An introduction to TRIGONOMETRY

Definitions and basics

Trigonometric circle and angles

Take an x-axis and an y-axis (orthonormal) and let O be the origin.
A circle centered in O and with radius = 1, is called a trigonometric circle or unit circle.
Turning counterclockwise is the positive orientation in trigonometry.
Angles are measured starting from the x-axis.
Two units to measure an angle are degrees and radians
An orthogonal angle = 90 degrees = pi/2 radians
In this theory we use mainly radians.
With each real number t corresponds just one angle, and just one point P on the unit circle, when we start measuring on the x-axis. We call that point the image point of t.

Examples:

Trigonometric numbers of a real number t

With t radians corresponds exactly one point p on the unit circle.
The line with equation sin(t).x - cos(t).y = 0
contains the origin and point P(cos(t),sin(t)). So this line is OP.
On this line we take the intersection point S(1,?) with the line x = 1.
It is easy to see that ? = tan(t).
So tan(t) is the y-coordinate of the point S.

Analogous cotan(t) is the x-coordinate of the intersection point S' of the line OP with the line y = 1.

Basic formulas

With t radians corresponds exactly one point p(cos(t),sin(t)) on the unit circle. The square of the distance [OP] = 1. Calculating this distance with the coordinates of P we have for each t :
 
cos2(t) + sin2(t) = 1

                    sin2(t)
1 + tan2(t) = 1 + ----------
                    cos2(t)

             cos2(t)+sin2(t)
           = -----------------
                 cos2(t)


                  1
           = ----------- = sec2(t)
               cos2(t)

Analogous :

1 + cotan2(t) = 1/ sin2(t) = csc2(t)

 
       cos2(t) + sin2(t) = 1

       1 + tan2(t) = sec2(t)

       1 + cotan2(t) = csc2(t)

Related values

supplementary values

t and t' are supplementary values <=> t+t' = pi.

With the help of a unit circle we see that the corresponding image points are symmetric with respect to the Y-axis. Hence, we have :
If t and t' are supplementary values then 
 
       sin(t) = sin(t')
 
       cos(t) = -cos(t')
 
       tan(t) = -tan(t')
 
       cot(t) = -cot(t')

complementary values

t and t' are complementary values <=> t+t' = pi/2.

The corresponding image points on a unit circle are symmetric with respect to the line y = x . Hence, we have :
If t and t' are complementary values then 
 
       sin(t) = cos(t')
 
       cos(t) = sin(t')
 
       tan(t) = cot(t')
 
       cot(t) = tan(t')

Opposite values

t and t' are opposite values <=> t+t' = 0.

Now, the corresponding image points are symmetric with respect to the X-axis. Hence, we have :
If t and t' are opposite values then 
 
       sin(t) = -sin(t')
 
       cos(t) =  cos(t')
 
       tan(t) = -tan(t')
 
       cot(t) = -cot(t')

Anti supplementary values

t and t' are anti supplementary values <=> t-t' = pi.

The corresponding image points are symmetric with respect to the origin O . Hence, we have :
If t and t' are anti-supplementary values then 
 
       sin(t) = -sin(t')
 
       cos(t) = -cos(t')
 
       tan(t) =  tan(t')
 
       cot(t) =  cot(t')

The right-angled triangle

Say the angle A is the right angle of the triangle ABC. The distances |AB|, |BC| and |CA| are usualy denoted by c, a and b. Take point B in a suitable way as center of a trigonometric circle (see figure).

Now sin(B),cos(B) and 1 are directly propertional with b, c and a. 

 
        sin(B)   cos(B)    1
        ------ = ------ = ---
          b        c       a

=>      sin(B) = b/a   cos(B) = c/a  tan(B) = b/c

and since the angles B and C are complementary angles

        cos(C) = b/a   sin(C) = c/a   tan(C) = c/b
In each right-angled triangle ABC, with A as right angle, we have 
 
        sin(B) = b/a   cos(B) = c/a  tan(B) = b/c

        cos(C) = b/a   sin(C) = c/a   tan(C) = c/b

Area of a triangle

The area of the triangle is a.h/2 .
But in triangle BAH, we have sin(B) = h/c .
Hence the area of the triangle is a.c.sin(B)/2.
Similarly we have that the area of the triangle
= b.c.sin(A)/2 = a.b.sin(C)/2

