Definitions and Basics TRIGNOMETRY

7/31/2019 Definitions and Basics TRIGNOMETRY

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Definitions and basics

Trigonometric circle and angles

Choose an x-axis and a y-axis (orthonormal) and let O be the origin.

A circle of radius one centered at O is called 'the' trigonometric circle or 'the' unit circle.Turning counterclockwise is the positive orientation in trigonometry.

Angles are measured starting from the x-axis.

The units used to measure an angle are 'degree' and 'radian'.

A right angle is an angle whose measure is exactly 90 degrees or pi/2 radians.In this theory we use mainly radians.

Each real number t corresponds to exactly one angle, and to exactly one point P on the unit

circle.We call that point the 'image point' of t.

Examples:

pi/6 corresponds to the angle t and to point P on the circle.

-pi/2 corresponds to the angle u and to point Q on the circle.

Trigonometric numbers of a real number t

The real number t corresponds to exactly one point P on the unit circle.

The x-coordinate of P is called the cosine of t. We write cos(t).

The y-coordinate of P is called the sine of t. We write sin(t).

The number sin(t)/cos(t) is called the tangent of t. We write tan(t).

The number cos(t)/sin(t) is called the cotangent of t. We write cot(t).

The number 1/cos(t) is called the secant of t. We write sec(t)

The number 1/sin(t) is called the cosecant of t. We write csc(t) or cosec(t)

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.


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It is easy to see that ? = tan(t).

So tan(t) is the y-coordinate of the point S.

In an analogous manner we find that 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 squareof the distance [OP] = 1. Calculating |OP|2, using the coordinates of P, we find 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)

In the same way :

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

cos2

(t) + sin2

(t) = 1

1 + tan2(t) = sec2(t)

1 + cot2(t) = csc2(t)

Usage examples:


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sin2(t) = 1 - cos2(t)

cos2(4t) = 1 - sin2(4t)

1 + tan2(t/2) = sec2(t/2)

csc2(t2) - cot2(t2) = 1

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 symmetricwith 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')

Usage examples:

sin(t + pi/2) = sin(pi/2 - t)

tan(2t + 0.2) = - tan(pi -0.2 - 2t)

- tan(pi -t) = tan(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


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sin(t) = cos(t')

cos(t) = sin(t')

tan(t) = cot(t')

cot(t) = tan(t')

Usage examples:

tan(pi/4 +3t) = cot(pi/4 -3t)

cos(3pi/2 -t) = sin( t - pi) = sin(-t + 2pi) = sin(-t)

cot(3x - pi/2) = tan(-3x + pi ) = - tan(3x)

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)

Usage examples:

cos(-pi/2 + x) = cos(pi/2 - x) = sin (x)

sin(6x - pi) = - sin(pi - 6x) = - sin(6x)

cot(-x + 4pi) = cot(-x) = - cot(x)

Anti-supplementary values


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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 the 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 =(1/2) a.c.sin(B) = (1/2) b.c.sin(A) = (1/2) a.b.sin(C)

You can also use Heron's formula to calculate the area of a triangle.

Let s = half the circumference of the triangle = (a +b + c)/2.

The area of a triangle ABC =

______________________________

V s (s - a) (s - b) (s - c)


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Sine rule

The area of a triangle ABC = 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

a b c------ = ------ = ------sin(A) sin(B) sin(C)

This formula is called the sine rule in a triangle ABC.

Let R be the radius of the circle with center O through the points A,B and C. Let B' be the

second intersection point of BO with the circle. The angle B' in triangle BB'C is equal to, orsupplementary with, 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.

In any triangle ABC we have

a b c------ = ------ = ------ = 2Rsin(A) sin(B) sin(C)

Homogeneous expression in a, b and c

Note:A relation is called homogeneous in a, b and c if and only if this relation remains valid

when we replace a, b and c by a multiple r.a, r.b and r.c (r not 0).

