### Name of presentation

```TRIGONOMETRIC FUNCTIONS
OF ACUTE ANGLES
By
M. Jaya krishna Reddy
Mentor in mathematics,
India.
Acute Angle:
ACUTE
ANGLE
An angle whose measure is greater than zero but
less than 90 is called an “acute angle”
T
E
R
M
I
N
A
L
R
A
Y
o
Initial ray
sin  
opp

b
hyp
cos  
tan  

cos ec  

opp
c
a
hyp
sec  
hyp

a
opp
b
a
a
opp

c
c
c
cot  
A
b
hyp

c
b
b
Ѳ
C
a
Commonly used mnemonic for these ratios :
Some Old Houses
Can’t Always Hide
Their Old Age
B
History:
•Trigonometric functions(also called circular
functions) are functions of an angle.
•They are used to relate the angles of a
triangle to the lengths of the sides of a triangle.
•Sumerian astronomers introduced angle
measure, using a division of circles into 360
degrees.
•The sine function was first defined in the
“surya siddhanta” and its properties were
further documented by the fifth century Indian
mathematician and astronomer “Aryabhatta”.
•By 10th century the six trigonometric functions
were used.
Applications:
•In 240 B.C. a mathematician named “Eratosthenes”
discovered the radius of the earth as 4212.48 miles
using trigonometric functions..
•In 2001 a group of European astronomers did
an experiment by using trigonometric functions
and they got all the measurement, they
calculate the Venus was about 105,000,000
km away from the sun and the earth was about
150, 000, 000 km away.
•Optics and statics are 2 early fields of
Physics that use trigonometry.
•It is also the foundation of the practical art of
surveying
1. Prove that cos  . tan   sin 
Sol: c o s  . ta n   c o s  .
2. Prove that
cos 
 s in 
s in   c o s 
 s in  . c o s 
s e c   c o s e c
s in   c o s 
Sol:
s in 
s e c   c o s e c

s in   c o s 
1
cos 


1
s in 
s in   c o s 
s in   c o s 
s in  . c o s 
 sin  . co s 
Fundamental Relations:
opp
sin  

hyp
A
b
c
c
b
co s  

h yp
a
c
Ѳ
C
Squaring and adding both the equations
(s in  )
2
 (c o s  )
2
 b 


 c 
2
 a 
 

 c 
2
a
2
 b2  a2 
 c2 

   2   1
2
c


 c 
s in
B
  cos 2   1
From the above
diagram, By
Pythagorean Rule,
a
2
b
2
 c
2
A
tan  
opp
sec  

b
a
h yp
c

b
a
Squaring and subtracting the equations,
we get
(s e c  )
2
 (ta n  )
c
 c 


 a 
2
2
 b 
 

 a 
C
2
 c2  b2 
 a2 

   2  1
2
a


 a 
sec
2
Ѳ
  ta n 2   1
a
From the above
diagram, By
Pythagorean Rule,
a
Similarly,
co sec   cot   1
2
2
B
2
b
2
 c
2
  cos 2   1
2
2
s e c   ta n   1
2
2
co sec   cot   1
s in
2
sec   1
2
Example: Prove that
sec 
 sin 
sec   1
2
sol: G iven
that
sec 
tan 
2


sec 
sin 
cos 


tan 
sec 
cos 
1
 sin 
Ex: Prove that sec2Ѳ - cosec2Ѳ = tan2Ѳ - cot2Ѳ
Sol: We know that sec2Ѳ - tan2Ѳ = 1 = cosec2Ѳ - cot2Ѳ
sec2Ѳ - tan2Ѳ = cosec2Ѳ - cot2Ѳ
sec2Ѳ - cosec2Ѳ = tan2Ѳ - cot2Ѳ
co s B  co s A
2
Example: Prove that tan A  tan B 
2
2
2
2
2
co s A . co s B
2
2
Sol: Given that tan A  tan B
 sec A  1  (sec B  1)
2
2
 sec A  sec B
2

2
1
2
co s A

1
2
co s B
co s B  co s A
2

2
2
2
co s A . co s B
Values of the trigonometrical ratios :
Sin
00
300
450
600
900
0
1
1
3
1
2
Cos
Tan
1
0
3
2
1
2
1
2
1
3
Cosec
∞
2
2
2
1
0
2
3
2
∞
1
3
Sec
1
2
3
Cot
∞
3
2
1
2
∞
1
0
3
Example: find the value of tan450.sec300 - cot900.cosec450
Sol: Given that tan450.sec300 -- cot900.cosec450
=1.
2
-- 0.
2 =
3
2
3
Example: If cosѲ = 3/5, find the value of the other ratios
Sol: Given that cosѲ = 3/5 = adj / hyp
thus using reference triangle adj = 3, hyp = 5,by Pythagorean
principle opp = 4
opp 4
hyp 5
sin  

sec  

hyp 5
5
cos ec  
tan  
hyp

5
opp
4
opp
4

cot  