There are several equivalent ways for defining trigonometric functions, and the proofs of the trigonometric identities between them depend on the chosen definition. The oldest and most elementary definitions are based on the geometry of right triangles and the ratio between their sides. The proofs given in this article use these definitions, and thus apply to non-negative angles not greater than a right angle. For greater and negative angles, see Trigonometric functions. Other definitions, and therefore other proofs are based on the Taylor series of sine and cosine, or on the differential equation f ″ + f = 0 {\displaystyle f''+f=0} to which they are solutions.
Elementary trigonometric identities
Definitions
The six trigonometric functions are defined for every real number, except, for some of them, for angles that differ from 0 by a multiple of the right angle (90°). Referring to the diagram at the right, the six trigonometric functions of θ are, for angles smaller than the right angle:
sin θ = o p p o s i t e h y p o t e n u s e = a h {\displaystyle \sin \theta ={\frac {\mathrm {opposite} }{\mathrm {hypotenuse} }}={\frac {a}{h}}}
cos θ = a d j a c e n t h y p o t e n u s e = b h {\displaystyle \cos \theta ={\frac {\mathrm {adjacent} }{\mathrm {hypotenuse} }}={\frac {b}{h}}}
tan θ = o p p o s i t e a d j a c e n t = a b {\displaystyle \tan \theta ={\frac {\mathrm {opposite} }{\mathrm {adjacent} }}={\frac {a}{b}}}
cot θ = a d j a c e n t o p p o s i t e = b a {\displaystyle \cot \theta ={\frac {\mathrm {adjacent} }{\mathrm {opposite} }}={\frac {b}{a}}}
sec θ = h y p o t e n u s e a d j a c e n t = h b {\displaystyle \sec \theta ={\frac {\mathrm {hypotenuse} }{\mathrm {adjacent} }}={\frac {h}{b}}}
csc θ = h y p o t e n u s e o p p o s i t e = h a {\displaystyle \csc \theta ={\frac {\mathrm {hypotenuse} }{\mathrm {opposite} }}={\frac {h}{a}}}
Ratio identities In the case of angles smaller than a right angle, the following identities are direct consequences of above definitions through the division identity
a b = ( a h ) ( b h ) . {\displaystyle {\frac {a}{b}}={\frac {\left({\frac {a}{h}}\right)}{\left({\frac {b}{h}}\right)}}.}
They remain valid for angles greater than 90° and for negative angles.
tan θ = o p p o s i t e a d j a c e n t = ( o p p o s i t e h y p o t e n u s e ) ( a d j a c e n t h y p o t e n u s e ) = sin θ cos θ {\displaystyle \tan \theta ={\frac {\mathrm {opposite} }{\mathrm {adjacent} }}={\frac {\left({\frac {\mathrm {opposite} }{\mathrm {hypotenuse} }}\right)}{\left({\frac {\mathrm {adjacent} }{\mathrm {hypotenuse} }}\right)}}={\frac {\sin \theta }{\cos \theta }}}
cot θ = a d j a c e n t o p p o s i t e = ( a d j a c e n t a d j a c e n t ) ( o p p o s i t e a d j a c e n t ) = 1 tan θ = cos θ sin θ {\displaystyle \cot \theta ={\frac {\mathrm {adjacent} }{\mathrm {opposite} }}={\frac {\left({\frac {\mathrm {adjacent} }{\mathrm {adjacent} }}\right)}{\left({\frac {\mathrm {opposite} }{\mathrm {adjacent} }}\right)}}={\frac {1}{\tan \theta }}={\frac {\cos \theta }{\sin \theta }}}
sec θ = 1 cos θ = h y p o t e n u s e a d j a c e n t {\displaystyle \sec \theta ={\frac {1}{\cos \theta }}={\frac {\mathrm {hypotenuse} }{\mathrm {adjacent} }}}
csc θ = 1 sin θ = h y p o t e n u s e o p p o s i t e {\displaystyle \csc \theta ={\frac {1}{\sin \theta }}={\frac {\mathrm {hypotenuse} }{\mathrm {opposite} }}}