If an expression between the sides of a triangle is homogeneous in a, b and c, we obtain an

equivalent expression by replacing a,b and c by sin(A), sin(B) and 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


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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:

We'll prove that a2 = b2 + c2 - 2 b c cos(A)

If the angle A is a right angle, then the proof is obvious.

Now, suppose the angle A is an acute angle.

a2 = h2 + p2 (*)

b2 = h2 + q2

= h2 + (c - p)2

so,h2 = b2 - (c - p)2 (**)

From (*) and (**)

a2 = b2 - (c - p)2 + p2

= b2 - (c2 - 2 p c + p2) + p2

= b2 - c2 + 2 p c

= b2 + c2 + 2 p c - 2 c2

= b2

+ c2

+ 2 c (p - c)

= b2 + c2 - 2 c (c - p)

= b2 + c2 - 2 c q

= b2 + c2 - 2 c b cos(A)

Now, suppose the angle A is an obtuse angle.


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The proof proceeds in the same manner as above.

Draw a new picture and work this out as an exercise.

This cosine rule can also be proved using the dot product of vectors.See Proof cosine rule

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 positive X-axis.

The triangle 'OVE' is equilateral. 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.

tan(pi/3) = sqrt(3)

pi/4

Let V be the image point corresponding with the angle pi/4 on the unit circle. From this, itis 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)

tan(pi/4) = 1

pi/6

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

1/2.

tan(pi/6) = 1/sqrt(3).

Solving Triangles

Case SSS

Three sides.

Substitute all the sides in de Cosine Rule to compute the angles.

Example: a=4 b=5 c=7
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The Cosine Rule gives

58 = 70 cos(A)40 = 56 cos(B)-8 = 40 cos(C)

A = 34.05 B = 44.41 C = 101.53

Test : A + B + C = ...

Case ASA or AAS

Two angles and a side.

Calculate the third angle and then the sides with the Sine Rule.

Example: a=4 A=34 B=45

The third angle is C =101From the Sine Rule

4 sin(45)b = -------------- = 5.06

sin(34)

4 sin(101)c = ------------- = 7.02

sin(34)

Test : draw a figure of the triangle

Case SAS

Two sides and an included angle.

Use the Cosine Rule.

Example: b=5 c=7 A=34.05

From the Cosine Rule

a2 = 25 + 49 - 70 cos(34.05) => a = 4

The other two formulas of the cosine rule give

40 = 56 cos(B)-8 = 40 cos(C)


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B = 44.41 C = 101.53

Test : A + B + C = ...

Case SSA

Two sides and a non-included angle.

Draw a figure. There are three cases.

1) no solutions

2) one solution3) two solutions

1. A=60 b=5 a=1

From a figure we see that there are no solutions.

2. A=60 b=5 a=7

From a figure we see that there is one solution. We use the Sine Rule.

7 5 c--------- = -------- = ---------sin(60) sin(B) sin(C)

So, sin(B)= 0.6186 and this gives us two supplementary solutions for B.But from our figure, we know what value to choose. B = 38.21.

Then C = 180 - 38.21 - 60 = 81.79

and c = 8

3. A=60 b=5 a=4.5

From a figure we see that there are two solutions for B. We use the Sine Rule.

4.5 5 c--------- = -------- = ---------sin(60) sin(B) sin(C)

So, sin(B)= 0.96225 and this gives us two supplementary solutions for B.

B = 74.2 of 105.8First choose B = 74.2 and first compute C and then c with the Sine Rule.

Then choose B = 105.8 and first compute C and then c with the Sine Rule.

Check the results using your figure.

Trigonometric functions


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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 see that the range of the function is [-1,1].

The cosine function


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


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.piThe range or image is R.


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The cotangent function


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.

Related functions and period

We can submit previous functions to all kinds of transformations. We obtain relatedfunctions. ( seeInfluence of a transformation on the graph of a function )

Example 1

y = sin(4x)The graph of this function arises from the graph of sin(x). We compress the graph of sin(x)towards the y-axis with a factor 4. From this it follows that the period of sin(4x) is pi/2.