The area of a triangle ABC = a.c.sin(B)/2 = b.c.sin(A)/2 = a.b.sin(C)/2

Sine rule

In a triangle ABC we have seen that the area = 
 
        a.c.sin(B)/2 =  b.c.sin(A)/2 = a.b.sin(C)/2
=>
        a.c.sin(B) =  b.c.sin(A) = a.b.sin(C)

dividing through by  a.b.c, we get
In any triangle ABC we have 
 
          a         b        c
        ------ =  ------ = ------
        sin(A)    sin(B)   sin(C)

This formule is called the sine rule in a triangle ABC. Say R is the radius of the circle with center O through the points A,B and C. Let B' be the second intersection point of BO and the circle. The angle B' in triangle BB'C is equal to A. In the right-angled triangle BB'C we see that a = 2R sin(B') = 2R sin(A). Thus, The fractions in the sinus rule are all equal to 2R.

Homogeneous expression in a, b and c

If an expression between the sides of a triangle is homogeneous in a, b and c, we have an equivalent expression by replacing a,b and c with sin(A), sin(B), sin(C).

Example:
In a triangle 

 
        b.sin(A-C) = 3.c.cos(A+C)
<=>
        sin(B).sin(A-C) = 3.sin(C).cos(A+C)

Cosine rule

In any triangle ABC we have 
 
        a2  = b2  + c2  - 2 b c cos(A)

        b2  = c2  + a2  - 2 c a cos(B)

        c2  = a2  + b2  - 2 a b cos(C)

Proof:
This rule is deduced using the dot product of vectors.
See Proof cosine rule

Trigonometric functions

The sine function

The function defined by : 
 
sin : R -> R : x -> sin(x)
is called, the sine function.
The images are bounded in [-1,1] and the period is 2.pi .
We say that the range of the function is [-1,1].

The cosine function

The function defined by : 
 
cos : R -> R : x -> cos(x)
is called, the cosine function.
The images are bounded in [-1,1] and the period is 2.pi .
The range of the function is [-1,1].

The tangent function

The function defined by : 
 
tan : R -> R : x -> tan(x)
is called, the tangent function.
Now, the period is pi and the images are not defined in x = (pi/2) + k.pi
The range or image is R.

The cotangent function

The function defined by : 
 
cot : R -> R : x -> cot(x)
is called, the cotangent function.
The period is pi and that the images are not defined in x = k.pi
The range or image is R.

Inverse Trigonometric Functions

The arcsin function

We restrict the domain of the sine function to [-pi/2 , pi/2].
Now this restriction is invertible because each image value in [-1,1] has just one origin in [-pi/2 , pi/2].
The inverse function of that restricted sine function is called the arcsine function.
We write arcsin(x) or asin(x).
The graph y = arcsin(x) is the mirror image of the restricted sine graph with respect to the line y = x.
The domain is [-1,1] and the range is [-pi/2 , pi/2].

The arccos function

We restrict the domain of the cosine function to [0 , pi].
Now this restriction is invertible because each image value in [-1,1] has just one origin in [0 , pi].
The inverse function of that restricted cosine function is called the arccosine function.
We write arccos(x) or acos(x) .
The graph y = arccos(x) is the mirror image of the restricted cosine graph with respect to the line y = x.
The domain is [-1,1] and the range is [0 , pi].

The arctan function

We restrict the domain of the tangent function to [-pi/2 , pi/2].
Now this restriction is invertible because each image value in has just one origin in [-pi/2 , pi/2].
The inverse function of that restricted tangent function is called the arctangent function. We write arctan(x) or atan(x) . The graph y = arctan(x) is the mirror image of the restricted tangent graph with respect to the line y = x.
The domain is R and the range is [-pi/2 , pi/2].

The arccot function

We restrict the domain of the cotangent function to [0 , pi].
Now this restriction is invertible because each image value in has just one origin in [0 , pi].
The inverse function of that restricted cotangent function is called the arccotangent function.
We write arccot(x) or acot(x) .
The graph y = arccot(x) is the mirror image of the restricted cotangent graph with respect to the line y = x.
The domain is R and the range is [0 , pi].

