এখানে,
sin
(
A
)
=
a
/
h
{\displaystyle \sin(A)=a/h}
cos
(
A
)
=
b
/
h
{\displaystyle \cos(A)=b/h}
একটি ত্রিভুজ
A
B
C
{\displaystyle ABC}
A
C
=
b
{\displaystyle AC=b}
A
B
=
h
{\displaystyle AB=h}
B
C
=
a
{\displaystyle BC=a}
∠
A
{\displaystyle \angle A}
a
{\displaystyle a}
b
{\displaystyle b}
h
{\displaystyle h}
x
{\displaystyle x}
sin
(
x
)
{\displaystyle \sin(x)}
cos
(
x
)
{\displaystyle \cos(x)}
tan
(
x
)
{\displaystyle \tan(x)}
csc
(
x
)
{\displaystyle \csc(x)}
sec
(
x
)
{\displaystyle \sec(x)}
cot
(
x
)
{\displaystyle \cot(x)}
x
{\displaystyle x}
y
{\displaystyle y}
tan
(
x
)
=
sin
(
x
)
cos
(
x
)
;
{\displaystyle \tan(x)={\frac {\sin(x)}{\cos(x)}};}
csc
(
x
)
=
1
sin
(
x
)
;
{\displaystyle \csc(x)={\frac {1}{\sin(x)}};}
csc
(
x
)
=
1
sin
(
x
)
;
{\displaystyle \csc(x)={\frac {1}{\sin(x)}};}
sec
(
x
)
=
1
cos
(
x
)
;
{\displaystyle \sec(x)={\frac {1}{\cos(x)}};}
sin
2
(
x
)
+
cos
2
(
x
)
=
1
{\displaystyle \sin ^{2}(x)+\cos ^{2}(x)=1}
csc
2
(
x
)
−
cot
2
(
x
)
=
1
{\displaystyle \csc ^{2}(x)-\cot ^{2}(x)=1}
sec
2
(
x
)
−
tan
2
(
x
)
=
1
{\displaystyle \sec ^{2}(x)-\tan ^{2}(x)=1}
sin
(
x
+
y
)
=
sin
(
x
)
⋅
cos
(
y
)
+
cos
(
x
)
⋅
sin
(
y
)
{\displaystyle \sin(x+y)=\sin(x)\cdot \cos(y)+\cos(x)\cdot \sin(y)}
cos
(
x
+
y
)
=
cos
(
x
)
⋅
cos
(
y
)
−
sin
(
x
)
⋅
sin
(
y
)
{\displaystyle \cos(x+y)=\cos(x)\cdot \cos(y)-\sin(x)\cdot \sin(y)}
sine (সাইন) ফাংশনের অন্তরজ
(
sin
(
x
)
)
′
=
cos
(
x
)
{\displaystyle (\sin(x))'=\cos(x)}
প্রমাণ:
অন্তরজের সংজ্ঞা অনুসারে,
(
sin
(
x
)
)
′
{\displaystyle (\sin(x))'}
=
lim
Δ
x
→
0
sin
(
x
+
Δ
x
)
−
sin
(
x
)
Δ
x
{\displaystyle =\lim _{\Delta x\to 0}{\frac {\sin(x+\Delta x)-\sin(x)}{\Delta x}}}
=
lim
Δ
x
→
0
sin
(
x
)
⋅
cos
(
Δ
x
)
+
cos
(
x
)
⋅
sin
(
Δ
x
)
−
sin
(
x
)
Δ
x
{\displaystyle =\lim _{\Delta x\to 0}{\frac {\sin(x)\cdot \cos(\Delta x)+\cos(x)\cdot \sin(\Delta x)-\sin(x)}{\Delta x}}}
=
lim
Δ
x
→
0
sin
(
x
)
(
cos
(
Δ
x
)
−
1
)
+
cos
(
x
)
⋅
sin
(
Δ
x
)
Δ
x
{\displaystyle =\lim _{\Delta x\to 0}{\frac {\sin(x)(\cos(\Delta x)-1)+\cos(x)\cdot \sin(\Delta x)}{\Delta x}}}
=
lim
Δ
x
→
0
sin
(
x
)
(
cos
(
Δ
x
)
−
1
)
Δ
x
+
lim
Δ
x
→
0
cos
(
x
)
⋅
sin
(
Δ
x
)
Δ
x
{\displaystyle =\lim _{\Delta x\to 0}{\frac {\sin(x)(\cos(\Delta x)-1)}{\Delta x}}+\lim _{\Delta x\to 0}{\frac {\cos(x)\cdot \sin(\Delta x)}{\Delta x}}}
=
[
lim
Δ
x
→
0
sin
(
x
)
]
⋅
[
lim