The function y = sin(ax) has a period 2.pi/a.

Similar rules apply to the other trigonometric functions. Thus the period of tan(x/3) is 3.pi.

Example 2
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y = sin(x+5)

The graph of this function comes about by moving the graph of sin(x) five units to the left.

The period does not change.

Example 3

y = tan(x)+5

The graph of this function is obtained by moving the graph of tan(x) five units upwards.

The period does not change.

Example 4

We start with y = tan(x). We compress the graph towards the y-axis with a factor 3. The

new function is y = tan(3x). We move the graph two units to the right. The new function is

y = tan(3(x-2)) . Finally, we move the last graph two units downwards. We obtain y =

tan(3x -6)-2. The period is pi/3.

Generalization:

The period of A sin(a x + b ) is 2 pi/|a|

The period of A cos(a x + b ) is 2 pi/|a|

The period of A tan(a x + b ) is pi/|a|The period of A cot(a x + b ) is pi/|a|

The period of A / sin(a x + b ) is 2 pi/|a|

The period of A / cos(a x + b ) is 2 pi/|a|The period of A / tan(a x + b ) is pi/|a|

The period of A / cot(a x + b ) is pi/|a|

Period of a sum of two functions

If the function f(x) has a as period and g(x) has b as period, then f(x)+g(x) has a period c if

and only if there are strictly positive integers m and n such that c = m.a = n.b

Examples

sin(2x) has pi as period and cos(3x) has 2pi/3 as period .Now, c = 2.(pi) = 3.(2pi/3). So, 2 pi is a period of sin(2x) + cos(3x)

sin(pi x) has 2 as period and tan(2 pi x/7) has 7/2 as period.Now, c = 7.(2) = 4.(7/2). So, 14 is a period of sin(pi x) + tan(2 pi x/7)

sin(sqrt(2) x) has pi.sqrt(2) as period and cos(2x) has pi as period .There are no strictly positive integers m and n such that

m.(pi.sqrt(2)) = n.(pi). So, sin(sqrt(2) x) + cos(2x) has NO period!


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sin(x) has 2pi as period and cos(pi x) has 2 as period.

There are no strictly positive integers m and n such that

m.(2pi) = n.(2). So, sin(x) + cos(pi x) has NO period!

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] corresponds to exactly

one original value 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] corresponds to exactly

one original value 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


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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].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 Rand the range is [-pi/2 , pi/2].

The arccot function

We restrict the domain of the cotangent function to [0 , pi].

The inverse function of that restricted cotangent function is called the arccotangentfunction.

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 Rand the range is [0 , pi].

No period

The inverse trigonometric functions have no period!

Transformations

As with the trigonometric functions, we can create related functions using simple

transformations.

y = 2.arcsin(x-1) comes about by moving the graph of arcsin(x) one unit to the right, and

then by multiplying all the images by two. The domain is [0,2] and the range is [-pi,pi].

Sum formulas

cos(u - v)


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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 circleWith v corresponds one point Q(cos(v),sin(v)) on the unit circle

The angle, corresponding with the arc QP on the circle, has a value u - v .

The dot product 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)

Example:

cos(pi/3-2x) = cos(pi/3)cos(2x) + sin(pi/3)sin(2x) = 0.5 cos(2x) + 0.5

sqrt(3) sin(2x)

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)

Example:

cos(x + x/2) + cos(x - x/2) = cos(x)cos(x/2) + sin(x)sin(x/2) +cos(x)cos(x/2) - sin(x)sin(x/2)

= 2 cos(x)cos(x/2)

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)

Example:

sin(x - pi/4) = sin(x) cos(pi/4) - cos(x) sin(pi/4) = (sin(x)-cos(x))/sqrt(2)
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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)

Example:

tan(u) + tan(pi/4) tan(u) + 1 1 + tan(u)tan(u+pi/4) = -------------------- = -------------- = -------------

