The Pythagorean trigonometric identity, also called simply the Pythagorean identity, is an identity expressing the Pythagorean theorem in terms of trigonometric functions.
Identitas Pythagoras, atau identitas trigonometri Pythagoras (bahasa Inggris: Pythagorean trigonometric identitycode: en is deprecated ), adalah identitas yang menyatakan teorema Pythagoras dalam fungsi trigonometri.
The identity is sin 2 ⁡ θ + cos 2 ⁡ θ1 \theta +\cos ^{2}\theta1} , where sin 2 ⁡ θ \theta } means ( sin ⁡ θ ) 2 {\textstyle (\sin \theta )^{2}} and cos 2 ⁡ θ \theta } means ( cos ⁡ θ ) 2 {\textstyle (\cos \theta )^{2}} .
Identitas tersebut dirumuskan sebagai sin 2 ⁡ θ + cos 2 ⁡ θ1 , \theta +\cos ^{2}\theta1,} untuk setiap bilangan riil θ , dengan sin2 θ yang berarti ( sin ⁡ θ ) 2 {\textstyle (\sin \theta )^{2}} .
cos 2 ⁡ t + sin 2 ⁡ t1. \theta +\cos ^{2}\theta\sin ^{2}\left(t+{\tfrac {\pi }{2}}\right)+\cos ^{2}\left(t+{\tfrac {\pi }{2}}\right)=\cos ^{2}t+\sin ^{2}t=1.} Finally, it remains is to prove the formula for −π <
={\left(-\sin \theta \right)}^{2}+{\left(-\cos \theta \right)}^{2}=\sin ^{2}\theta +\cos ^{2}\theta1.} Karena fungsi sinus dan kosinus merupakan fungsi periodik, maka persamaan sin 2 ⁡ x + cos 2 ⁡ x1 x+\cos ^{2}x=1} berlaku untuk setiap bilangan riil x , dan selesailah pembuktiannya.
θ < 0; this can be done by squaring the symmetry identities to get sin 2 ⁡ θsin 2 ⁡ ( − θ ) and cos 2 ⁡ θcos 2 ⁡ ( − θ ) .
\theta \end{aligned}}} sehingga diperoleh sin 2 ⁡ β + cos 2 ⁡ β( − sin ⁡ θ ) 2 + ( − cos ⁡ θ ) 2sin 2 ⁡ θ + cos 2 ⁡ θ1. \beta +\cos ^{2}\beta
{\text{ and }}\cos ^{2}\theta\cos ^{2}(-\theta ).}
\theta \right)}^{2}=\cos ^{2}\theta +\sin ^{2}\theta
In that way, this trigonometric identity involving the cotangent and the cosecant also follows from the Pythagorean theorem.
Dengan cara ini, identitas trigonometri ini melibatkan kotangen dan kosekan juga, yang didapatkan dari teorema Pythagoras.
In the figure, the point P has a negative x-coordinate, and is appropriately given by xcos θ, which is a negative number: cos θ−cos(π − θ).
Pada gambar berikut, titik P memiliki koordinat- x yang negatif, dan itu didapatkan dari xcos ⁡ θ , yang merupakan bilangan negatif: cos ⁡ θ− cos ⁡ ( π − θ ) .
Point P has a positive y-coordinate, and sin θsin(π − θ) > 0.
Titik P memiliki koordinat- y yang positif, lantaran sin ⁡ θsin ⁡ ( π − θ ) > 0 .
As θ increases from zero to the full circle θ2π, the sine and cosine change signs in the various quadrants to keep x and y with the correct signs.
Saat θ bertambah dari nol menuju satu lingkaran penuh ( θ2 π ), nilai sinus dan kosinusnya berganti tanda di berbagai kuadran agar tanda x dan y nya benar.
x 2 n . \sin x&=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n+1)!}}x^{2n+1},\\\cos x&=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n)!}}x^{2n}.\end{aligned}}} Using the multiplication formula for power series at Multiplication and division of power series (suitably modified to account for the form of the series here) we obtain sin 2 ⁡ x∑ i0 ∞ ∑ j0 ∞ ( − 1 )
x 2 n , \sin x&=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n+1)!}}x^{2n+1},\\\cos x&=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n)!}}x^{2n},\end{aligned}}} dengan asumsi satuan x adalah radian.
