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Physic notes from
Chapter 1 to 25
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Physics HandBook CH APTER
TRI GONOMETRY
2 radian = 360° 1 rad = 57.3°
perpendic ular base perpendicular
sin cos = tan =
hypoten use hypoten use base
a 2+b 2
base hypotenus e hypoten use a
cot = sec = cosec =
perpendi cular base perpendi cular
a b a b
sin = cos = tan =
a 2 b2 2
a b 2 b
1 1 1
cosec = sec = cot =
sin cos tan
sin 2 + cos 2 = 1 1 + tan 2 = sec 2 1 + cot 2 = cosec 2
90°
sin(A±B) = sinAcosB cosAsinB cos(A±B) = cosAcosB sinAsinB II I
tan A tan B S in All
tan A B sin2A = 2sinAcosA 0°
1 tan A tan B 180° 360°
cos2A = cos 2 A–sin 2 A = 1–2sin 2 A = 2c os 2 A–1 T an C os
2 tan A III IV
tan2 A sin3 = 3sin – 4sin 3
1 tan 2 A EN 270°
cos3 = 4c os 3 – 3c os 2sinAsinB = cos(A–B) – cos(A+B)
2cosAcosB = cos(A–B) + cos(A+B) 2sinAcosB = sin(A+B) + sin(A–B)
0° 30° 45° 60° 90° 120° 135° 150° 180° 270° 360°
(0) /6) /4) /3) /2) /3) /4) /6) /2) )
1 1 3 3 1 1
LL
sin 0 1 0 1 0
2 2 2 2 2 2
3 1 1 1 1 3
cos 1 0 -1 0 1
2 2 2 2 2 2
1 1
tan 0 1 3 3 -1 0 0
3 3
A
sin (90 ° + ) = c os sin (180° – ) = sin sin (– = –sin sin (90° – ) = cos
cos (90°+ ) = –sin cos (180° – ) =– cos cos (–) = cos cs 90° – ) = sin
tan (90°+ ) =– cot tan(180°– ) =– tan tan (–) = – tan tan(90°– ) = cot
sin (18 0° + ) = – sin sin (270°– ) = – cos sin (27 0°+ ) = – cos sin (360°– ) = – sin
cos (180° + ) = – cos cos(270° – )= – sin cos (270° + ) = sin cos (360° – ) = cos
tan (180° + ) = tan tan (270° – ) = cot tan (270° + ) = – cot tan (360° – ) = – tan
A
sine law For smal l
A sin cos 1 tan sin tan
c b
sin A sinB sinC
B C
a b c
B C
a
cosine law
2 2 2
b +c -a c 2 + a 2 - b2 a 2 + b 2 - c2
cos A , cosB , cos C
2bc 2ca 2a b
2 E
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CHAPTER
Physics HandBook
Differentiat ion Maxima & Minima of a function y=f(x)
dy dy 1
• yx
n
nx n 1 • y nx dy d2 y
dx dx x • For maximum value 0& ve
dx dx 2
dy dy
• y sin x cos x • y cos x sin x dy d2 y
dx dx • For minimum value 0& ve
dx dx 2
x dy dy dv du
• ye e x • y uv u v
dx dx dx dx
Average o f a varying quantity
dy df g x d g x
• y f g x x2 x2
dx dg x dx
dy
ydx
x1
ydx
x1
• y=k(const ant) 0 If y = f(x) then y y x2
dx x 2 x1
du dv
dx
x1
v u
• y u dy dx dx
v dx v2
Integration EN
C = Arbitrary constant, k = constant
• f(x)dx g(x) C
d
• (g(x)) f(x)
dx
• kf(x)dx k f(x)dx
LL
• (u v w)dx udx vdx wdx
• e x dx e x C
x n 1
• x n dx C,n 1
A
n 1
1
• dx nx C
x
• sin xdx cos x C
• cos xdx sin x C
1 x
• e x dx e C
n x n 1
• x dx n 1
C
Definite integration
b
b
f(x)dx
a
g(x)a g(b) g(a)
Area under the curve y = f(x) from x =a to a = b is
b
A f(x)dx
a
E 3
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Physics HandBook CH APTER
FORMULAE FOR DETERMINATION OF AREA
FO RMULAE FOR
• Area of a square = (side) 2 DETERMINATION OF
• Area of rectangle = length ×breadth VOLUME
1
• Area of a triangle = ×base × height t
2
• Area of a trapezoid a
1
= × (distance between parallel sides) × (sum of parallel sides) b
2
• Area enclosed by a circle = r2 (r = radius) • Volume of a rectangular
• Surface area of a sphere = 4 r2 (r = radius) slab
• Area of a parallelogram = base × height = length × breadth × height
• Area of curved surface of cylinder = r = abt
where r = radius and = length
• Volume of a cube = (side) 3
• Area of whole surface of cylinder = 2 r (r + ) where = length
• Area of ellipse = ab 4 3
• Volume of a sphere = r
(a & b are semi major and semi minor axis respectively) 3
• Surface area of a cube = 6(side) 2 EN (r = radius)
• Tot al surface area of a cone = r2 + r • Volume of a cylinder = r2
(r = radius and = length)
where r = r r 2 h 2 = lateral area
• Arc length s = r. r
1
• Volume of a cone = r2 h
3
r2
• Area of sector = s (r = radius and h = height)
2
LL
s r
• Plane angle, radian
r
A
• Solid angle, steradian
r2
A
• To convert an angle from degree to radian, we have to multiply it by and to convert an angle
180
180
from radian to degree, we have to multiply it by .
dy dy
KEY POINTS
• By help of differentiation, if y is given, we can find and by help of integration, if is given,
dx dx
we can find y.
• The maximum and minimum values of function
A cos B sin are A2 B2 and A2 B2 respectively.
• (a+b)2 = a 2 + b 2 + 2ab (a–b)2 = a 2 + b 2 – 2ab
(a+b) (a–b) = a 2 – b 2 (a+b)3 = a 3 + b 3 + 3ab (a+b)
(a–b)3 = a 3 – b 3 – 3ab (a–b)
4 E