Δ
x
→
0
cos
(
Δ
x
)
−
1
Δ
x
]
+
[
lim
Δ
x
→
0
cos
(
x
)
]
⋅
[
lim
Δ
x
→
0
sin
(
Δ
x
)
Δ
x
]
{\displaystyle =\left[\lim _{\Delta x\to 0}{\sin(x)}\right]\cdot \left[\lim _{\Delta x\to 0}{\frac {\cos(\Delta x)-1}{\Delta x}}\right]+\left[\lim _{\Delta x\to 0}\cos(x)\right]\cdot \left[\lim _{\Delta x\to 0}{\frac {\sin(\Delta x)}{\Delta x}}\right]}
=
[
lim
Δ
x
→
0
sin
(
x
)
]
⋅
[
0
]
+
[
lim
Δ
x
→
0
cos
(
x
)
⋅
]
⋅
[
1
]
{\displaystyle =\left[\lim _{\Delta x\to 0}{\sin(x)}\right]\cdot \left[0\right]+\left[\lim _{\Delta x\to 0}\cos(x)\cdot \right]\cdot \left[1\right]}
=
cos
(
x
)
{\displaystyle =\cos(x)}
◻
{\displaystyle \square }
cosine (কোসাইন) ফাংশনের অন্তরজ
(
sin
(
x
)
)
′
=
cos
(
x
)
{\displaystyle (\sin(x))'=\cos(x)}
প্রমাণ:
অন্তরজের সংজ্ঞা অনুসারে,
(
cos
(
x
)
)
′
{\displaystyle (\cos(x))'}
=
lim
Δ
x
→
0
cos
(
x
+
Δ
x
)
−
cos
(
x
)
Δ
x
{\displaystyle =\lim _{\Delta x\to 0}{\frac {\cos(x+\Delta x)-\cos(x)}{\Delta x}}}
=
lim
Δ
x
→
0
cos
(
x
)
⋅
cos
(
Δ
x
)
−
sin
(
x
)
⋅
sin
(
Δ
x
)
−
cos
(
x
)
Δ
x
{\displaystyle =\lim _{\Delta x\to 0}{\frac {\cos(x)\cdot \cos(\Delta x)-\sin(x)\cdot \sin(\Delta x)-\cos(x)}{\Delta x}}}
=
lim
Δ
x
→
0
−
sin
(
x
)
⋅
sin
(
Δ
x
)
+
cos
(
x
)
(
cos
(
Δ
x
)
−
1
)
Δ
x
{\displaystyle =\lim _{\Delta x\to 0}{\frac {-\sin(x)\cdot \sin(\Delta x)+\cos(x)(\cos(\Delta x)-1)}{\Delta x}}}
=
lim
Δ
x
→
0
cos
(
x
)
(
cos
(
Δ
x
)
−
1
)
Δ
x
−
lim
Δ
x
→
0
sin
(
x
)
⋅
sin
(
Δ
x
)
Δ
x
{\displaystyle =\lim _{\Delta x\to 0}{\frac {\cos(x)(\cos(\Delta x)-1)}{\Delta x}}-\lim _{\Delta x\to 0}{\frac {\sin(x)\cdot \sin(\Delta x)}{\Delta x}}}
=
[
lim
Δ
x
→
0
cos
(
x
)
]
⋅
[
lim
Δ
x
→
0
cos
(
Δ
x
)
−
1
Δ
x
]
−
[
lim
Δ
x
→
0
sin
(
x
)
]
⋅
[
lim
Δ
x
→
0
sin
(
Δ
x
)
Δ
x
]
{\displaystyle =\left[\lim _{\Delta x\to 0}\cos(x)\right]\cdot \left[\lim _{\Delta x\to 0}{\frac {\cos(\Delta x)-1}{\Delta x}}\right]-\left[\lim _{\Delta x\to 0}{\sin(x)}\right]\cdot \left[\lim _{\Delta x\to 0}{\frac {\sin(\Delta x)}{\Delta x}}\right]}
=
[
lim
Δ
x
→
0
cos
(
x
)
]
⋅
[
0
]
−
[
lim
Δ
x
→
0
sin
(
x
)
⋅
]
⋅
[
1
]
{\displaystyle =\left[\lim _{\Delta x\to 0}{\cos(x)}\right]\cdot \left[0\right]-\left[\lim _{\Delta x\to 0}\sin(x)\cdot \right]\cdot \left[1\right]}
=
−
sin
(
x
)
{\displaystyle =-\sin(x)}
◻
{\displaystyle \square }
সাইন এবং কোসাইন ফাংশন দুইটির অন্তরজ জানা থাকলে অন্য সকল ত্রিকোণমিতিক ফাংশনের অন্তরজ নির্ণয় করা সম্ভব। নিম্নে কয়েকটি উদাহরণ লিপিবদ্ধ করা হলো:
