इस लेख में प्रमुख त्रिकोणमितीय सर्वसमिकाओं की उपपतियां (सिद्धि) दी गयीं हैं।
यह सम्पूर्ण पृष्ठ या इसके कुछ अनुभाग हिन्दी के अतिरिक्त अन्य भाषा(ओं) में भी लिखे गए हैं। आप इनका अनुवाद
Trigonometric functions specify the relationships between side lengths and interior angles of a right triangle. For example, the sine of angle θ is defined as being the length of the opposite side divided by the length of the hypotenuse. Referring to the diagram at the right, the six trigonometric functions of θ are:
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}}}
The following identities are trivial algebraic consequences of these definitions and the division identity.
a
b
=
(
a
c
)
(
b
c
)
{\displaystyle {\frac {a}{b}}={\frac {\left({\frac {a}{c}}\right)}{\left({\frac {b}{c}}\right)}}}
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 }}}
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
)
=
sec
θ
{\displaystyle \sec \left(\pi /2-\theta \right)=\sec \theta }
csc
(
π
/
2
−
θ
)
=
csc
θ
{\displaystyle \csc \left(\pi /2-\theta \right)=\csc \theta }
Identity 1:
sin
2
(
x
)
+
cos
2
(
x
)
=
1
{\displaystyle \sin ^{2}(x)+\cos ^{2}(x)=1\,}
Proof 1:
Refer to the triangle diagram above. Note that
a
2
+
b
2
=
h
2
{\displaystyle a^{2}+b^{2}=h^{2}}
Pythagorean theorem .
sin
2
(
x
)
+
cos
2
(
x
)
=
a
2
h
2
+
b
2
h
2
=
a
2
+
b
2
h
2
=
h
2
h
2
=
1.
{\displaystyle \sin ^{2}(x)+\cos ^{2}(x)={\frac {a^{2}}{h^{2}}}+{\frac {b^{2}}{h^{2}}}={\frac {a^{2}+b^{2}}{h^{2}}}={\frac {h^{2}}{h^{2}}}=1.\,}
The following two results follow from this and the ratio identities. To obtain the first, divide both sides of
sin
2
(
x
)
+
cos
2
(
x
)
=
1
{\displaystyle \sin ^{2}(x)+\cos ^{2}(x)=1}
cos
2
(
x
)
{\displaystyle \cos ^{2}(x)}
sin
2
(
x
)
{\displaystyle \sin ^{2}(x)}
tan
2
(
x
)
+
1
=
sec
2
(
x
)
{\displaystyle \tan ^{2}(x)+1\ =\sec ^{2}(x)}
sec
2
(
x
)
−
tan
2
(
x
)
=
1
{\displaystyle \sec ^{2}(x)-\tan ^{2}(x)=1\ }
Similarly
1
+
cot
2
(
x
)
=
csc
2
(
x
)
{\displaystyle 1\ +\cot ^{2}(x)=\csc ^{2}(x)}
csc
2
(
x
)
−
cot
2
(
x
)
=
1
{\displaystyle \csc ^{2}(x)-\cot ^{2}(x)=1\ }
Proof 2:
Differentiating the left-hand side of the identity yields:
2
sin
x
⋅
cos
x
−
2
sin
x
⋅
cos
x
=
0
{\displaystyle 2\sin x\cdot \cos x-2\sin x\cdot \cos x=0}
Integrating this shows that the original identity is equal to a constant, and this constant can be found by plugging in any arbitrary value of x.
Identity 2:
The following accounts for all three reciprocal functions.
csc
2
(
x
)
+
sec
2
(
x
)
−
cot
2
(
x
)
=
2
+
tan
2
(
x
)
{\displaystyle \csc ^{2}(x)+\sec ^{2}(x)-\cot ^{2}(x)=2\ +\tan ^{2}(x)}
Proof 1:
Refer to the triangle diagram above. Note that
a
2
+
b
2
=
h
2
{\displaystyle a^{2}+b^{2}=h^{2}}
Pythagorean theorem .
