A 
c b
B a C
The hypotenuse
is the side opposite to the 90 degree angle in a right angle triangle; it is
the longest side of the triangle. The perpendicular is the other side
that is adjacent to angle A.
The base is the side that is opposite to angle A.
9.2 Angles: when
a ray OA starting from its original position, rotates about its fixed point O,
to a final position OB, then an angle AOB is formed. Point O is called the
Vertex of the angle.
B
O A
9.2.1 Measure of the angle: It is defined as the amount of rotation
from initial side to final side.
9.2.1.1 Degree measure (Sexagesimal System): The number of degrees on the
circumference of circle between initial and final sides of the angle is called
its degree measure. Each degree is divided into 60 equal parts called minutes
and each minute is divided into another 60 equal parts named as seconds.
Degree
is denoted by the symbol (o),
minute is denoted by the symbol (‘)
and second is denoted by the symbol (“).
Example: 35o
34’ 23’’ means 35 degrees 34 minutes and 23 seconds.
Types of angles: For an angle
θ
Acute
angle:- 0o ≤ θ <90o
Obtuse angle:
90o < θ <180o
Reflex
angle: 180o < θ <360o
Right
angle: θ = 90o
Straight
angle: θ = 180o
9.2.1.2 Radian measure:
It is the measure of an angle subtended at the centre of a circle of radius r by an arc of the length r. To convert from degrees to radians, multiply by (π/180o). To convert from radians
to degrees, multiply by (180o/π).
B
1c
r
O
r
A
9.2.1.3 Grade
measure: Each
right angle is divided into 100 equal parts known as grades. Each grade is
subdivided into 100 equal parts called as minutes. Each minute is subdivided
into 100 equal parts known as seconds.
9.3 Trigonometric
functions:
i) SinA = Perpendicular = b
Hypotenuse c
ii) CosA
= Base = a
Hypotenuse c
iii) TanA
= Perpendicular
= b
Base a
iv) CosecA= Hypotenuse
= c
Perpendicular b
v) SecA =
Hypotenuse = c
Base
a
vi) CotA = Base =
a
Perpendicular b
POINTS TO REMEMBER:
For any angle θ
Cosecθ = 1 / Sinθ
Secθ = 1/ Cosθ
Cotθ = 1/Tanθ
9.4 Trigonometric identities and values:
i)
sin2θ
+ cos2θ = 1
ii)
1
+ tan2θ = Sec2θ
iii)
1
+ cot2θ = cosec2θ
iv)
tanθ
= sinθ/ cosθ
v)
cotθ
= cosθ/sinθ
vi)
tanθ.cotθ
=1
|
Ratio
θ |
0o |
30o |
45o |
60o |
90o |
|
Sinθ |
0 |
1/
2 |
1/√2 |
√3/2 |
1 |
|
Cosθ |
1 |
√3/2 |
1/√2 |
1/
2 |
0 |
|
Tanθ |
0 |
1/√3 |
1 |
√3 |
Not
defined |
|
Cosecθ |
Not
defined |
2 |
√2 |
2/√3 |
1 |
|
Secθ |
1 |
2/√3 |
√2 |
2 |
Not
defined |
|
Cotθ |
Not
defined |
√3 |
1 |
1/√3 |
0 |
+ +
Only
Sinθ and Cosecθ is +ve
All are +ve
IInd quadrant Ist
quadrant
- +
- +
IIIrd
quadrant IVth
quadrant
Only
Tanθ and Cotθ are +ve Only Cosθ and Secθ are +ve
- -
9.5 Periodic function: A function f(x) is called periodic if there is a non
zero number p such that f(x+p) is
defined and f(x+p) = f(x) for all values of x in the domain of f(x). The number p is called the period of
f(x).
The
function sinx has periods 2p,
4p,6p,....since sin (x+2p), sin (x+4p), sin (x+6p),....all equal sin x.
However, 2p is the least period or the period of sinx.
The
function cosx has periods 2p, 4p, 6p, ....... since cos(x+2p), cos(x+4p),
cos(x+6p), ....... all equal cosx.
Here also 2p is the period, i.e., the least period or the period of cos x.
The
period of tanx is p.
9.6 Even and Odd functions: A function f(x)
is said to be even if f(-x) = f(x)
Example: f(x) = x2,
f(-x) = x2
f(x) = x2 is
even
A
function f(x) is said to be odd if
f(-x) = -f(x)
e.g., f(x) = x3, f(-x) = (-x)3 = -x3 = -f(x)
f(x) = cos x is even for f(-x) = cos
(-x) = cos q = f(x)
f(x) = x
cos x is odd for f(-x) = (-x) cos (-x) = -x cos x = -f(x)
f(x) = sin x is odd whereas f(x) = x sin x is even.
