⁡ x . (n-1)x{\bigr )}+\tan x}{1-\tan {\bigl (}(n-1)x{\bigr )}\tan x}}\,.}
x ) tan ⁡ ( n x )tan ⁡ ( ( n − 1 ) x ) + tan ⁡ x 1 − tan ⁡ ( ( n − 1 ) x ) tan ⁡ x \sin(nx)&=2\cos x\sin((n-1)x)-\sin((n-2)x)\\\cos(nx)&=2\cos x\cos((n-1)x)-\cos((n-2)x)\\\tan(nx)&={\frac {\tan((n-1)x)+\tan x}{1-\tan((n-1)x)\tan x}}\end{aligned}}}
sin ⁡ θ 2sgn ⁡ ( sin ⁡ θ 2 ) 1 − cos ⁡ θ 2 cos ⁡ θ 2sgn ⁡ ( cos ⁡ θ 2 ) 1 + cos ⁡ θ 2 tan ⁡ θ 21 − cos ⁡ θ sin ⁡ θsin ⁡ θ 1 + cos ⁡ θcsc ⁡ θ − cot ⁡ θtan ⁡ θ 1 + sec ⁡ θsgn ⁡ ( sin ⁡ θ ) 1 − cos ⁡ θ 1 + cos ⁡ θ− 1 + sgn ⁡ ( cos ⁡ θ ) 1 + tan 2 ⁡ θ tan ⁡ θ cot ⁡ θ 21 + cos ⁡ θ sin ⁡ θsin ⁡ θ 1 − cos ⁡ θcsc ⁡ θ + cot ⁡ θsgn ⁡ ( sin ⁡ θ ) 1 + cos ⁡ θ 1 − cos ⁡ θ sec ⁡ θ 2sgn ⁡ ( cos ⁡ θ 2 ) 2 1 + cos ⁡ θ csc ⁡ θ 2sgn ⁡ ( sin ⁡ θ 2 ) 2 1 − cos ⁡ θ \sin {\frac {\theta }{2}}&=\operatorname {sgn} \left(\sin {\frac {\theta }{2}}\right){\sqrt {\frac {1-\cos \theta }{2}}}\\[3pt]\cos {\frac {\theta }{2}}&=\operatorname {sgn} \left(\cos {\frac {\theta }{2}}\right){\sqrt {\frac {1+\cos \theta }{2}}}\\[3pt]\tan {\frac {\theta }{2}}&={\frac {1-\cos \theta }{\sin \theta }}={\frac {\sin \theta }{1+\cos \theta }}=\csc \theta -\cot \theta{\frac {\tan \theta }{1+\sec {\theta }}}\\[6mu]&=\operatorname {sgn}(\sin \theta ){\sqrt {\frac {1-\cos \theta }{1+\cos \theta }}}={\frac {-1+\operatorname {sgn}(\cos \theta ){\sqrt {1+\tan ^{2}\theta }}}{\tan \theta }}\\[3pt]\cot {\frac {\theta }{2}}&={\frac {1+\cos \theta }{\sin \theta }}={\frac {\sin \theta }{1-\cos \theta }}=\csc \theta +\cot \theta\operatorname {sgn}(\sin \theta ){\sqrt {\frac {1+\cos \theta }{1-\cos \theta }}}\\\sec {\frac {\theta }{2}}&=\operatorname {sgn} \left(\cos {\frac {\theta }{2}}\right){\sqrt {\frac {2}{1+\cos \theta }}}\\\csc {\frac {\theta }{2}}&=\operatorname {sgn} \left(\sin {\frac {\theta }{2}}\right){\sqrt {\frac {2}{1-\cos \theta }}}\\\end{aligned}}} Also tan ⁡ η ± θ 2sin ⁡ η ± sin ⁡ θ cos ⁡ η + cos ⁡ θ tan ⁡ ( θ 2 + π 4 )
Berikut adalah sudut setengah rangkap, antara lain sin ⁡ θ 2± 1 − cos ⁡ θ 2 cos ⁡ θ 2± 1 + cos ⁡ θ 2 tan ⁡ θ 2csc ⁡ θ − cot ⁡ θ± 1 − cos ⁡ θ 1 + cos ⁡ θsin ⁡ θ 1 + cos ⁡ θ1 − cos ⁡ θ sin ⁡ θ− 1 ± 1 + tan 2 ⁡ θ tan ⁡ θtan ⁡ θ 1 + sec ⁡ θ cot ⁡ θ 2csc ⁡ θ + cot ⁡ θ± 1 + cos ⁡ θ 1 − cos ⁡ θsin ⁡ θ 1 − cos ⁡ θ1 + cos ⁡ θ sin ⁡ θ \sin {\frac {\theta }{2}}&=\pm {\sqrt {\frac {1-\cos \theta }{2}}}\\[3pt]\cos {\frac {\theta }{2}}&=\pm {\sqrt {\frac {1+\cos \theta }{2}}}\\[3pt]\tan {\frac {\theta }{2}}&=\csc \theta -\cot \theta\pm \,{\sqrt {\frac {1-\cos \theta }{1+\cos \theta }}}={\frac {\sin \theta }{1+\cos \theta }}\\[3pt]&={\frac {1-\cos \theta }{\sin \theta }}={\frac {-1\pm {\sqrt {1+\tan ^{2}\theta }}}{\tan \theta }}={\frac {\tan \theta }{1+\sec {\theta }}}\\[3pt]\cot {\frac {\theta }{2}}&=\csc \theta +\cot \theta\pm \,{\sqrt {\frac {1+\cos \theta }{1-\cos \theta }}}={\frac {\sin \theta }{1-\cos \theta }}={\frac {1+\cos \theta }{\sin \theta }}\end{aligned}}}
sec ⁡ θ + tan ⁡ θ 1 − sin ⁡ θ 1 + sin ⁡ θ| 1 − tan ⁡ θ 2 | | 1 + tan ⁡ θ 2 | \tan {\frac {\eta \pm \theta }{2}}&={\frac {\sin \eta \pm \sin \theta }{\cos \eta +\cos \theta }}\\[3pt]\tan \left({\frac {\theta }{2}}+{\frac {\pi }{4}}\right)&=\sec \theta +\tan \theta \\[3pt]{\sqrt {\frac {1-\sin \theta }{1+\sin \theta }}}&={\frac {\left|1-\tan {\frac {\theta }{2}}\right|}{\left|1+\tan {\frac {\theta }{2}}\right|}}\end{aligned}}} These can be shown by using either the sum and difference identities or the multiple-angle formulae.
sec ⁡ θ + tan ⁡ θ 1 − sin ⁡ θ 1 + sin ⁡ θ| 1 − tan ⁡ θ 2 | | 1 + tan ⁡ θ 2 | \tan {\frac {\eta \pm \theta }{2}}&={\frac {\sin \eta \pm \sin \theta }{\cos \eta +\cos \theta }}\\[3pt]\tan \left({\frac {\theta }{2}}+{\frac {\pi }{4}}\right)&=\sec \theta +\tan \theta \\[3pt]{\sqrt {\frac {1-\sin \theta }{1+\sin \theta }}}&={\frac {\left|1-\tan {\frac {\theta }{2}}\right|}{\left|1+\tan {\frac {\theta }{2}}\right|}}\end{aligned}}} Suatu penjumlahan fungsi trigonometri dapat dikonversikan menjadi perkalian fungsi trigonometri.
− sin ⁡ ( θ − φ ) .
sin ⁡ 3 θsin ⁡ ( 2 θ + θ )
^{n-1}\sin \left({\frac {k\pi }{n}}\right)={\frac {n}{2^{n-1}}}.} More generally for an integer n > 0 sin ⁡ ( n x )
∑ k0 n ( n k ) cos k ⁡ x sin n − k ⁡ x sin ⁡ ( π 2 ( n − k ) ) ^{n}{\binom {n}{k}}\cos ^{k}x\sin ^{n-k}x\sin \left({\frac {\pi }{2}}(n-k)\right)}
2 n − 1 ∏ k0 n − 1 sin ⁡ ( k π n + x )2 n − 1 ∏ k1 n sin ⁡ ( k π n − x ) .
∑ k0 n ( n k ) cos k ⁡ x sin n − k ⁡ x sin ⁡ ( π 2 ( n − k ) ) ^{n}{\binom {n}{k}}\cos ^{k}x\sin ^{n-k}x\sin \left({\frac {\pi }{2}}(n-k)\right)}
c sin ⁡ ( x + φ ) )+b\sin(x+\theta _{b})=c\sin(x+\varphi )} where c and φ satisfy: c 2a 2 + b 2 + 2 a b cos ⁡ ( θ a − θ b ) , tan ⁡ φa sin ⁡ θ a + b sin ⁡ θ b a cos ⁡ θ a + b cos ⁡ θ b .