[
tan
(
x
)
]
′
{\displaystyle \left[\tan(x)\right]'}
=
[
sin
(
x
)
cos
(
x
)
]
′
{\displaystyle =\left[{\frac {\sin(x)}{\cos(x)}}\right]'}
=
cos
(
x
)
⋅
(
sin
(
x
)
)
′
−
sin
(
x
)
⋅
(
cos
(
x
)
)
′
cos
2
(
x
)
{\displaystyle ={\frac {\cos(x)\cdot (\sin(x))'-\sin(x)\cdot (\cos(x))'}{\cos ^{2}(x)}}}
[দুইটি ফাংশনের ভাগফলের অন্তরজ নির্ণয়ের সূত্র প্রয়োগ করে]
=
cos
(
x
)
⋅
cos
(
x
)
+
sin
(
x
)
⋅
sin
(
x
)
cos
2
(
x
)
{\displaystyle ={\frac {\cos(x)\cdot \cos(x)+\sin(x)\cdot \sin(x)}{\cos ^{2}(x)}}}
=
cos
2
(
x
)
+
sin
2
(
x
)
cos
2
(
x
)
{\displaystyle ={\frac {\cos ^{2}(x)+\sin ^{2}(x)}{\cos ^{2}(x)}}}
=
1
cos
2
(
x
)
{\displaystyle ={\frac {1}{\cos ^{2}(x)}}}
=
sec
2
(
x
)
{\displaystyle =\sec ^{2}(x)}
[
cos
(
1
x
)
]
′
{\displaystyle \left[\cos \left({\frac {1}{x}}\right)\right]'}
=
cos
′
(
u
)
⋅
u
′
(
x
)
{\displaystyle =\cos '(u)\cdot u'(x)}
মনে করি,
u
(
x
)
=
1
x
{\displaystyle u(x)={\frac {1}{x}}}
=
−
sin
(
u
)
⋅
u
′
(
x
)
{\displaystyle =-\sin(u)\cdot u'(x)}
=
−
sin
(
1
x
)
⋅
(
1
x
)
′
{\displaystyle =-\sin \left({\frac {1}{x}}\right)\cdot \left({\frac {1}{x}}\right)'}
=
−
sin
(
1
x
)
(
−
1
x
2
)
{\displaystyle =-\sin \left({\frac {1}{x}}\right)\left({\frac {-1}{x^{2}}}\right)}
=
sin
(
x
−
1
)
x
2
{\displaystyle ={\frac {\sin(x^{-1})}{x^{2}}}}
মনে করি,
f
(
x
,
y
)
=
g
(
x
,
y
)
{\displaystyle f(x,y)=g(x,y)}
x
{\displaystyle x}
y
{\displaystyle y}
x
2
+
y
2
=
1
{\displaystyle x^{2}+y^{2}=1}
(
0
,
0
)
{\displaystyle (0,0)}
1
{\displaystyle 1}
y
{\displaystyle y}
x
{\displaystyle x}
y
{\displaystyle y}
y
=
±
1
−
x
2
{\displaystyle y=\pm {\sqrt {1-x^{2}}}}
x
∈
[
−
1
,
1
]
{\displaystyle x\in [-1,1]}
y
′
{\displaystyle y'}
x
{\displaystyle x}
y
{\displaystyle y}
y
′
{\displaystyle y'}
y
≥
0
{\displaystyle y\geq 0}
যদি
u
=
sin
2
(
x
)
{\displaystyle u=\sin ^{2}(x)}
v
=
cos
2
(
x
)
{\displaystyle v=\cos ^{2}(x)}
u
2
+
v
2
=
1
{\displaystyle u^{2}+v^{2}=1}
(
0
,
0
)
{\displaystyle (0,0)}
1
{\displaystyle 1}
x
{\displaystyle x}
x
2
+
y
2
=
1
{\displaystyle x^{2}+y^{2}=1}
x
{\displaystyle x}
y
{\displaystyle y}
h
{\displaystyle h}
x
2
+
y
2
=
h
2
{\displaystyle x^{2}+y^{2}=h^{2}}
⟹
(
x
h
)
2
+
(
y
h
)
2
=
1
{\displaystyle \implies \left({\frac {x}{h}}\right)^{2}+\left({\frac {y}{h}}\right)^{2}=1}
⟹
sin
2
(
θ
)
+
cos
2
(
θ
)
=
1