9.7 Tangent
function
The
figure is a unit circle, with origin O as centre cuts the x-axis at A (1, 0) and let a variable point moving on the
circumference move through an arc length q. i.e., AP = p (q). The coordinates
at the position of p (q) are p(x, y)
= (cos q, sin q).
Then the
tangent function is defined in the form as
Tanθ
= y/x where x ≠ 0
Tanθ
= sinθ/ cosθ, where cosθ ≠ 0
X = 0
or cosθ = 0 for θ = (nπ + π/2), where n is an integer.
Therefore tanθ is defined for all θ€R except
θ = (nπ + π/2), n€ Z
Therefore,
the domain of the tanθ is R’ = R – (nπ+π/2); n€ Z
Theorem1. Sin2θ
+ Cos2θ = 1
Proof: Let X’OX and Y’OY be the coordinate axes and O
being the centre. Draw a ircle of unit radius cutting OX at A. Let a moving
point starts from A, moving along the circumference of the circle. Let its
final position be P (x, y), and arc
AP=θ.
We know
that the equation of the circle is x2
+ y2 = 1
Since
the point P (Cosθ, Sinθ) lies on it, therefore
Cos2θ + Sin2θ
= 1
Theorem3. Let
XOX' and YOY' be the rectangular coordinate axes. Taking O as centre and radius
= 1, draw a circle to cut x-axis at A
and A' and y-axis at B and B'. Lethe rotating line start from OA revolving in
anticlockwise direction and taking the final position p (x,y) so that arc AP = θ.
On
the other hand, if the point starts from A and moves in clockwise direction
through arc length AP’ equal arc length AP, then
AP’ =
-θ
therefore,
AOP = θ and AOP’ = θ
cosθ
P
P(x,
y) SinθSinθ
P
X’ X
X X’
S
Cos-θ
P’(x,
y) Sin(-θ)
P’
Join P
and P'. Triangles POM and P'OM are congruent.
\ PM =
MP'. If PM = y then P'M = -y.
From the
DPOM, sin q = y and cos q = x
In D
P'OM, sin (-q) = -y and cos (-q) = x
sin(-θ) = -sin θ
and
cos(-θ) = cos θ
Theorem4. Let x and y be any two real numbers and let P(x)and Q(y) be the corresponding trigonometric points on the unit
circle. In the above figures we have taken x
and y so that
π/2 < y< x <π
The
coordinates of P(x) are (cosx, sin x) and that of Q(y) are (cos y, sin y) by the definitions of cosine
and sine functions.
Now
choose a point R on the unit circle so that arc AR has a measure of (x-y)
units. Then the trigonometric point with respect to (x-y) is R(x-y) and the
corresponding coordinates of R are (cos(x-y),sin(x-y)). The arc length of AR is the same
as arc PQ and hence the chord lengths of PQ and AR are same.
i)
|AR|
= distance from R to A.
Using
distance formula
[d2
= (x1 – x2)2 + (y1
– y2)2]
|AR|2
= [cos(x – y) – 1]2+ [sin(x – y) – 0]2
= cos2(x – y) – 2 cos(x – y) + 1
+ sin2 (x – y)
=
[cos2 (x – y) + sin2
(x – y)] – 2 cos(x – y) + 1
= 1 – 2 cos(x – y) +1
= 2 – 2cos(x – y) …..(i)
|PQ|2 = (cosx – cosy)2 + (sinx – siny)2
=
cos2x – 2cosxcosy+cos2y +sin2x +2sinxsiny + sin2y
= (cos2x + sin2x) + (cos2y + sin2y)
– 2[cosxcosy + sinxsiny]
= 1 + 1 - 2[cosxcosy + sinxsiny]
= 2 - 2[cosxcosy + sinxsiny] ….(ii)
Since AR
= PQ, therefore,
2
– 2cos(x – y) = 2 -2[cosxcosy + sinxsiny]
-2
cos(x – y) = -2[cosxcosy + sinxsiny]
Therefore,
cos(x
– y) = cosxcosy +sinxsiny
ii)
cos(x+y)
= cos[x – (-y)]
= (cosx) {cos(-y)} + (sinx)
{sin(-y)}
= cosxcosy + [-sinxsiny] [since cos(-x) = cosx & sin(-y) = -siny]
= cosxcosy
- sinxsiny
Theorem5. Proof: i) cos( π/2 – x) = sinx
cos (π/2 – x) = cos (π/2)cosx +
sin(π/2)sinx
= 0 +(1) sinx
= sinx
ii)
sin(π/2
– x) = cosx
sin(π/2
– x) = cos[π/2 – (π/2 – x)] [since cos(π/2 – θ) = sinθ]
= cosx
iii)
cos(π/2 +x) = -sinx
cos(π/2
+ x) = cos π/2cosx - sin π/2sinx
=
0 – (1) sinx
=
- sinx
iv)
sin(π/2
+ x) = cosx
sin(π/2
+ x) = sin[π/2 – (-x)]
=
cos(-x)
= cosx
Theorem 8
For
all real values of x.