+ sin ⁡ θ − 2 sin 3 ⁡ θ2 sin ⁡ θ − 2 sin 3 ⁡ θ + sin ⁡ θ − 2 sin 3 ⁡ θ3 sin ⁡ θ − 4 sin 3 ⁡ θ \sin 3\theta &=\sin(2\theta +\theta )\\&=\sin 2\theta \cos \theta +\sin \theta \cos 2\theta \\&=2\sin \theta \cos ^{2}\theta +\sin \theta (1-2\sin ^{2}\theta )\\&=2\sin \theta (1-\sin ^{2}\theta )
1 + 2 ∑ k1 n cos ⁡ k θsin ⁡ ( ( n + 1 2 ) θ ) sin ⁡ 1 2 θ .
sin ⁡ 2 θ cos ⁡ θ + sin ⁡ θ cos ⁡ 2 θ2 sin ⁡ θ cos 2 ⁡ θ + sin ⁡ θ ( 1 − 2 sin 2 ⁡ θ )2 sin ⁡ θ ( 1 − sin 2 ⁡ θ )
=1+2\sum _{k=1}^{n}\cos k\theta{\frac {\sin \left(\left(n+{\tfrac {1}{2}}\right)\theta \right)}{\sin {\tfrac {1}{2}}\theta }}.}
={\sqrt {(1-\cos \theta )^{2}+\sin ^{2}\theta }}={\sqrt {2-2\cos \theta }}=2\sin \left({\frac {\theta }{2}}\right)} .
( sin ⁡ 2 α − sin ⁡ 0 ) + ( sin ⁡ 4 α − sin ⁡ 2 α ) + ( sin ⁡ 6 α − sin ⁡ 4 α ) + ⋯ + ( sin ⁡ ( 2 n α )
sin ⁡ 2 θ cos ⁡ θ + sin ⁡ θ cos ⁡ 2 θ2 sin ⁡ θ cos 2 ⁡ θ + sin ⁡ θ ( 1 − 2 sin 2 ⁡ θ )2 sin ⁡ θ ( 1 − sin 2 ⁡ θ )
( cos ⁡ ( α + β ) ) x − sin ⁡ ( α + β ) ( sin ⁡ ( α + β ) ) x + cos ⁡ ( α + β ) .
2 cos ⁡ x sin ⁡ ( ( n − 1 ) x ) − sin ⁡ ( ( n − 2 ) x )
1 1 + x 2 tan ⁡ ( arctan ⁡ x )x sin ⁡ ( arccsc ⁡ x )1 x cos ⁡ ( arccsc ⁡ x )1 − 1 x 2 tan ⁡ ( arccsc ⁡ x )1 x 1 − 1 x 2 sin ⁡ ( arcsec ⁡ x )
1 1 + x 2 tan ⁡ ( arctan ⁡ x )x sin ⁡ ( arccsc ⁡ x )1 x cos ⁡ ( arccsc ⁡ x )x 2 − 1 x tan ⁡ ( arccsc ⁡ x )1 x 2 − 1 sin ⁡ ( arcsec ⁡ x )x 2 − 1 x cos ⁡ ( arcsec ⁡ x )
1 x \sin(\arcsin x)&=x&\cos(\arcsin x)&={\sqrt {1-x^{2}}}&\tan(\arcsin x)&={\frac {x}{\sqrt {1-x^{2}}}}\\\sin(\arccos x)&={\sqrt {1-x^{2}}}&\cos(\arccos x)&=x&\tan(\arccos x)&={\frac {\sqrt {1-x^{2}}}{x}}\\\sin(\arctan x)&={\frac {x}{\sqrt {1+x^{2}}}}&\cos(\arctan x)&={\frac {1}{\sqrt {1+x^{2}}}}&\tan(\arctan x)&=x\\\sin(\operatorname {arccsc} x)&={\frac {1}{x}}&\cos(\operatorname {arccsc} x)&={\sqrt {1-{\frac {1}{x^{2}}}}}&\tan(\operatorname {arccsc} x)&={\frac {1}{x{\sqrt {1-{\frac {1}{x^{2}}}}}}}\\\sin(\operatorname {arcsec} x)&={\sqrt {1-{\frac {1}{x^{2}}}}}&\cos(\operatorname {arcsec} x)&={\frac {1}{x}}&\tan(\operatorname {arcsec} x)&=x{\sqrt {1-{\frac {1}{x^{2}}}}}\\\sin(\operatorname {arccot} x)&={\frac {1}{\sqrt {1+x^{2}}}}&\cos(\operatorname {arccot} x)&={\frac {x}{\sqrt {1+x^{2}}}}&\tan(\operatorname {arccot} x)&={\frac {1}{x}}\\\end{aligned}}} Taking the multiplicative inverse of both sides of the each equation above results in the equations for csc1 sin , sec1 cos , and cot1 tan .
