Download Basic Trigonometry

Basic Trigonometry.
Angles in trigonometry. How do we measure them?
http://www.sosmath.com/trig/Trig1/trig1/trig1.html
The unit of measurement for angles: radian, and degree.
If r is the radius of the circle above, the length of the string
s around it (from the point P to Q) is
s  r when  is expressed in radians. When  is 1
radian, the length of s is r .
Conventionally, we have  radians  180 (degrees)
Therefore,
2 radians = 360

radians = 90
2
etc.
The trigonometric functions.
Consider a right angled triangle below.
h
a

b
By Pythagoras theorem, we have
a 2  b2  h2
… (1)
We also define
a
h
b
cos( )  cos 
h
sin  a
tan 

cos b
cos b
cot  

sin  a
1
h
cosec =

sin  a
sin( )  sin  
… (2.1)
… (2.2)
… (2.3)
… (2.4)
… (2.5)
sec 
1
h

cos b
… (2.6)
Given these we see:
a. The | sin  | and | cos | are bounded above by 1.0
and bounded below by 0.0.
a 2 b2 a 2  b2
2
2
1
b. sin   cos   2  2 
2
h
h
h
c. 1  tan2   sec2 
d. 1  cot 2   cosec2
Some trigonometric results for specific angles.
Consider the triangle below.
2
1
45
1

1
4
2

1
cos 45  cos 
4
2
sin 45  sin

Now, on this equilateral triangle, we have
60
1
1
60
60
1
If we split this triangle into two equal parts, we would have
1
1
0.5
0.5
Assume length of the perpendicular drop = x
Then 1  x 2  0.52
sin 60  sin

3


 x 2 
3
2
3
3
i.e. x 
2
4
cos 60  cos

3

1
2
3
 1
and sin 30  sin 
6
2
6 2
The graphs of trigonometric functions.
Similarly, cos 30  cos


This is a plot of y  sin  for  2    2
Next is the plot of both y  sin  and y  cos over the
same domain.
Next is the plot of y  tan , again on the same domain
http://oolong.co.uk/trig.htm
To get values for trigonometric functions at some
“traditional” angles (in degrees), we construct a table in the
following way:
angles 0 30 45 60 90
sin
0 1 2 3 4
cos
4 3 2 1 0
Next divide each entry in the second and third rows by 4
angles 0 30 45 60 90
sin
0 1/4 1/2 3/4 1
cos
1 3/4 1/2 1/4 0
Next take square root of each element. Now the elements
correspond to the actual values of the functions at these
angles (expressed in degrees).
angles 0 30
45
60
90
Sin
Cos
0 1/2
1 / 2 3 /2 1
1 3 /2 1 / 2 1/2
0
Other identities.
a. sin(  x)   sin x
b. cos( x)  cos x
… (3.1)
… (3.2)
Note the asymmetry/symmetry of the two functions.
Sum/Difference identities.
a. sin(   )  sin  cos   cos sin 
b. sin(   )  sin  cos   cos sin 
c. cos(   )  cos cos   sin  sin 
d. cos(   )  cos cos   sin  sin 
… (4.1)
… (4.2)
… (4.3)
… (4.4)
How do we prove (4.1)? Consider the diagram below.
http://arc.iki.rssi.ru/mirrors/stern/stargaze/Strig5.htm
Here sin(   ) 
But
DF DE  EF DE  CB


R
R
R
DE
 cos Therefore, DE  DC cos . Also,
DC
DC
 sin  . Therefore, DC  R sin  . Thus,
R
DE  R sin  cos
Now,
CB
AC
 sin  . Thus, CB  AC sin  . But,
 cos 
AC
R
Therefore, CB  R cos  sin  .
Thus, sin(   )  sin  cos   cos sin 
(4.1)
How about proving (4.3)?
AF
 cos(   )
R
But AF  AB  FB  AB  EC
EC
 sin  . Thus,
CD
EC  CD sin   R sin  sin 
Similarly, AB  AC cos  R cos  cos
Therefore, cos(   )  cos cos   sin  sin  (4.3)
Given these, we can find out others.
Example.
1. Expand tan(x  y)
tan( x  y ) 
sin( x  y ) sin x cos y  cos x sin y

cos( x  y ) cos x cos y  sin x sin y
Divide both numerator and denominator by cos xcos y . We
get,
tan( x  y ) 
tan x  tan y
1  tan x tan y
… (5.1)
2. How do we get expansions of sin 2 x , cos 2 x ?
sin 2 x  sin( x  x)  sin x cos x  cos x sin x
 2 sin xcos x … (5.2a)
cos 2 x  cos( x  x)  cos x cos x  sin x sin x
 cos2 x  sin 2 x … (5.2b)
3. Compute the values of cos(   ), cos(   )
cos(   )  cos cos  sin  sin    cos
4. What is the value of sin 135 ?
135   

4



sin(  )  sin  cos  cos sin
4
4
4
1

2
Inverse functions.
If sin   x then the angle  in radian is
  sin 1 x
(6.1)
Similarly,   cos1 y 
 cos   y (6.2)
And,   tan1 z implies tan  z (6.3)
These inverse functions are also known as arcsin, arcos,
arctan, respectively.
Thus,   sin 1   arcsin    sec1 x  arc sec x , etc.
Ok. Given this definition, how do we add to inverse
functions which are basically angles?
We want to find the sum
sin 1 x  sin 1 y .
Let sin   x . This means cos  (1  x 2 ) . Similarly,
let sin   y implying cos   (1  y 2 ) .
Now, sin(   )  sin  cos   cos sin 
 x 1  y2  y 1  x2

Therefore,     sin 1 x 1  y 2  y 1  x 2

But the LHS is sin 1 x  sin 1 y . Therefore, we conclude,

sin 1 x  sin 1 y  sin 1 x 1  y 2  y 1  x 2

Examples.
sin  / 6  0.5 Therefore, sin 1 0.5 

6
Notice that there are many angles  for which sin  are
half. We choose the smallest angle within the positive
domain that would satisfy the result. If we know the angle
is going to be negative we choose the largest negative angle
on the given domain.
2. The shadow of a tree around afternoon appears to be 30
ft long with the sun at an angle of 300 to the horizon. What
is the height of the tree?
tree
shadow
Now, tan

6

1
tree

3 shadow
Length of the tree = length of the shadow  1/ 3
= 17.32ft