1 - tan(u).tan(pi/4) 1 - tan(u) 1 - tan(u)

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)


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Examples

sin(x) = 2 sin(x/2).cos(x/2)

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

12 sin(8x) cos(8x) = 6 sin(16x)

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)

Example:

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

tan(2x)

1 - tan2(x)= -------------

2 tan(x)

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)


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

Applications:

From the carnot formulas, it follows that :

The period of cos(2u) = the period of cos2 (u) = the period of sin2 (u)

Factorize the expression 1 + 2 cos(x) + cos(2x) 1 + 2 cos(x) + cos(2x)

= 2 cos(x) + ( 1 + cos(2x))

= 2 cos(x) + 2 cos2 (x)

= 2 cos(x) (1 + cos(x))

= 2 cos(x) 2 cos2 (x/2)

= 4 cos(x) cos2 (x/2)

Since 2 pi is the period of (1 + 2 cos(x) + cos(2x)), it follows that the period of

cos(x) cos2 (x/2) is 2pi.

Find the period of tan2(4x)

The period of tan2

(4x) is equal to the period of 1+tan2

(4x).The period of 1+tan2(4x) is equal to the period of 1/ cos2(4x).

The period of 1/ cos2(4x) is equal to the period of cos2(4x).

The period of cos2(4x) is equal to the period of 0.5(1+cos(8x)).The period of 0.5(1+cos(8x)) is equal to the period of cos(8x).

And this period is pi/4.

In triangle ABC the sides a, b, c are such that 3a = 7c en 3b = 8c.

Find tan2(A/2) without calculating A or A/2.

Solution:

About the three edges we know :

a b c--- = --- = ---7 8 3

Since similar triangles have the same angles, we can use


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a = 7 , b = 8 and c = 3 as edges of the triangle.

From the cosine rule we can write

b2 + c2 - a2

cos(A) = ------------------ = 1/22 b c

Now we use the Carnot formulas

1 - cos(A) 2 sin2(A/2)---------- = -------------- = tan2(A/2) = 1/31 + cos(A) 2 cos2(A/2)

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

2tsin(2u) = -------- ;


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1 + t2

2ttan(2u) = ------- ;

1 - t2

These 3 formulas are called the t-formulas.

Simpson formulas

We know that

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

and from this, we have

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

Let x = u + v and y = u - vthen u = (1/2)(x + y) and v = (1/2)(x - y)

Now we have

cos(x) + cos(y) = 2 cos((1/2)(x + y)) cos((1/2)(x - y))cos(x) - cos(y) = -2 sin((1/2)(x + y)) sin((1/2)(x - y))sin(x) + sin(y) = 2 sin((1/2)(x + y)) cos((1/2)(x - y))sin(x) - sin(y) = 2 cos((1/2)(x + y)) sin((1/2)(x - y))

Simpson formulas

x + y x - ycos(x) + cos(y) = 2 cos ------ cos -------

2 2

x + y x - ycos(x) - cos(y) = -2 sin ------ sin -------

2 2

x + y x - ysin(x) + sin(y) = 2 sin ------ cos -------

2 2

x + y x - ysin(x) - sin(y) = 2 cos ------ sin -------

2 2


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Example:

cos(2x) - cos(2y)-----------------cos(2x) + cos(2y)

-2 sin(x+y) sin(x-y)= --------------------

2 cos(x+y) cos(x-y)

= - tan(x+y) tan(x-y)

= tan(y+x) tan(y-x)

Period of the product of two related functions

We know that

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

Thus

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

or

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

The period of cos(u).cos(v) is equal to the period of cos(u + v) + cos(u - v)The period of sin(u).sin(v) is equal to the period of cos(u + v) - cos(u - v)

The period of sin(u).cos(v) is equal to the period of sin(u + v) + sin(u - v)

The period of cos(u).sin(v) is equal to the period of sin(u + v) - sin(u - v)

Examples:

The period of cos(2x).sin(x+3) is equal to the period of sin(3x+3) - sin(x-3)and this period is 2 pi.