csc
2
(
x
)
+
sec
2
(
x
)
=
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}(x)+\sec ^{2}(x)={\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
(
x
)
+
cot
2
(
x
)
{\displaystyle 2\ +{\frac {b^{2}}{a^{2}}}+{\frac {a^{2}}{b^{2}}}=2\ +\tan ^{2}(x)+\cot ^{2}(x)}
Rearranging gives:
csc
2
(
x
)
+
sec
2
(
x
)
−
cot
2
(
x
)
=
2
+
tan
2
(
x
)
{\displaystyle \csc ^{2}(x)+\sec ^{2}(x)-\cot ^{2}(x)=2\ +\tan ^{2}(x)}
यह सम्पूर्ण पृष्ठ या इसके कुछ अनुभाग हिन्दी के अतिरिक्त अन्य भाषा(ओं) में भी लिखे गए हैं। आप इनका अनुवाद
Illustration of the sum formula. Draw a horizontal line (the x -axis); mark an origin O. Draw a line from O at an angle
α
{\displaystyle \alpha }
β
{\displaystyle \beta }
x -axis is
α
+
β
{\displaystyle \alpha +\beta }
Place P on the line defined by
α
+
β
{\displaystyle \alpha +\beta }
Let PQ be a line perpendicular to line defined by angle
α
{\displaystyle \alpha }
∴
{\displaystyle \therefore }
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 }
Draw QR parallel to the x -axis.
Now angle
R
P
Q
=
α
{\displaystyle RPQ=\alpha }
O
Q
A
=
90
−
α
{\displaystyle OQA=90-\alpha }
R
Q
O
=
α
,
R
Q
P
=
90
−
α
{\displaystyle RQO=\alpha ,RQP=90-\alpha }
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 \,}
A
Q
=
sin
α
cos
β
{\displaystyle AQ=\sin \alpha \cos \beta }
P
R
P
Q
=
cos
α
{\displaystyle {\frac {PR}{PQ}}=\cos \alpha \,}
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 }
β
{\displaystyle \beta }
Symmetry , 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 }
Another simple "proof" can be given using Euler's formula known from complex analysis:
Euler's formula is:
e
i
φ
=
cos
φ
+
i
sin
φ
{\displaystyle e^{i\varphi }=\cos \varphi +i\sin \varphi }
Although it is more precise to say that Euler's formula entails the trigonometric identities, it follows that for angles
α
{\displaystyle \alpha }
β
{\displaystyle \beta }
e
i
(
α
+
β
)
=
cos
(
α
+
β
)
+
i
sin
(
α
+
β
)
{\displaystyle e^{i(\alpha +\beta )}=\cos(\alpha +\beta )+i\sin(\alpha +\beta )}
Also using the following properties of exponential functions:
e
i
(
α
+
β
)
=
e
i
α
e
i
β
=
(
cos
α
+
i
sin
α
)
(
cos
β
+
i
sin
β
)
{\displaystyle e^{i(\alpha +\beta )}=e^{i\alpha }e^{i\beta }=(\cos \alpha +i\sin \alpha )(\cos \beta +i\sin \beta )}
Evaluating the product:
e
i
(
α
+
β
)
=
(
cos
α
cos
β
−
sin
α
sin
β
)
+
i
(
sin
α
cos
β
+
sin
β
cos
α
)
{\displaystyle e^{i(\alpha +\beta )}=(\cos \alpha \cos \beta -\sin \alpha \sin \beta )+i(\sin \alpha \cos \beta +\sin \beta \cos \alpha )}
Equating real and imaginary parts:
cos
(
α
+
β
)
=
cos
α
cos
β
−
sin
α
sin
β
{\displaystyle \cos(\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta }
sin
(
α
+
β
)
=
sin
α
cos
β
+
sin
β
cos
α
{\displaystyle \sin(\alpha +\beta )=\sin \alpha \cos \beta +\sin \beta \cos \alpha }
यह सम्पूर्ण पृष्ठ या इसके कुछ अनुभाग हिन्दी के अतिरिक्त अन्य भाषा(ओं) में भी लिखे गए हैं। आप इनका अनुवाद
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 \,}
O
A
=
cos
α
cos
β
{\displaystyle OA=\cos \alpha \cos \beta \,}
R
Q
P
Q
=
sin