tan θ = o p p o s i t e a d j a c e n t = ( o p p o s i t e × h y p o t e n u s e o p p o s i t e × a d j a c e n t ) ( a d j a c e n t × h y p o t e n u s e o p p o s i t e × a d j a c e n t ) = ( h y p o t e n u s e a d j a c e n t ) ( h y p o t e n u s e o p p o s i t e ) = sec θ csc θ {\displaystyle \tan \theta ={\frac {\mathrm {opposite} }{\mathrm {adjacent} }}={\frac {\left({\frac {\mathrm {opposite} \times \mathrm {hypotenuse} }{\mathrm {opposite} \times \mathrm {adjacent} }}\right)}{\left({\frac {\mathrm {adjacent} \times \mathrm {hypotenuse} }{\mathrm {opposite} \times \mathrm {adjacent} }}\right)}}={\frac {\left({\frac {\mathrm {hypotenuse} }{\mathrm {adjacent} }}\right)}{\left({\frac {\mathrm {hypotenuse} }{\mathrm {opposite} }}\right)}}={\frac {\sec \theta }{\csc \theta }}}
Or
tan θ = sin θ cos θ = ( 1 csc θ ) ( 1 sec θ ) = ( csc θ sec θ csc θ ) ( csc θ sec θ sec θ ) = sec θ csc θ {\displaystyle \tan \theta ={\frac {\sin \theta }{\cos \theta }}={\frac {\left({\frac {1}{\csc \theta }}\right)}{\left({\frac {1}{\sec \theta }}\right)}}={\frac {\left({\frac {\csc \theta \sec \theta }{\csc \theta }}\right)}{\left({\frac {\csc \theta \sec \theta }{\sec \theta }}\right)}}={\frac {\sec \theta }{\csc \theta }}}
cot θ = csc θ sec θ {\displaystyle \cot \theta ={\frac {\csc \theta }{\sec \theta }}}
Complementary angle identities Two angles whose sum is π/2 radians (90 degrees) are complementary. In the diagram, the angles at vertices A and B are complementary, so we can exchange a and b, and change θ to π/2 − θ, obtaining:
sin ( π / 2 − θ ) = cos θ {\displaystyle \sin \left(\pi /2-\theta \right)=\cos \theta }
cos ( π / 2 − θ ) = sin θ {\displaystyle \cos \left(\pi /2-\theta \right)=\sin \theta }
tan ( π / 2 − θ ) = cot θ {\displaystyle \tan \left(\pi /2-\theta \right)=\cot \theta }
cot ( π / 2 − θ ) = tan θ {\displaystyle \cot \left(\pi /2-\theta \right)=\tan \theta }
sec ( π / 2 − θ ) = csc θ {\displaystyle \sec \left(\pi /2-\theta \right)=\csc \theta }
csc ( π / 2 − θ ) = sec θ {\displaystyle \csc \left(\pi /2-\theta \right)=\sec \theta }
Pythagorean identities
Identity 1:
sin 2 θ + cos 2 θ = 1 {\displaystyle \sin ^{2}\theta +\cos ^{2}\theta =1}
The following two results follow from this and the ratio identities. To obtain the first, divide both sides of sin 2 θ + cos 2 θ = 1 {\displaystyle \sin ^{2}\theta +\cos ^{2}\theta =1} by cos 2 θ {\displaystyle \cos ^{2}\theta } ; for the second, divide by sin 2 θ {\displaystyle \sin ^{2}\theta } .
tan 2 θ + 1 = sec 2 θ {\displaystyle \tan ^{2}\theta +1\ =\sec ^{2}\theta }
sec 2 θ − tan 2 θ = 1 {\displaystyle \sec ^{2}\theta -\tan ^{2}\theta =1}
Similarly
1 + cot 2 θ = csc 2 θ {\displaystyle 1\ +\cot ^{2}\theta =\csc ^{2}\theta }
csc 2 θ − cot 2 θ = 1 {\displaystyle \csc ^{2}\theta -\cot ^{2}\theta =1}
Identity 2: The following accounts for all three reciprocal functions.
csc 2 θ + sec 2 θ − cot 2 θ = 2 + tan 2 θ {\displaystyle \csc ^{2}\theta +\sec ^{2}\theta -\cot ^{2}\theta =2\ +\tan ^{2}\theta }
Proof 2: Refer to the triangle diagram above. Note that a 2 + b 2 = h 2 {\displaystyle a^{2}+b^{2}=h^{2}} by Pythagorean theorem.
csc 2 θ + sec 2 θ = h 2 a 2 + h 2 b 2 = a 2 + b 2 a 2 + a 2 + b 2 b 2 = 2 + b 2 a 2 + a 2 b 2 {\displaystyle \csc ^{2}\theta +\sec ^{2}\theta ={\frac {h^{2}}{a^{2}}}+{\frac {h^{2}}{b^{2}}}={\frac {a^{2}+b^{2}}{a^{2}}}+{\frac {a^{2}+b^{2}}{b^{2}}}=2\ +{\frac {b^{2}}{a^{2}}}+{\frac {a^{2}}{b^{2}}}}
Substituting with appropriate functions -
2 + b 2 a 2 + a 2 b 2 = 2 + tan 2 θ + cot 2 θ {\displaystyle 2\ +{\frac {b^{2}}{a^{2}}}+{\frac {a^{2}}{b^{2}}}=2\ +\tan ^{2}\theta +\cot ^{2}\theta }
Rearranging gives:
csc 2 θ + sec 2 θ − cot 2 θ = 2 + tan 2 θ {\displaystyle \csc ^{2}\theta +\sec ^{2}\theta -\cot ^{2}\theta =2\ +\tan ^{2}\theta }
Angle sum identities
Sine
Draw a horizontal line (the x-axis); mark an origin O. Draw a line from O at an angle α {\displaystyle \alpha } above the horizontal line and a second line at an angle β {\displaystyle \beta } above that; the angle between the second line and the x-axis is α + β . {\displaystyle \alpha +\beta .}