1 1 + x 2 cos ⁡ ( arccot ⁡ x )x 1 + x 2 tan ⁡ ( arccot ⁡ x )1 x \sin(\arcsin x)&=x&\cos(\arcsin x)&={\sqrt {1-x^{2}}}&\tan(\arcsin x)&={\frac {x}{\sqrt {1-x^{2}}}}\\\sin(\arccos x)&={\sqrt {1-x^{2}}}&\cos(\arccos x)&=x&\tan(\arccos x)&={\frac {\sqrt {1-x^{2}}}{x}}\\\sin(\arctan x)&={\frac {x}{\sqrt {1+x^{2}}}}&\cos(\arctan x)&={\frac {1}{\sqrt {1+x^{2}}}}&\tan(\arctan x)&=x\\\sin(\operatorname {arccsc} x)&={\frac {1}{x}}&\cos(\operatorname {arccsc} x)&={\frac {\sqrt {x^{2}-1}}{x}}&\tan(\operatorname {arccsc} x)&={\frac {1}{\sqrt {x^{2}-1}}}\\\sin(\operatorname {arcsec} x)&={\frac {\sqrt {x^{2}-1}}{x}}&\cos(\operatorname {arcsec} x)&={\frac {1}{x}}&\tan(\operatorname {arcsec} x)&={\sqrt {x^{2}-1}}\\\sin(\operatorname {arccot} x)&={\frac {1}{\sqrt {1+x^{2}}}}&\cos(\operatorname {arccot} x)&={\frac {x}{\sqrt {1+x^{2}}}}&\tan(\operatorname {arccot} x)&={\frac {1}{x}}\\\end{aligned}}} Berikut adalah penyelesaian persamaan trigonometri, dengan nilai θ dan x .
1 sin ⁡ ( arccos ⁡ x )1 1 − x 2 and sec ⁡ ( arccos ⁡ x )
1 − x 2 x sin ⁡ ( arctan ⁡ x )x 1 + x 2 cos ⁡ ( arctan ⁡ x )
Combining these gives us ∏ k1 n − 1 tan ⁡ k π nn sin ⁡ π n 2 ^{n-1}\tan {\frac {k\pi }{n}}={\frac {n}{\sin {\frac {\pi n}{2}}}}}
∑ k0 n ( n k ) cos k ⁡ x sin n − k ⁡ x sin ⁡ ( π 2 ( n − k ) ) ^{n}{\binom {n}{k}}\cos ^{k}x\sin ^{n-k}x\sin \left({\frac {\pi }{2}}(n-k)\right)}
J 0 ( t ) + 2 ∑ k1 ∞ J 2 k ( t ) cos ⁡ ( 2 k x ) (t)+2\sum _{k=1}^{\infty }J_{2k}(t)\cos(2kx)} sin ⁡ ( t sin ⁡ x )
∑ k0 n ( n k ) cos k ⁡ x sin n − k ⁡ x sin ⁡ ( π 2 ( n − k ) ) ^{n}{\binom {n}{k}}\cos ^{k}x\sin ^{n-k}x\sin \left({\frac {\pi }{2}}(n-k)\right)}
cos ⁡ ( t cos ⁡ x )J 0 ( t ) + 2 ∑ k1 ∞ ( − 1 ) k J 2 k ( t ) cos ⁡ ( 2 k x ) (t)+2\sum _{k=1}^{\infty }(-1)^{k}J_{2k}(t)\cos(2kx)} sin ⁡ ( t cos ⁡ x )
cos ⁡ ( n x )∑ k0 n ( n k ) cos k ⁡ x sin n − k ⁡ x cos ⁡ ( π 2 ( n − k ) ) ^{n}{\binom {n}{k}}\cos ^{k}x\sin ^{n-k}x\cos \left({\frac {\pi }{2}}(n-k)\right)}