i)
cos(π – x) = -cosx
Proof: cos(π
– x) = cosπ cosx + sinπ sinx
= (-1)cosx + 0 sinx
= -
cosx
ii)
sin(π – x) = sinx
Proof: sin(π
– x) = sin[(π/2) + (π/2) – x]
= sin[(π/2) + (π/2 – x)]
= cos(π/2 – x)
= sinx
iii)
cos(π+x) = -cosx
Proof: cos(π+x) = cosπcosx – sinπsinx
= (-1)cosx
– (0)sinx
= -cosx
iv)
sin(π + x) = -sinx
Proof: sin(π + x) = sin [(π/2) + (π/2 + x)]
= cos(π/2 +x)
= -
sinx
Theorem 9
For
real values of x
i)
cos(2π+x) = cosx
Proof: cos(2π+x) = cos2πcosx – sin2πsinx
= (1)cosx
– (0)sinx
= cosx
ii)
sin(2π+ x) = sinx
Proof : sin(2π+
x) = sin [π + (π + x)]
= - [-sin(π+x)]
= - [-sinx]
= sinx
Theorem 10
i)
sin(x+y) =sinxcosy + cosxsiny
Proof: sin(
x+y) = cos[π/2 – (x +y)]
= cos[(π/2 – x) – y]
= cos(π/2 – x) cosy + sin(π/2 – x)
siny
= sinxcosy + cosxsiny
ii)
sin(x – y) = sinxcosy - cosxsiny
Proof: sin(x – y) = sin[x + (-y)]
= sinxcos(-y)
+ cosxsin(-y)
= sinxcosy – cosxsiny
Theorem 11
i)
sin2x = 2sinxcosx
Proof: Consider
sin(x+y) = sinxcosy + cosxsiny
Now
put x = y, then
Sin(x + x) = sinxcosx
+ cosxsinx
Sin2x = 2 sinx
cosx
ii)
cos2x = cos2x –
sin2x = 2cos2x – 1 = 1 – 2sin2x
Proof; take,
cos(x+y) = cosxcosy – sinxsiny
now
place x = y,
cos(x+x) = cosxcosx
– sinxsinx
= cos2x – sin2x
= cos2x – (1 - cos2x)
= 2cos2x – 1
= 2(1 – sin2x) – 1
= 1 -
2 sin2x
iii)
sin3x = 3sinx – 4 sin3x
Proof: we
can write,
Sin3x = sin(2x
+ x)
= sin2xcosx
+ cos2xsinx
= (2sinxcosx)
cosx + (1 – 2sin2x) sinx
= 2sinx cos2x + sinx – 2 sin3x
= 2sinx – 2sin3x + sinx – 2 sin3x
= 3sinx – 4sin3x
iv)
cos3x = 4 cos3x –
3cosx
Proof: cos3x = cos(2x
+ x)
= cos2xcosx
– sin2xsinx
= (2cos2x – 1) cosx – 2sinxcosxsinx
= 2 cos3x – cosx – 2cosxsin2x
= 2 cos3x – cosx – 2cosx(1 - cos2x)
= 2cos3x – cosx – 2cosx + 2cos3x
= 4cos3x – 3cosx
Theorem 12
i)
2sinx cosy = sin(x + y) +
sin(x – y)
ii)
2cosx siny = sin(x + y) –
sin(x – y)
iii)
2cosx cosy = cos(x + y) +
cos(x – y)
iv)
2sinx siny = cos(x – y) – cos(x + y)
Proof:
The
domain of the tangent function is R’ = R - nπ +π/2 : nϵ Z
Adding
(1) and (2)
sin(x+y) + sin(x – y) = 2 sinx cosy ….(i)
Subtracting
2 from 1
sin(x+y) - sin(x – y) = 2 cosx siny ….(ii)
Consider
cos(x+y) = cosx cosy – sinx siny …..(3)
And,
cos(x – y) = cosxcosy +sinxsiny ….(4)
Adding
(3) & (4)
cos(x + y) + cos(x – y) = 2cosxcosy ….(iii)
Subtracting
(4) from (3)
cos(x + y) – cos(x – y) = -
2sinxsiny
Or
cos(x – y) – cos(x + y) = 2sinxsiny …(iv)
i)
1
– cosx = 2sin2(x/2)
Proof: Consider
cosx = cos(x/2
+ x/2) = cos [2(x/2)]
=
cos2 (x/2) – sin2(x/2)
= 1 – sin2(x/2) – sin2(x/2)
cosx = 1 – 2sin2(x/2)
1
– cosx = 2sin2(x/2)
ii)
1 + cosx = 2cos2(x/2)
Proof: cosx
= cos(x/2 + x/2) = 2cos[2(x/2)]
= cos2 (x/2) – sin2(x/2)
= cos2(x/2) – [1 – cos2(x/2)]