{\displaystyle \implies \sin ^{2}(\theta )+\cos ^{2}(\theta )=1}
θ
{\displaystyle \theta }
x
2
−
y
2
=
1
{\displaystyle x^{2}-y^{2}=1}
x
≥
1
{\displaystyle x\geq 1}
(
0
,
0
)
{\displaystyle (0,0)}
x
{\displaystyle x}
a
{\displaystyle a}
x
{\displaystyle x}
sinh
(
a
)
{\displaystyle \sinh(a)}
y
{\displaystyle y}
cosh
(
a
)
{\displaystyle \cosh(a)}
x
{\displaystyle x}
a
{\displaystyle a}
sinh
(
x
)
{\displaystyle \sinh(x)}
cosh
(
x
)
{\displaystyle \cosh(x)}
tanh
(
x
)
=
sinh
(
x
)
cosh
(
x
)
{\displaystyle \tanh(x)={\frac {\sinh(x)}{\cosh(x)}}}
cosh
2
(
x
)
−
sinh
2
(
x
)
=
1
{\displaystyle \cosh ^{2}(x)-\sinh ^{2}(x)=1}
অধিবৃত্তীয় ফাংশনের অন্তরীকরণের জন্য সর্বপ্রথম ফাংশনসমূহের জন্য একটি বীজগাণিতির সমীকরণের প্রয়োজন। এক্ষেত্রে, সূচকীয় সংজ্ঞায়নের মাধ্যমে পাই,
অধিবৃত্তীয় সাইন,
sinh
(
x
)
=
e
x
−
e
−
x
2
{\displaystyle \sinh(x)={\frac {e^{x}-e^{-x}}{2}}}
অধিবৃত্তীয় কোসাইন,
cosh
(
x
)
=
e
x
+
e
−
x
2
{\displaystyle \cosh(x)={\frac {e^{x}+e^{-x}}{2}}}
অধিবৃত্তীয় সাইনের অন্তরজ
(
sinh
(
x
)
)
′
=
cosh
(
x
)
{\displaystyle (\sinh(x))'=\cosh(x)}
প্রমাণ:
(
sinh
(
x
)
)
′
{\displaystyle (\sinh(x))'}
=
(
e
x
−
e
−
x
2
)
{\displaystyle =\left({\frac {e^{x}-e^{-x}}{2}}\right)}
=
1
2
[
(
e
x
)
′
−
(
e
−
x
)
′
]
{\displaystyle ={\frac {1}{2}}[(e^{x})'-(e^{-x})']}
=
1
2
[
e
x
−
(
−
x
)
′
⋅
(
e
−
x
)
′
(
−
x
)
]
{\displaystyle ={\frac {1}{2}}[e^{x}-(-x)'\cdot (e^{-x})'(-x)]}
=
1
2
[
e
x
+
e
−
x
]
{\displaystyle ={\frac {1}{2}}[e^{x}+e^{-x}]}
=
cosh
(
x
)
{\displaystyle =\cosh(x)}
অধিবৃত্তীয় কোসাইনের অন্তরজ
(
cosh
(
x
)
)
′
=
sinh
(
x
)
{\displaystyle (\cosh(x))'=\sinh(x)}
প্রমাণ:
(
cosh
(
x
)
)
′
{\displaystyle (\cosh(x))'}
=
(
e
x
+
e
−
x
2
)
{\displaystyle =\left({\frac {e^{x}+e^{-x}}{2}}\right)}
=
1
2
[
(
e
x
)
′
+
(
e
−
x
)
′
]
{\displaystyle ={\frac {1}{2}}[(e^{x})'+(e^{-x})']}
=
1
2
[
e
x
+
(
−
x
)
′
⋅
(
e
−
x
)
′
(
−
x
)
]
{\displaystyle ={\frac {1}{2}}[e^{x}+(-x)'\cdot (e^{-x})'(-x)]}
=
1
2
[
e
x
−
e
−
x
]
{\displaystyle ={\frac {1}{2}}[e^{x}-e^{-x}]}
=
sinh
(
x
)
{\displaystyle =\sinh(x)}
![{\displaystyle =\left[\lim _{\Delta x\to 0}{\sin(x)}\right]\cdot \left[\lim _{\Delta x\to 0}{\frac {\cos(\Delta x)-1}{\Delta x}}\right]+\left[\lim _{\Delta x\to 0}\cos(x)\right]\cdot \left[\lim _{\Delta x\to 0}{\frac {\sin(\Delta x)}{\Delta x}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/48edc918c7e2f7a37c3b24c645a5642d7e1373c8)