α
{\displaystyle {\frac {RQ}{PQ}}=\sin \alpha \,}
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 }
β
{\displaystyle \beta }
Symmetry , 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}}}
[ संपादित करें ]
यह सम्पूर्ण पृष्ठ या इसके कुछ अनुभाग हिन्दी के अतिरिक्त अन्य भाषा(ओं) में भी लिखे गए हैं। आप इनका अनुवाद
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 }}}
Dividing both numerator and denominator by
cos
α
cos
β
{\displaystyle \cos \alpha \cos \beta }
tan
(
α
+
β
)
=
tan
α
+
tan
β
1
−
tan
α
tan
β
{\displaystyle \tan(\alpha +\beta )={\frac {\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }}}
Subtracting
β
{\displaystyle \beta }
α
{\displaystyle \alpha }
tan
(
−
β
)
=
−
tan
β
{\displaystyle \tan(-\beta )=-\tan \beta }
tan
(
α
−
β
)
=
tan
α
+
tan
(
−
β
)
1
−
tan
α
tan
(
−
β
)
=
tan
α
−
tan
β
1
+
tan
α
tan
β
{\displaystyle \tan(\alpha -\beta )={\frac {\tan \alpha +\tan(-\beta )}{1-\tan \alpha \tan(-\beta )}}={\frac {\tan \alpha -\tan \beta }{1+\tan \alpha \tan \beta }}}
Similarly from the sine and cosine formulae, we get
cot
(
α
+
β
)
=
cos
(
α
+
β
)
sin
(
α
+
β
)
=
cos
α
cos
β
−
sin
α
sin
β
sin
α
cos
β
+
cos
α
sin
β
{\displaystyle \cot(\alpha +\beta )={\frac {\cos(\alpha +\beta )}{\sin(\alpha +\beta )}}={\frac {\cos \alpha \cos \beta -\sin \alpha \sin \beta }{\sin \alpha \cos \beta +\cos \alpha \sin \beta }}}
Then by dividing both numerator and denominator by
sin
α
sin
β
{\displaystyle \sin \alpha \sin \beta }
cot
(
α
+
β
)
=
cot
α
cot
β
−
1
cot
α
+
cot
β
{\displaystyle \cot(\alpha +\beta )={\frac {\cot \alpha \cot \beta -1}{\cot \alpha +\cot \beta }}}
Or, using
cot
θ
=
1
tan
θ
{\displaystyle \cot \theta ={\frac {1}{\tan \theta }}}
cot
(
α
+
β
)
=
1
−
tan
α
tan
β
tan
α
+
tan
β
=
1
tan
α
tan
β
−
1
1
tan
α
+
1
tan
β
=
cot
α
cot
β
−
1
cot
α
+
cot
β
{\displaystyle \cot(\alpha +\beta )={\frac {1-\tan \alpha \tan \beta }{\tan \alpha +\tan \beta }}={\frac {{\frac {1}{\tan \alpha \tan \beta }}-1}{{\frac {1}{\tan \alpha }}+{\frac {1}{\tan \beta }}}}={\frac {\cot \alpha \cot \beta -1}{\cot \alpha +\cot \beta }}}
Using
cot
(
−
β
)
=
−
cot
β
{\displaystyle \cot(-\beta )=-\cot \beta }
cot
(
α
−
β
)
=
cot
α
cot
(
−
β
)
−
1
cot
α
+
cot
(
−
β
)
=
cot
α
cot
β
+
1
cot
β
−
cot
α
{\displaystyle \cot(\alpha -\beta )={\frac {\cot \alpha \cot(-\beta )-1}{\cot \alpha +\cot(-\beta )}}={\frac {\cot \alpha \cot \beta +1}{\cot \beta -\cot \alpha }}}
यह सम्पूर्ण पृष्ठ या इसके कुछ अनुभाग हिन्दी के अतिरिक्त अन्य भाषा(ओं) में भी लिखे गए हैं। आप इनका अनुवाद
From the angle sum identities, we get
sin
(
2
θ
)
=
2
sin
θ
cos
θ
{\displaystyle \sin(2\theta )=2\sin \theta \cos \theta \,}
and
cos
(
2
θ
)
=
cos
2
θ
−
sin
2
θ
{\displaystyle \cos(2\theta )=\cos ^{2}\theta -\sin ^{2}\theta \,}
The Pythagorean identities give the two alternative forms for the latter of these:
cos
(
2
θ
)
=
2
cos
2
θ
−
1
{\displaystyle \cos(2\theta )=2\cos ^{2}\theta -1\,}
cos
(
2
θ
)
=
1
−
2
sin
2
θ
{\displaystyle \cos(2\theta )=1-2\sin ^{2}\theta \,}
The angle sum identities also give
tan
(
2
θ
)
=
2