cosx = 2 cos2(x/2) – 1
1 + cosx = 2 cos2(x/2)
Theorem 14
i)
sinx + siny = 2 sin (x + y)
cos(x – y)
2 2
ii)
sinx + siny = 2 cos (x + y)
sin(x – y)
2 2
iii)
cosx + cosy = 2 cos(x+y)sin(x – y)
2 2
iv)
cosx - cosy = - 2 sin(x+y)sin(x – y)
2 2
Proof:
Sin(A+B) = sinAcosB +cosAsinB ….(a)
Sin(A – B) = sinAcosB – cosAsinB ….(b)
Adding
(a) and(b)
Sin(A+B) +sin(A – B) = 2sinAcosB
Let
A+B = x and A – B = y
Then,
A
= x + y and
B = x – y
2 2
Therefore,
sinx + siny = 2 sin (x + y) cos(x – y) ….(i)
2 2
Subtracting
(b) from (a)
sin(A + B) – sin(A – B) = 2cosAsinB
sinx
+ siny = 2 cos (x + y) sin(x – y) ….(ii)
2 2
Now
take,
cos(A + B) = cosA cosB – sinA sinB ….(c)
and,
cos(A – B) = cosA cosB + sinA sinB …(d)
adding ( c) and (d)
cos(A + B) + cos(A – B) = 2 cosA cosB
cosx + cosy = 2 cos(x+y)sin(x – y) ….(iii)
2 2
Subtracting (d) from ( c)
cos(A + B) – cos(A – B) =
-2sinAsinB
cosx - cosy = - 2 sin(x+y)sin(x – y) ….(iv)
2 2
Theorem 15
i)
Sin(x +y) sin(x – y) = sin2x – sin2y
Proof:
sin(x + y)sin(x – y)
= (sinx cosy + siny cosx)(sinx cosy – cosx siny)
= sin2x cos2y – sin2y
cos2x
= sin2x (1 – sin2y) – sin2y(1
– sin2x)
= sin2x – sin2x sin2y – sin2y + sin2x sin2y
= sin2x – sin2y
ii)
cos(x + y) cos(x – y) = cos2x – cos2y
Proof:
cos(x+y) cos(x – y)
= (cosx cosy – sinx siny) (cosx cosy + sinx siny)
= cos2x cos2y – sin2x sin2y
= cos2x (1 – sin2y) – sin2y(1
– cos2x)
= cos2x – cos2x sin2y – sin2y + sin2y cos2x
= cos2x – cos2y
9.8 Trigonometric Formulae:
9.8.1 Sum formulae
For any
angle A and B
Sin
(A+B) = sinAcosB + cosAsinB
Cos
(A+B) = cosAcosB – sinAsinB
Tan (A+B) = tanA + tanB
1 – tanAtanB
9.8.2 Difference
formulae:
For any
angle A and B
Sin (A –
B) = sinAcosB – cosAsinB
Cos (A –
B) = cosAcosB + sinAsinB
Tan
(A – B) = tanA – tanB
1 + tanAtanB
For any
angle A and B
Sin
(A+B)sin (A – B) = (sin2A – sin2B)
Cos
(A+B)cos (A – B) = (cos2A – sin2B)
9.8.3 A,
B formulae:
For any
angles A and B
2 sinA
cosB = sin (A+B) + sin (A-B)
2 cosA
cosB = sin (A+B) – sin (A-B)
2 cosA
cosB = cos (A+B) + cos (A-B)
2 sinA
sinB = cos (A-B) – cos (A+B)
9.8.4 C, D formulae:
For any
angles C and D
sinC + sinD = 2sin(C+D) cos(C+D)
2 2
sinC – sinD = 2cos(C+D) sin(C+D)
2 2
cosC + cosD = 2sin(C+D) sin(C+D)
2
2
9.9 T- Ratios of multiple and sub-multiple
angles:
i) sin2A = 2sinAcosA
ii) cos2A
= cos2A – sin2A = 1 – 2sin2A = 2cos2A
– 1
iii) tan2A = 2tanA
1
– tan2A
iv) 1
– cos2A = 2sin2A
v) 1 + cos2A = 2cos2A
vi) sin2A = 2tanA
1+
tan2A
vii) cos2A = 1 - tan2A
1+ tan2A
viii) sin3A = (3sinA – 4sin3A)
ix) cos3A = (4cos3A – 3cosA)
x) tan3A = 3tanA – tan3A
1 –3 tan2A
9.9.1 T-Ratios of sub-multiple angles
Replacing
A by A/2 in above results.