The period of cos(4x).cos(x/2) is equal to the period of cos(9x/2) + cos(7x/2)

and this period is 4pi


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Trigonometric equations

Base equations

cos(u) = cos(v)

With the help of the 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)


sin(u) = sin(v)

(u = v + 2.k.pi) or (u = pi - v + 2.k.pi)

tan(u) = tan(v)


tan(u) = tan(v)(u = v + k.pi) on condition that tan(u) and tan(v) exist

cot(u) = cot(v)


cot(u) = cot(v)(u = v + k.pi) on condition that cot(u) and cot(v) exist

Reducing to base equations

Example 1

cos(2x) = cos(pi-3x)2x = (pi-3x) + 2.k.pi or 2x = -(pi-3x) + 2.k'.pi5x = pi + 2.k.pi or -x = -pi + 2.k'.pix = pi/5 + 2.k.pi/5 or x = pi - 2.k'.pi

Example 2


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tan(x-pi/2) = tan(2x)(x-pi/2) = 2x + k.pi-x = pi/2 + k.pix = -pi/2 - k.pi ( for these values tan(x-pi/2) and tan(2x)) exist)

Example 3

cos(x) = -1/3cos(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'.pix = pi/6 + pi/9 + 2.k.pi/3 or x = pi/2 - pi/3 - 2.k'.pix = 5pi/18 + 2.k.pi/3 or x = pi/6 - 2.k'.pi

Example 5

3 sin(2x) = cos(2x)3 tan(2x) = 1tan(2x) = 1/3tan(2x) = tan(0.32)2x = 0.32 + k pix = 0.16 + k pi/2

Example 6

tan(2x) . cot( x + pi/2) = 1

tan(2x) = tan( x + pi/2)

2x = x + pi/2 + k.pi

x = pi/2 + k.pi (on condition that tan(2x) and cot( x + pi/2) exist)

But cot( x + pi/2) does not exist for x = pi/2 + k.pi !!!!!


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So, tan(2x) . cot( x + pi/2) = 1 has no solutions !

The expression " on condition that ...." is not redundant!

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 or2x = -pi/2 +2.k.pi or 2x = pi + pi/2 +2.k.pi

x = pi/12 + k.pi or x = 5pi/12 + k.pi orx = -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 =02 cos2 5x - 8 cos 5x + 6 =0

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

Let t = cos 5x

t2 - 4t + 3 = 0


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t = 3 or t = 1

cos 5x = 1

cos 5x = cos 0

5x = 2kpi

x = 2kpi / 5

ExamplesIn 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

Check your results by plotting graphs.

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(7x)-sin(x) = sin(3x)

cos(4x) + cos(2x) + cos(x) = 0


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

Check your results by plotting graphs.

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

First Method

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')

Example

3 sin(x) - 2 cos(x)

= 3( sin(x) - (2/3) cos(x) )

Let tan(uo) = 2/3 ; take uo = 0.588

= 3( sin(x) - tan(uo) cos(x) )

= (3/cos(uo)) ( sin(x) cos(uo) - cos(x) sin(uo) )

= 3.6055 sin( x - 0.588)

of ook

= 3.6055 cos( x - 2.1598)

Plot the graph of 3 sin(x) - 2 cos(x) and the graph of 3.6055 sin( x - 0.588)

With this method we can solve the equation

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

Example


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

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

....

Second Method

Using the t-formules

Example

3 sin(2x) + 4 cos(2x) = 2Let tan(x) = t

2 t 1 - t2

3 ------- + 4 -------- = 21 + t2 1 + t2

6 t + 4 - 4 t2 = 2 + 2 t2

6 t2 - 6 t - 2 = 0

3 t2 - 3 t -1 = 0

t = 1.26 or t = -0.26


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tan(x) = 1.26 or tan(x) = -0.26

x = 0.9 + k pi or x = -0.26 + k pi

Homogeneous equations

An equation is homogeneous in a and b if and only if we obtain an equivalent equationwhen we replace a and b by ra and rb (r is not 0). Example: a3 x2 +5 a.b2 x +3 a2.b = 0 is anequation in x which is homogeneous in a en b.