tan
θ
1
−
tan
2
θ
=
2
cot
θ
−
tan
θ
{\displaystyle \tan(2\theta )={\frac {2\tan \theta }{1-\tan ^{2}\theta }}={\frac {2}{\cot \theta -\tan \theta }}\,}
cot
(
2
θ
)
=
cot
2
θ
−
1
2
cot
θ
=
cot
θ
−
tan
θ
2
{\displaystyle \cot(2\theta )={\frac {\cot ^{2}\theta -1}{2\cot \theta }}={\frac {\cot \theta -\tan \theta }{2}}\,}
It can also be proved using Euler's formula
e
i
φ
=
cos
φ
+
i
sin
φ
{\displaystyle e^{i\varphi }=\cos \varphi +i\sin \varphi }
Squaring both sides yields
e
i
2
φ
=
(
cos
φ
+
i
sin
φ
)
2
{\displaystyle e^{i2\varphi }=(\cos \varphi +i\sin \varphi )^{2}}
But replacing the angle with its doubled version, which achieves the same result in the left side of the equation, yields
e
i
2
φ
=
cos
2
φ
+
i
sin
2
φ
{\displaystyle e^{i2\varphi }=\cos 2\varphi +i\sin 2\varphi }
It follows that
(
cos
φ
+
i
sin
φ
)
2
=
cos
2
φ
+
i
sin
2
φ
{\displaystyle (\cos \varphi +i\sin \varphi )^{2}=\cos 2\varphi +i\sin 2\varphi }
Expanding the square and simplifying on the left hand side of the equation gives
i
(
2
sin
φ
cos
φ
)
+
cos
2
φ
−
sin
2
φ
=
cos
2
φ
+
i
sin
2
φ
{\displaystyle i(2\sin \varphi \cos \varphi )+\cos ^{2}\varphi -\sin ^{2}\varphi \ =\cos 2\varphi +i\sin 2\varphi }
Because the imaginary and real parts have to be the same, we are left with the original identities
cos
2
φ
−
sin
2
φ
=
cos
2
φ
{\displaystyle \cos ^{2}\varphi -\sin ^{2}\varphi \ =\cos 2\varphi }
and also
2
sin
φ
cos
φ
=
sin
2
φ
{\displaystyle 2\sin \varphi \cos \varphi =\sin 2\varphi }
यह सम्पूर्ण पृष्ठ या इसके कुछ अनुभाग हिन्दी के अतिरिक्त अन्य भाषा(ओं) में भी लिखे गए हैं। आप इनका अनुवाद
The two identities giving the alternative forms for cos 2θ lead to the following equations:
cos
θ
2
=
±
1
+
cos
θ
2
,
{\displaystyle \cos {\frac {\theta }{2}}=\pm \,{\sqrt {\frac {1+\cos \theta }{2}}},\,}
sin
θ
2
=
±
1
−
cos
θ
2
.
{\displaystyle \sin {\frac {\theta }{2}}=\pm \,{\sqrt {\frac {1-\cos \theta }{2}}}.\,}
The sign of the square root needs to be chosen properly—note that if π is added to θ, the quantities inside the square roots are unchanged, but the left-hand-sides of the equations change sign. Therefore the correct sign to use depends on the value of θ.
For the tan function, the equation is:
tan
θ
2
=
±
1
−
cos
θ
1
+
cos
θ
.
{\displaystyle \tan {\frac {\theta }{2}}=\pm \,{\sqrt {\frac {1-\cos \theta }{1+\cos \theta }}}.\,}
Then multiplying the numerator and denominator inside the square root by (1 + cos θ) and using Pythagorean identities leads to:
tan
θ
2
=
sin
θ
1
+
cos
θ
.
{\displaystyle \tan {\frac {\theta }{2}}={\frac {\sin \theta }{1+\cos \theta }}.\,}
Also, if the numerator and denominator are both multiplied by (1 - cos θ), the result is:
tan
θ
2
=
1
−
cos
θ
sin
θ
.
{\displaystyle \tan {\frac {\theta }{2}}={\frac {1-\cos \theta }{\sin \theta }}.\,}
This also gives:
tan
θ
2
=
csc
θ
−
cot
θ
.
{\displaystyle \tan {\frac {\theta }{2}}=\csc \theta -\cot \theta .\,}
Similar manipulations for the cot function give:
cot
θ
2
=
±
1
+
cos
θ
1
−
cos
θ
=
1
+
cos
θ
sin
θ
=
sin
θ
1
−
cos
θ
=
csc
θ
+
cot
θ
.