![{\displaystyle =\left[\lim _{\Delta x\to 0}{\sin(x)}\right]\cdot \left[0\right]+\left[\lim _{\Delta x\to 0}\cos(x)\cdot \right]\cdot \left[1\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1453015302cad3ab5b133ee927b7af3efc8ae515)
![{\displaystyle =\left[\lim _{\Delta x\to 0}\cos(x)\right]\cdot \left[\lim _{\Delta x\to 0}{\frac {\cos(\Delta x)-1}{\Delta x}}\right]-\left[\lim _{\Delta x\to 0}{\sin(x)}\right]\cdot \left[\lim _{\Delta x\to 0}{\frac {\sin(\Delta x)}{\Delta x}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a6dfb91e2011176897efbe3913f4662e3a319482)
![{\displaystyle =\left[\lim _{\Delta x\to 0}{\cos(x)}\right]\cdot \left[0\right]-\left[\lim _{\Delta x\to 0}\sin(x)\cdot \right]\cdot \left[1\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0ccb48e1f58c7f804df10f0711af5effdfef2908)
![{\displaystyle \left[\tan(x)\right]'}](https://wikimedia.org/api/rest_v1/media/math/render/svg/458046d66901f5aeea055fde5eed016ca2cfce1f)
![{\displaystyle =\left[{\frac {\sin(x)}{\cos(x)}}\right]'}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c1908f57b382fd4c4dc7b9aa52bf8b8ee1b1c927)
![{\displaystyle \left[\cos \left({\frac {1}{x}}\right)\right]'}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d846aa218e99806f655235260a724e88d4e92a33)
![{\displaystyle x\in [-1,1]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1f9d0dda56ce3e01e14570ac9aef0021c6125722)
![{\displaystyle ={\frac {1}{2}}[(e^{x})'-(e^{-x})']}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5fa4ed7ae82de001a1c4cb9213acbfa946fc5717)
![{\displaystyle ={\frac {1}{2}}[e^{x}-(-x)'\cdot (e^{-x})'(-x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e4adc15c3ad5a5ab42a261e2597ab652f9c17491)
![{\displaystyle ={\frac {1}{2}}[e^{x}+e^{-x}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/80b213900317ab301f99bba0effaeddd29f1b1c8)
![{\displaystyle ={\frac {1}{2}}[(e^{x})'+(e^{-x})']}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fdccf3b33c05274975e0a1f3fb9c463a88bd0972)
![{\displaystyle ={\frac {1}{2}}[e^{x}+(-x)'\cdot (e^{-x})'(-x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b6fa61cce899304ca73589f780cfb8e3c58c18b0)
![{\displaystyle ={\frac {1}{2}}[e^{x}-e^{-x}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c60be33969b4e58707e84b98712b60eefed24e9b)