Sum formulas

cos(u - v)

We prove this formula using the concept of dot product of two vectors. (See theory about vectors) With u corresponds one point p(cos(u),sin(u)) on the unit circle
With v corresponds one point q(cos(v),sin(v)) on the unit circle
The angle, corresponding with the arc qp of the circle, has a value u - v .
Now : p.q = 1.1.cos(u-v) .
But using the coordinates we also have p.q = cos(u).cos(v)+sin(u).sin(v).
Hence,

cos(u-v) = cos(u).cos(v)+sin(u).sin(v)

cos(u + v)

cos(u + v) = cos(u - (-v)) = cos(u).cos(-v)+sin(u).sin(-v)

cos(u + v) = cos(u).cos(v)-sin(u).sin(v)

sin(u - v)

sin(u - v) = cos(pi/2-(u-v)) = cos( (pi/2-u) +v )
= cos(pi/2 - u).cos(v)-sin(pi/2 - u).sin(v)

sin(u - v) = sin(u).cos(v)-cos(u).sin(v)

sin(u + v)

sin(u + v) = cos(pi/2-(u+v)) = cos( (pi/2-u) -v )
= cos(pi/2 - u).cos(v)+sin(pi/2 - u).sin(v)

sin(u + v) = sin(u).cos(v)+cos(u).sin(v)

tan(u + v)

 
            sin(u + v)    sin(u).cos(v)+cos(u).sin(v)
tan(u+v) = ------------ = ---------------------------
            cos(u + v)    cos(u).cos(v)-sin(u).sin(v)
Dividing the dominator and denominator by cos(u).cos(v) we have 

 
           tan(u) + tan(v)
tan(u+v) = -----------------
           1 - tan(u).tan(v)

tan(u - v)

In the same way, we have 

 
           tan(u) - tan(v)
tan(u-v) = -----------------
           1 + tan(u).tan(v)

sin(2u)

sin(2u) = sin(u + u) = sin(u).cos(u)+cos(u).sin(u) = 2sin(u).cos(u)

sin(2u) = 2sin(u).cos(u)

cos(2u)

cos(2u) = cos(u+u) = cos(u).cos(u)-sin(u).sin(u) = cos2 (u) - sin2 (u)

cos(2u) = cos2 (u) - sin2 (u)

tan(2u)

 
           tan(u) + tan(u)            2 tan(u)
tan(2u) =   ------------------  =  ---------------
           1 - tan(u).tan(u)        1- tan(u)tan(u)

 
            2 tan(u)
tan(2u)  =  -----------
            1- tan2(u)

Carnot formulas

 
1 + cos(2u) =  1+cos2 (u)-sin2 (u) =  2 cos2 (u)

1 - cos(2u) =  1-cos2 (u)+sin2 (u) =  2 sin2 (u)


 
1 + cos(2u) = 2 cos2 (u)

1 - cos(2u) = 2 sin2 (u)

t-formulas

From the Carnot formulas we have 
 

cos(2u)  =  2 cos2(u) -1


       2
=  ------------ - 1
   1 + tan2 (u)


   1 -  tan2(u)
=  -------------
   1 + tan2 (u)


We know:
            2 tan(u)
tan(2u)=  -------------
          1 - tan2 (u)

Hence,

             2 tan(u)
sin(2u) =  -----------
           1 + tan2 (u)


 
Let t = tan(u) , then


           1 - t2
cos(2u) = ---------  ;
           1 + t2

             2t
sin(2u) =  -------- ;
            1 + t2

              2t
tan(2u) =  -------  ;
            1 - t2

These 3 formulas are called the t-formulas.

Special values

pi/3

Let v be the image point corresponding with the angle pi/3 on the unit circle and let e the intersection point of that circle with the X-axis.
The triangle 'ove' is regular. Hence cos(pi/3) = 1/2. 
 
sin2 (pi/3) = sqrt( 1 - cos2 (pi/3)) = sqrt(3)/2
So, sin(pi/3) = sqrt(3)/2 and cos(pi/3) = 1/2.

pi/4

Let v be the image point corresponding with the angle pi/4 on the unit circle. From this, it is obvious that cos(pi/4) = sin(pi/4) and tan(pi/4) = 1. 
 
cos2(pi/4)+sin2(pi/4) = 1  => 2cos2(pi/4) = 1 =>  cos (pi/4) = sqrt(1/2)

So, cos (pi/4) =  sin(pi/4) = sqrt(1/2)

pi/6

From properties for complementary angles we have: cos (pi/6) = sqrt(3)/2 and sin(pi/6) = 1/2.