{\displaystyle \cot {\frac {\theta }{2}}=\pm \,{\sqrt {\frac {1+\cos \theta }{1-\cos \theta }}}={\frac {1+\cos \theta }{\sin \theta }}={\frac {\sin \theta }{1-\cos \theta }}=\csc \theta +\cot \theta .\,}
If
ψ
{\displaystyle \psi }
θ
{\displaystyle \theta }
ϕ
{\displaystyle \phi }
ψ
+
θ
+
ϕ
=
π
=
{\displaystyle \psi +\theta +\phi =\pi =}
tan
(
ψ
)
+
tan
(
θ
)
+
tan
(
ϕ
)
=
tan
(
ψ
)
tan
(
θ
)
tan
(
ϕ
)
.
{\displaystyle \tan(\psi )+\tan(\theta )+\tan(\phi )=\tan(\psi )\tan(\theta )\tan(\phi ).}
Proof:[ 1]
ψ
=
π
−
θ
−
ϕ
tan
(
ψ
)
=
tan
(
π
−
θ
−
ϕ
)
=
−
tan
(
θ
+
ϕ
)
=
−
tan
θ
−
tan
ϕ
1
−
tan
θ
tan
ϕ
=
tan
θ
+
tan
ϕ
tan
θ
tan
ϕ
−
1
(
tan
θ
tan
ϕ
−
1
)
tan
ψ
=
tan
θ
+
tan
ϕ
tan
ψ
tan
θ
tan
ϕ
−
tan
ψ
=
tan
θ
+
tan
ϕ
tan
ψ
tan
θ
tan
ϕ
=
tan
ψ
+
tan
θ
+
tan
ϕ
{\displaystyle {\begin{aligned}\psi &=\pi -\theta -\phi \\\tan(\psi )&=\tan(\pi -\theta -\phi )\\&=-\tan(\theta +\phi )\\&={\frac {-\tan \theta -\tan \phi }{1-\tan \theta \tan \phi }}\\&={\frac {\tan \theta +\tan \phi }{\tan \theta \tan \phi -1}}\\(\tan \theta \tan \phi -1)\tan \psi &=\tan \theta +\tan \phi \\\tan \psi \tan \theta \tan \phi -\tan \psi &=\tan \theta +\tan \phi \\\tan \psi \tan \theta \tan \phi &=\tan \psi +\tan \theta +\tan \phi \\\end{aligned}}}
Miscellaneous -- the triple cotangent identity [ संपादित करें ] If
ψ
+
θ
+
ϕ
=
π
2
=
{\displaystyle \psi +\theta +\phi ={\tfrac {\pi }{2}}=}
cot
(
ψ
)
+
cot
(
θ
)
+
cot
(
ϕ
)
=
cot
(
ψ
)
cot
(
θ
)
cot
(
ϕ
)
{\displaystyle \cot(\psi )+\cot(\theta )+\cot(\phi )=\cot(\psi )\cot(\theta )\cot(\phi )}
Proof:
Replace each of
ψ
{\displaystyle \psi }
θ
{\displaystyle \theta }
ϕ
{\displaystyle \phi }
Given
ψ
+
θ
+
ϕ
=
π
2
{\displaystyle \psi +\theta +\phi ={\tfrac {\pi }{2}}\,}
∴
(
π
2
−
ψ
)
+
(
π
2
−
θ
)
+
(
π
2
−
ϕ
)
=
3
π
2
−
(
ψ
+
θ
+
ϕ
)
=
3
π
2
−
π
2
=
π
{\displaystyle \therefore ({\tfrac {\pi }{2}}-\psi )+({\tfrac {\pi }{2}}-\theta )+({\tfrac {\pi }{2}}-\phi )={\tfrac {3\pi }{2}}-(\psi +\theta +\phi )={\tfrac {3\pi }{2}}-{\tfrac {\pi }{2}}=\pi }
so the result follows from the triple tangent identity.