Place P on the line defined by α + β {\displaystyle \alpha +\beta } at a unit distance from the origin. Let PQ be a line perpendicular to line OQ defined by angle α {\displaystyle \alpha } , drawn from point Q on this line to point P. ∴ {\displaystyle \therefore } OQP is a right angle. Let QA be a perpendicular from point A on the x-axis to Q and PB be a perpendicular from point B on the x-axis to P. ∴ {\displaystyle \therefore } OAQ and OBP are right angles. Draw R on PB so that QR is parallel to the x-axis. Now angle R P Q = α {\displaystyle RPQ=\alpha } (because O Q A = π 2 − α {\displaystyle OQA={\frac {\pi }{2}}-\alpha } , making R Q O = α , R Q P = π 2 − α {\displaystyle RQO=\alpha ,RQP={\frac {\pi }{2}}-\alpha } , and finally R P Q = α {\displaystyle RPQ=\alpha } )
R P Q = π 2 − R Q P = π 2 − ( π 2 − R Q O ) = R Q O = α {\displaystyle RPQ={\tfrac {\pi }{2}}-RQP={\tfrac {\pi }{2}}-({\tfrac {\pi }{2}}-RQO)=RQO=\alpha }
O P = 1 {\displaystyle OP=1}
P Q = sin β {\displaystyle PQ=\sin \beta }
O Q = cos β {\displaystyle OQ=\cos \beta }
A Q O Q = sin α {\displaystyle {\frac {AQ}{OQ}}=\sin \alpha } , so A Q = sin α cos β {\displaystyle AQ=\sin \alpha \cos \beta }
P R P Q = cos α {\displaystyle {\frac {PR}{PQ}}=\cos \alpha } , so P R = cos α sin β {\displaystyle PR=\cos \alpha \sin \beta }
sin ( α + β ) = P B = R B + P R = A Q + P R = sin α cos β + cos α sin β {\displaystyle \sin(\alpha +\beta )=PB=RB+PR=AQ+PR=\sin \alpha \cos \beta +\cos \alpha \sin \beta }
By substituting − β {\displaystyle -\beta } for β {\displaystyle \beta } and using the reflection identities of even and odd functions, we also get:
sin ( α − β ) = sin α cos ( − β ) + cos α sin ( − β ) {\displaystyle \sin(\alpha -\beta )=\sin \alpha \cos(-\beta )+\cos \alpha \sin(-\beta )}
sin ( α − β ) = sin α cos β − cos α sin β {\displaystyle \sin(\alpha -\beta )=\sin \alpha \cos \beta -\cos \alpha \sin \beta }
Cosine Using the figure above,
O P = 1 {\displaystyle OP=1}
P Q = sin β {\displaystyle PQ=\sin \beta }
O Q = cos β {\displaystyle OQ=\cos \beta }
O A O Q = cos α {\displaystyle {\frac {OA}{OQ}}=\cos \alpha } , so O A = cos α cos β {\displaystyle OA=\cos \alpha \cos \beta }
R Q P Q = sin α {\displaystyle {\frac {RQ}{PQ}}=\sin \alpha } , so R Q = sin α sin β {\displaystyle RQ=\sin \alpha \sin \beta }
cos ( α + β ) = O B = O A − B A = O A − R Q = cos α cos β − sin α sin β {\displaystyle \cos(\alpha +\beta )=OB=OA-BA=OA-RQ=\cos \alpha \cos \beta \ -\sin \alpha \sin \beta }
By substituting − β {\displaystyle -\beta } for β {\displaystyle \beta } and using the reflection identities of even and odd functions, we also get:
cos ( α − β ) = cos α cos ( − β ) − sin α sin ( − β ) , {\displaystyle \cos(\alpha -\beta )=\cos \alpha \cos(-\beta )-\sin \alpha \sin(-\beta ),}
cos ( α − β ) = cos α cos β + sin α sin β {\displaystyle \cos(\alpha -\beta )=\cos \alpha \cos \beta +\sin \alpha \sin \beta }
Also, using the complementary angle formulae,
cos ( α + β ) = sin ( π / 2 − ( α + β ) ) = sin ( ( π / 2 − α ) − β ) = sin ( π / 2 − α ) cos β − cos ( π / 2 − α ) sin β = cos α cos β − sin α sin β {\displaystyle {\begin{aligned}\cos(\alpha +\beta )&=\sin \left(\pi /2-(\alpha +\beta )\right)\\&=\sin \left((\pi /2-\alpha )-\beta \right)\\&=\sin \left(\pi /2-\alpha \right)\cos \beta -\cos \left(\pi /2-\alpha \right)\sin \beta \\&=\cos \alpha \cos \beta -\sin \alpha \sin \beta \\\end{aligned}}}
Tangent and cotangent From the sine and cosine formulae, we get
tan ( α + β ) = sin ( α + β ) cos ( α + β ) = sin α cos β + cos α sin β cos α cos β − sin α sin β {\displaystyle \tan(\alpha +\beta )={\frac {\sin(\alpha +\beta )}{\cos(\alpha +\beta )}}={\frac {\sin \alpha \cos \beta +\cos \alpha \sin \beta }{\cos \alpha \cos \beta -\sin \alpha \sin \beta }}}
Divid