Trigonometric equations

Base equations

cos(u) = cos(v)

With the help of a unit circle it is easy to see that
cos(u) = cos(v) <=> (u = v + k.2pi) or (u = -v + k.2pi)

sin(u) = sin(v)

With the help of a unit circle it is easy to see that
sin(u) = sin(v) <=> (u = v + 2.k.pi) or (u = pi - v + 2.k.pi)

tan(u) = tan(v)

With the help of a unit circle it is easy to see that
tan(u) = tan(v) <=> (u = v + k.pi)

cot(u) = cot(v)

With the help of a unit circle it is easy to see that
cot(u) = cot(v) <=> (u = v + k.pi)

Reducing to base equations

Example 1 
 

cos(2x) = cos(pi-3x)
<=>
2x = (pi-3x) + 2.k.pi  or 2x = -(pi-3x) + 2.k'.pi
<=>
5x = pi + 2.k.pi  or    -x = -pi + 2.k'.pi
<=>
x = pi/5 + 2.k.pi/5  or   x = pi - 2.k'.pi
Example 2 
 
tan(x-pi/2) = tan(2x)
<=>
(x-pi/2) = 2x + k.pi
<=>
-x = pi/2 + k.pi
<=>
x = -pi/2 - k.pi
Example 3 
 
cos(x) = -1/3
<=>
cos(x) = cos(1.91)
<=>
x = 1.91 +2.k.pi  or x = -1.91 - 2.k.pi
Example 4 
 
sin(2x) = cos(x-pi/3)
<=>
cos(pi/2 - 2x) = cos(x-pi/3)
<=>
pi/2 - 2x = x - pi/3 + 2.k.pi   or  pi/2 - 2x = - x + pi/3 + 2.k'.pi
<=>
-3x = - pi/2 - pi/3 + 2.k.pi   or  -x =  -pi/2 + pi/3 + 2.k'.pi
<=>
x = pi/6 + pi/9 + 2.k.pi/3   or  x = pi/2 -  pi/3 - 2.k'.pi
<=>
x = 5pi/18  + 2.k.pi/3   or  x = pi/6 - 2.k'.pi

Using an additional unknown

Example 1 
 
2sin2 (2x)+sin(2x)-1=0

<=>             (let t = sin(2x) )

2t2 + t - 1 = 0

<=>

t = 0.5  or t = -1

<=>

sin(2x) = 0.5  or sin(2x) = -1

<=>

sin(2x) = sin(pi/6)   or  sin(2x) = sin(-pi/2)

<=>

2x = pi/6 +2.k.pi or 2x = pi - pi/6 +2.k.pi or
        2x = -pi/2 +2.k.pi or 2x = pi + pi/2  +2.k.pi
<=>

x = pi/12 + k.pi  or x = 5pi/12 + k.pi or
        x = -pi/4 + k.pi  or x = 3pi/4 + k.pi
Sometimes it is convenient to view these solutions on the unit circle. Example 2 
 
     cos 10x + 7 = 8 cos 5x
<=>
     cos 10x - 8 cos 5x + 7 =0
<=>
     1 +  cos 10x - 8 cos 5x + 6 =0
<=>
     2 cos2 5x - 8 cos 5x + 6 =0
<=>

     cos2  5x - 4 cos 5x + 3 = 0

say  t = cos 5x

    t2 - 4t + 3 = 0
<=>
     t = 3  or  t = 1
<=>
     cos 5x  = 1
<=>
     cos 5x = cos 0
<=>
     5x = 2kpi
<=>
     x =  2kpi / 5
Examples
In the same way, the following equations can be solved using an additional unknown. 
 
tan2 (3x)+tan(3x)=0

sin2 (x)(sin(x)+1)-0.25(sin(x)+1) = 0

cos(2x)+sin2 (x) = 0.5


tan(2x)-cot(2x) = 1

Using factorization

Example 1 
 
3.sin(2x)-2.sin(x) = 0

<=>

6sin(x)cos(x)-2.sin(x) = 0

<=>

2.sin(x).(3cos()-1) = 0

<=>

sin(x) = 0  or cos(x) = 1/3

<=>

x = k.pi  or x = 1.23 + 2.k.pi or x = -1.23 + 2.k'.pi
Examples
In the same way, the following equations can be solved using factorization. 
 