sin
θ
±
sin
ϕ
=
2
sin
(
θ
±
ϕ
2
)
cos
(
θ
∓
ϕ
2
)
{\displaystyle \sin \theta \pm \sin \phi =2\sin \left({\frac {\theta \pm \phi }{2}}\right)\cos \left({\frac {\theta \mp \phi }{2}}\right)}
cos
θ
+
cos
ϕ
=
2
cos
(
θ
+
ϕ
2
)
cos
(
θ
−
ϕ
2
)
{\displaystyle \cos \theta +\cos \phi =2\cos \left({\frac {\theta +\phi }{2}}\right)\cos \left({\frac {\theta -\phi }{2}}\right)}
cos
θ
−
cos
ϕ
=
−
2
sin
(
θ
+
ϕ
2
)
sin
(
θ
−
ϕ
2
)
{\displaystyle \cos \theta -\cos \phi =-2\sin \left({\frac {\theta +\phi }{2}}\right)\sin \left({\frac {\theta -\phi }{2}}\right)}
First, start with the sum-angle identities:
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 }
By adding these together,
sin
(
α
+
β
)
+
sin
(
α
−
β
)
=
sin
α
cos
β
+
cos
α
sin
β
+
sin
α
cos
β
−
cos
α
sin
β
=
2
sin
α
cos
β
{\displaystyle \sin(\alpha +\beta )+\sin(\alpha -\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta +\sin \alpha \cos \beta -\cos \alpha \sin \beta =2\sin \alpha \cos \beta }
Similarly, by subtracting the two sum-angle identities,
sin
(
α
+
β
)
−
sin
(
α
−
β
)
=
sin
α
cos
β
+
cos
α
sin
β
−
sin
α
cos
β
+
cos
α
sin
β
=
2
cos
α
sin
β
{\displaystyle \sin(\alpha +\beta )-\sin(\alpha -\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta -\sin \alpha \cos \beta +\cos \alpha \sin \beta =2\cos \alpha \sin \beta }
Let
α
+
β
=
θ
{\displaystyle \alpha +\beta =\theta }
α
−
β
=
ϕ
{\displaystyle \alpha -\beta =\phi }
∴
α
=
θ
+
ϕ
2
{\displaystyle \therefore \alpha ={\frac {\theta +\phi }{2}}}
β
=
θ
−
ϕ
2
{\displaystyle \beta ={\frac {\theta -\phi }{2}}}
Substitute
θ
{\displaystyle \theta }
ϕ
{\displaystyle \phi }
sin
θ
+
sin
ϕ
=
2
sin
(
θ
+
ϕ
2
)
cos
(
θ
−
ϕ
2
)
{\displaystyle \sin \theta +\sin \phi =2\sin \left({\frac {\theta +\phi }{2}}\right)\cos \left({\frac {\theta -\phi }{2}}\right)}
sin
θ
−
sin
ϕ
=
2
cos
(
θ
+
ϕ
2
)
sin
(
θ
−
ϕ
2
)
=
2
sin
(
θ
−
ϕ
2
)
cos
(
θ
+
ϕ
2
)
{\displaystyle \sin \theta -\sin \phi =2\cos \left({\frac {\theta +\phi }{2}}\right)\sin \left({\frac {\theta -\phi }{2}}\right)=2\sin \left({\frac {\theta -\phi }{2}}\right)\cos \left({\frac {\theta +\phi }{2}}\right)}
Therefore,
sin
θ
±
sin
ϕ
=
2
sin
(
θ
±
ϕ
2
)
cos
(
θ
∓
ϕ
2
)
{\displaystyle \sin \theta \pm \sin \phi =2\sin \left({\frac {\theta \pm \phi }{2}}\right)\cos \left({\frac {\theta \mp \phi }{2}}\right)}
Similarly for cosine, start with the sum-angle identities:
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 }
Again, by adding and substracting
cos
(
α
+
β
)
+
cos
(
α
−
β
)
=
cos
α
cos
β
−
sin
α
sin
β
+
cos
α
cos
β
+
sin
α
sin
β
=
2
cos
α
cos
β
{\displaystyle \cos(\alpha +\beta )+\cos(\alpha -\beta )=\cos \alpha \cos \beta \ -\sin \alpha \sin \beta +\cos \alpha \cos \beta +\sin \alpha \sin \beta =2\cos \alpha \cos \beta \ }
cos
(
α
+
β
)
−
cos
(
α
−
β
)
=
cos
α
cos
β
−
sin
α
sin
β
−
cos
α
cos
β
−
sin
α
sin
β
=
−
2
sin
α
sin
β
{\displaystyle \cos(\alpha +\beta )-\cos(\alpha -\beta )=\cos \alpha \cos \beta \ -\sin \alpha \sin \beta -\cos \alpha \cos \beta -\sin \alpha \sin \beta =-2\sin \alpha \sin \beta }
Substitute
θ
{\displaystyle \theta }
ϕ
{\displaystyle \phi }
cos
θ
+
cos
ϕ
=
2
cos
(
θ
+
ϕ
2
)
cos
(
θ
−
ϕ
2
)
{\displaystyle \cos \theta +\cos \phi =2\cos \left({\frac {\theta +\phi }{2}}\right)\cos \left({\frac {\theta -\phi }{2}}\right)}
cos
θ
−
cos
ϕ
=
−
2
sin
(
θ
+
ϕ
2
)
sin
(
θ
−
ϕ
2
)
{\displaystyle \cos \theta -\cos \phi =-2\sin \left({\frac {\theta +\phi }{2}}\right)\sin \left({\frac {\theta -\phi }{2}}\right)}
Illustration of the sine and tangent inequalities. The figure at the right shows a sector of a circle with radius 1. The sector is θ/(2π) of the whole circle, so its area is θ/2.