x ( 2 i + 1 ) + ( 2 j + 1 )∑ n1 ∞ ( ∑ i0 n − 1 ( − 1 ) n − 1 ( 2 i + 1 ) ! ( 2 ( n − i − 1 ) + 1 ) ! )
x ( 2 i + 1 ) + ( 2 j + 1 )∑ n0 ∞ ( ∑ i0 n ( − 1 ) n ( 2 i + 1 ) ! ( 2 ( n − i ) + 1 ) ! )
x 2 n , cos 2 ⁡ x∑ i0 ∞ ∑ j0 ∞ ( − 1 ) i ( 2 i ) ! ( − 1 ) j ( 2 j ) !
x 2 n cos 2 ⁡ x∑ i0 ∞ ∑ j0 ∞ ( − 1 ) i ( 2 i ) ! ( − 1 ) j ( 2 j ) !
x 2 n . \sin ^{2}x&=\sum _{i=0}^{\infty }\sum _{j=0}^{\infty }{\frac {(-1)^{i}}{(2i+1)!}}{\frac {(-1)^{j}}{(2j+1)!}}x^{(2i+1)+(2j+1)}\\&=\sum _{n=1}^{\infty }\left(\sum _{i=0}^{n-1}{\frac {(-1)^{n-1}}{(2i+1)!(2(n-i-1)+1)!}}\right)x^{2n}\\&=\sum _{n=1}^{\infty }\left(\sum _{i=0}^{n-1}{2n \choose 2i+1}\right){\frac {(-1)^{n-1}}{(2n)!}}x^{2n},\\\cos ^{2}x&=\sum _{i=0}^{\infty }\sum _{j=0}^{\infty }{\frac {(-1)^{i}}{(2i)!}}{\frac {(-1)^{j}}{(2j)!}}x^{(2i)+(2j)}\\&=\sum _{n=0}^{\infty }\left(\sum _{i=0}^{n}{\frac {(-1)^{n}}{(2i)!(2(n-i))!}}\right)x^{2n}\\&=\sum _{n=0}^{\infty }\left(\sum _{i=0}^{n}{2n \choose 2i}\right){\frac {(-1)^{n}}{(2n)!}}x^{2n}.\end{aligned}}} In the expression for sin2, n must be at least 1, while in the expression for cos2, the constant term is equal to 1.
x 2 n \sin ^{2}x&=\sum _{i=0}^{\infty }\sum _{j=0}^{\infty }{\frac {(-1)^{i}}{(2i+1)!}}{\frac {(-1)^{j}}{(2j+1)!}}x^{(2i+1)+(2j+1)}\\&=\sum _{n=0}^{\infty }\left(\sum _{i=0}^{n}{\frac {(-1)^{n}}{(2i+1)!(2(n-i)+1)!}}\right)x^{2n+2}\\&=\sum _{n=1}^{\infty }\left(\sum _{i=0}^{n-1}{\frac {(-1)^{n-1}}{(2i+1)!(2(n-i-1)+1)!}}\right)x^{2n}\\&=\sum _{n=1}^{\infty }\left(\sum _{i=0}^{n-1}{\frac {(-1)^{n-1}}{(2i+1)!(2n-(2i+1))!}}\right)x^{2n}\\&=\sum _{n=1}^{\infty }\left(\sum _{i=0}^{n-1}{2n \choose 2i+1}\right){\frac {(-1)^{n-1}}{(2n)!}}x^{2n}\\\\\cos ^{2}x&=\sum _{i=0}^{\infty }\sum _{j=0}^{\infty }{\frac {(-1)^{i}}{(2i)!}}{\frac {(-1)^{j}}{(2j)!}}x^{(2i)+(2j)}\\&=\sum _{n=0}^{\infty }\left(\sum _{i=0}^{n}{\frac {(-1)^{n}}{(2i)!(2(n-i))!}}\right)x^{2n}\\&=\sum _{n=0}^{\infty }\left(\sum _{i=0}^{n}{2n \choose 2i}\right){\frac {(-1)^{n}}{(2n)!}}x^{2n}\end{aligned}}}
− ∑ i0 n − 1 ( 2 n 2 i + 1 )
− ∑ i1 n − 1 ( 2 n 2 i + 1 ) ) ) + ( 2 ⋅ 0 2 ⋅ 0 ) 1 0 !