i)
sinA
= 2sin(A/2) cos(A/2)
ii)
cosA
= (cos2(A/2) – sin2(A/2))
iii)
1
– cosA = 2sin2(A/2)
iv)
1
+ cosA = 2cos2(A/2)
v)
tanA = 2tan(A/2)
1 – tan2(A/2)
vi)
sinA = 2tan(A/2)
1 + tan2(A/2)
vii)
cosA = 1 – tan2(A/2)
1 + tan2(A/2)
9.9.2 T-Ratios of some special angles
i)
Sin22 ½
Solution: sin2A
= (1-cos2A)/2
Sin2 22 ½ = (1 –
cos45o)/2
= (1 – 1/√2)/2
= (√2 – 1)/2√2
Sin 22 ½ =
(√2 – 1)/2√2
ii)
Cos22 ½
Solution: cos2A = (1 + cos2A)/2
cos222 ½ = ( 1+ cos 45o)/2
= (1 +
1/√2)/2
= (√2 + 1)/2√2
cos
22 ½ =
(√2 +1)/2√2
iii)
Tan22 ½
Solution: tan22
½ =
sin22 ½
cos22 ½
put the values of sin and cos as
obtained above and solve.
Tan22 ½ = (√2 – 1)
iv)
Sin7 ½
Solution: sin2A
= (1 – cos2A)/2
sin27
½ = (1 – cos15o)/2 = 1 -
√3 + 1 = 4 - √6 -
√2
2
4√2 8
Sin7
½ = 4 - √6 - √2
8
Similarly we can solve cos7 ½ .
v) Tan11
½
Solution:
tanA = 2 tan(A/2)
1 – tan2(A/2)
Tan22 ½ = 2 tan11 ¼
1
– tan211 ¼
Take tan11 ¼ = x
Therefore,
2x = (√2 – 1)
1 – x2
(√2 – 1)x2 + 2x
– (√2 – 1) = 0
x =
- 2 ± 4 + 4(√2 – 1)2 = - 1±
( 4 – 2 √2)
2(√2 – 1) (√2 – 1)
So,
Tan11
¼ = - 1± ( 4 – 2 √2) × (√2 + 1)
(√2 – 1) (√2
+ 1)
Tan11 ¼ = - (√2 + 1) + (4+2√2)
9.10 Trigonometric or Conditional Identities:
i)
If A+B+C = π
Then,
A+B = π – C, B+C = π – A , A+C = π – B
So,
Sin
(A+B) = sin (π – C) = sinC
Cos
(A+B) = cos (π – C) = - cosC
Tan
(A+B) = tan (π – C) = - tanC
ii) If (A/2) + (B/2) + (C/2) = π/2
Then,
(A/2)+(B/2) =( π/2 –
C/2), (B/2)+(C/2) =(π/2) –(A/2)
, (A/2)+(C/2) =( π/2) – (B/2)
So,
Sin (A+B)/2 = sin
(π/2 – C/2) = cos(C/2)
Cos
(A+B)/2 = cos (π/2 – C/2) = sin (C/2)
Tan
(A+B)/2 = tan (π/2 – C/2) = cot (C/2)