O
A
=
O
D
=
1
{\displaystyle OA=OD=1\,}
A
B
=
sin
θ
{\displaystyle AB=\sin \theta \,}
C
D
=
tan
θ
{\displaystyle CD=\tan \theta \,}
The area of triangle OAD is AB/2, or sinθ/2. The area of triangle OCD is CD/2, or tanθ/2.
Since triangle OAD lies completely inside the sector, which in turn lies completely inside triangle OCD, we have
sin
θ
<
θ
<
tan
θ
{\displaystyle \sin \theta <\theta <\tan \theta \,}
This geometric argument applies if 0<θ<π/2. It relies on definitions of arc length and
area , which act as assumptions, so it is rather a condition imposed in construction of trigonometric functions than
a provable property.[ 2]
sin
θ
θ
<
1
i
f
0
<
θ
{\displaystyle {\frac {\sin \theta }{\theta }}<1\ \ \ \mathrm {if} \ \ \ 0<\theta \,}
For negative values of θ we have, by symmetry of the sine function
sin
θ
θ
=
sin
(
−
θ
)
−
θ
<
1
{\displaystyle {\frac {\sin \theta }{\theta }}={\frac {\sin(-\theta )}{-\theta }}<1\,}
Hence
sin
θ
θ
<
1
i
f
θ
≠
0
{\displaystyle {\frac {\sin \theta }{\theta }}<1\ \ \ \mathrm {if} \ \ \ \theta \neq 0\,}
tan
θ
θ
>
1
i
f
0
<
θ
<
π
2
{\displaystyle {\frac {\tan \theta }{\theta }}>1\ \ \ \mathrm {if} \ \ \ 0<\theta <{\frac {\pi }{2}}\,}
lim
θ
→
0
sin
θ
=
0
{\displaystyle \lim _{\theta \to 0}{\sin \theta }=0\,}
lim
θ
→
0
cos
θ
=
1
{\displaystyle \lim _{\theta \to 0}{\cos \theta }=1\,}
lim
θ
→
0
sin
θ
θ
=
1
{\displaystyle \lim _{\theta \to 0}{\frac {\sin \theta }{\theta }}=1}
Proof: From the previous inequalities, we have, for small angles
sin
θ
<
θ
<
tan
θ
{\displaystyle \sin \theta <\theta <\tan \theta \,}
Therefore,
sin
θ
θ
<
1
<
tan
θ
θ
{\displaystyle {\frac {\sin \theta }{\theta }}<1<{\frac {\tan \theta }{\theta }}\,}
Consider the right-hand inequality. Since
tan
θ
=
sin
θ
cos
θ
{\displaystyle \tan \theta ={\frac {\sin \theta }{\cos \theta }}}
∴
1
<
sin
θ
θ
cos
θ
{\displaystyle \therefore 1<{\frac {\sin \theta }{\theta \cos \theta }}}
Multply through by
cos
θ
{\displaystyle \cos \theta }
cos
θ
<
sin
θ
θ
{\displaystyle \cos \theta <{\frac {\sin \theta }{\theta }}}
Combining with the left-hand inequality:
cos
θ
<
sin
θ
θ
<
1
{\displaystyle \cos \theta <{\frac {\sin \theta }{\theta }}<1}
Taking
cos
θ
{\displaystyle \cos \theta }
θ
→
0
{\displaystyle \theta \to 0}
lim
θ
→
0
cos
θ
=
1
{\displaystyle \lim _{\theta \to 0}{\cos \theta }=1\,}
Therefore,
lim
θ
→
0
sin
θ
θ
=
1
{\displaystyle \lim _{\theta \to 0}{\frac {\sin \theta }{\theta }}=1}
lim
θ
→
0
1
−
cos
θ
θ
=
0
{\displaystyle \lim _{\theta \to 0}{\frac {1-\cos \theta }{\theta }}=0}
Proof:
1
−
cos
θ
θ
=
1
−
cos
2
θ
θ
(
1
+
cos
θ
)
=
sin
2
θ
θ
(
1
+
cos
θ
)
=
(
sin
θ
θ
)
×
sin
θ
×
(
1
1
+
cos
θ
)
{\displaystyle {\begin{aligned}{\frac {1-\cos \theta }{\theta }}&={\frac {1-\cos ^{2}\theta }{\theta (1+\cos \theta )}}\\&={\frac {\sin ^{2}\theta }{\theta (1+\cos \theta )}}\\&=\left({\frac {\sin \theta }{\theta }}\right)\times \sin \theta \times \left({\frac {1}{1+\cos \theta }}\right)\\\end{aligned}}}
The limits of those three quantities are 1, 0, and 1/2, so the resultant limit is zero.