tan(x)tan(4x)+tan2 (x) = 0

sin(5x)+sin(3x) = cos(2x)-cos(6x)

The equation a.sin(u)+b.cos(u) = c

First we'll show that a.sin(u)+b.cos(u) can be transformed in the form
A.sin(u-uo) or in the form A.cos(u-uo) . 

 
   a.sin(u) + b.cos(u)

=  a( sin(u) + (b/a) cos(u) )

      Take uo such that tan(uo) = - b/a

=  a( sin(u) -  tan(uo)  cos(u) )

=  (a/cos(uo)) . ( sin(u).cos(uo) - sin(uo).cos(u) )

      Let A =  (a/cos(uo))

=  A . sin(u - uo)

=  A . cos(pi/2 - u + uo)

=  A . cos(u - uo')

With this method we can solve the equation
a.sin(u)+b.cos(u) = c

Example 

 
3.sin(2x)+4.cos(2x) = 2

<=>

sin(2x) + 4/3 .cos(2x) =  2/3
                Let tan(t) =  4/3
<=>

sin(2x) + tan(t) .cos(2x) =  2/3

<=>

sin(2x)cos(t)+cos(2x)sin(t) = 2/3.cos(t)

<=>

sin(2x+t) = 2/3.cos(t)

                since 2/3.cos(t) =  0.4
<=>

sin(2x+0.927) = sin(0.39)

<=>

2x + 0.927  = 0.39 +2.k.pi  or 2x + 0.927  =  pi - 0.39 +2.k'.pi

<=>

....

Homogeneous equations

We have in view the equations, witch are homogeneous in sin(u) and cos(u).
Procedure
  1. If possible, use factorization and solve the simple parts
  2. Divide the remaining equation by a suitable power of cos(u), such that tan(u) appears everywhere.
  3. Let t = tan(u) and solve the algebraic equation.
  4. Return to tan(u)
Example 
 
2.cos3 (x)+2.sin2 (x)cos(x) = 5.sin(x)cos2 (x)

<=>

cos(x).(2.cos2 (x)+2.sin2 (x) - 5.sin(x)cos(x))  =  0

<=>

The first part  cos(x) = 0 gives us x = pi/2 + k.pi


In the second part, we divide both sides by cos2 (x). Then we have

 2.tan2 (x) - 5.tan(x) +2 = 0

                Let t = tan(x)
<=>

 2.t2  - 5 t + 2 = 0

<=>

t = 0.5  or t = 2

<=>

tan(x) = 0.5 or tan(x) = 2

<=>

x = 0.464 +k.pi or x = 1.107 +k.pi

Calculations with inverse trigonometric functions

Example 
 
                                  ________
                                 |      2
                                 | 1 - p
Show that :   cot(arcsin(p))  =  | ------
                                \|   p

        let b = arcsin(p) , then sin(b) = p with b in [-pi/2 , pi/2].


        So, cos(b) = 1 - p 2              and
                                      ________
                                     |      2
                                     | 1 - p
        cot(arcsin(p)) = cot(p) =    | ------
                                    \|   p
Example 
 
Solve : arcsin(2x) = pi/4 + arxsin(x)           (1)
Solution:
        Let arcsin(2x) = a and arcsin(x) = b .
        Then a and b are in [-pi/2 , pi/2].
We have to solve :
        a = pi/4 + b                    (2)

=>      sin(a) = sin(pi/4 + b)

                 cos(b) + sin(b)
=>      sin(a) = ---------------
                       ___
                      V 2

                                                      ________
       Since:  sin(a) = 2x, sin(b) = x and cos(b) =  V 1 - x2

                  ________
                 V 1 - x2   + x
=>       2x =   ------------------
                      ___
                     V 2

             ___             __________
=>      ( 2.V 2    - 1 )x = V 1 - x2

             ___
=>      (2. V 2  - 1)2 .x2   =  1-x2

=>      x = +0.4798   or x = -0.4798            (3)

Each solution of (1) is in (3), but the reverse is not true.
It is simple to calculate that -0.4798 is not a solution of (1),
and +0.4798 is.
So the only solution is 0.4798 .
You can find solved problems about trigonometry here

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