lim
θ
→
0
1
−
cos
θ
θ
2
=
1
2
{\displaystyle \lim _{\theta \to 0}{\frac {1-\cos \theta }{\theta ^{2}}}={\frac {1}{2}}}
Proof:
As in the preceding proof,
1
−
cos
θ
θ
2
=
sin
θ
θ
×
sin
θ
θ
×
1
1
+
cos
θ
.
{\displaystyle {\frac {1-\cos \theta }{\theta ^{2}}}={\frac {\sin \theta }{\theta }}\times {\frac {\sin \theta }{\theta }}\times {\frac {1}{1+\cos \theta }}.\,}
The limits of those three quantities are 1, 1, and 1/2, so the resultant limit is 1/2.
Proof of Compositions of trig and inverse trig functions [ संपादित करें ] All these functions follow from the Pythagorean trigonometric identity. We can prove for instance the function
sin
[
arctan
(
x
)
]
=
x
1
+
x
2
{\displaystyle \sin[\arctan(x)]={\frac {x}{\sqrt {1+x^{2}}}}}
Proof:
We start from
sin
2
θ
+
cos
2
θ
=
1
{\displaystyle \sin ^{2}\theta +\cos ^{2}\theta =1}
Then we divide this equation by
cos
2
θ
{\displaystyle \cos ^{2}\theta }
cos
2
θ
=
1
tan
2
θ
+
1
{\displaystyle \cos ^{2}\theta ={\frac {1}{\tan ^{2}\theta +1}}}
Then use the substitution
θ
=
arctan
(
x
)
{\displaystyle \theta =\arctan(x)}
1
−
sin
2
[
arctan
(
x
)
]
=
1
tan
2
[
arctan
(
x
)
]
+
1
{\displaystyle 1-\sin ^{2}[\arctan(x)]={\frac {1}{\tan ^{2}[\arctan(x)]+1}}}
Then we use the identity
tan
[
arctan
(
x
)
]
≡
x
{\displaystyle \tan[\arctan(x)]\equiv x}
sin
[
arctan
(
x
)
]
=
x
x
2
+
1
{\displaystyle \sin[\arctan(x)]={\frac {x}{\sqrt {x^{2}+1}}}}
E. T. Whittaker and G. N. Watson. A course of modern analysis , Cambridge University Press, 1952
↑ "संग्रहीत प्रति" . मूल से 29 अक्तूबर 2013 को पुरालेखित. अभिगमन तिथि 8 जून 2014 .↑ Richman, Fred (March 1993). . "A Circular Argument" . The College Mathematics Journal . 24 (2): 160–162. डीओआइ :10.2307/2686787 . मूल से 17 मार्च 2016 को पुरालेखित . अभिगमन तिथि 3 नवम्बर 2012 .
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![{\displaystyle 1-\sin ^{2}[\arctan(x)]={\frac {1}{\tan ^{2}[\arctan(x)]+1}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d0bb45d076e27715801c0437e364d4b1f45dfb1a)
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![{\displaystyle \sin[\arctan(x)]={\frac {x}{\sqrt {x^{2}+1}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5c5222bb21dc2d342d1363c855694bae8b967914)