Now, we have in view the equations which are homogeneous in sin(u) and cos(u).

Procedure

1. Reduce the equation to the form F = 0. If possible, use factorization to the left hand

side and solve the simple parts.2. Divide the remaining equation through 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 simple 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


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

Trigonometric inequalities

Conventions

k is an integer.

'==' means equal or greater than

Examples

sin(x/2) > 1/2

We draw the solutions for (x/2) on the unit circle.

Now, we see that:

sin(x/2) > 1/2

pi/6 + 2 k pi < x/2 < 5pi/6 + 2 k pi

pi/3 +4 k pi < x < 5pi/3 + 4 k pi

For each k, we have an open interval with solutions.The solution set V is the union of all these open intervals.


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V = { U (pi/3 +4 k pi , 5 pi/3 + 4 k pi) | k in Z }

k

tan(2x) < 1/3

We draw the solutions for (2x) on the unit circle.

Now, we see that:

tan(2x) < 1/3

-pi/2 + k pi < 2x < 0.32 + k pi

-pi/4 + k pi/2 < x < 0.16 + k pi/2

For each k, we have an open interval with solutions.

The solution set V is the union of all these open intervals.

V = { U (-pi/4 + k pi/2 , 0.16 + k pi/2) | k in Z }k

tan(2x + pi/5 ) < 1/3

This is a variation on the previous example.

The figure is the same as the previous one, but now it gives the solutions for (2x +

pi/5).Now we have :


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tan(2x + pi/5 ) < 1/3

-pi/2 + k pi < 2x + pi/5 < 0.32 + k pi

-pi/2 -pi/5 + k pi < 2x < 0.32 - pi/5 + k pi

-7 pi/20 + k pi/2 < x < 0.16 -pi/10 + k pi/2

The solution set V is:

V = { U (-7 pi/20 + k pi/2 , 0.16 - pi/10 + k pi/2) | k inZ }

k

2 sin2(x) - 3 sin(x) + 1 = < 0

Let t = sin(x) .

2 t2 - 3 t + 1 < 0

2 (t - 1)(t - 1/2) < 0

A sign study of the left hand side gives

1/2 =< t =< 1

1/2 =< sin(x) =< 1

Draw the solutions for x on the unit circle. We see that:

pi/6 + 2 k pi =< x =< 5pi/6 + 2 k pi

The solution set is

V = { U [pi/6 + 2 k pi ; 5pi/6 + 2 k pi] | k in Z }k

Another method to solve 2 sin2

(x) - 3 sin(x) + 1 = < 0

One can also factorize the left hand side directly and investigate the sign in aperiod-interval.

2 sin2(x) - 3 sin(x) + 1

2 (sin(x) - 1)(sin(x) - 1/2)


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We take a simple period-interval [0,2pi). We investigate the sign of each factor.

x 0 pi/6 pi/2 5pi/6 pi 2pi---------------------------------------------------------------sin(x)-1 - - - - - - 0 - - - - - - - - - - - - - -

---------------------------------------------------------------sin(x)-1/2 - - 0 + + + + + + 0 - - - - - - - - - ----------------------------------------------------------------product + + 0 - - - 0 - - 0 + + + + + + + + + +---------------------------------------------------------------

The solutions in the period-interval are

pi/6 =< x =< 5pi/6

The solution set is :

V = { U [pi/6 + 2 k pi ; 5pi/6 + 2 k pi] | k in Z }k

cosec(x) < sec(x)

We investigate first the inequality in the period-interval [0,2pi).

The values 0 ; pi/2 ; pi ; 3pi/2 are no solutions . We investigate the other values of x

in each quadrant.

o First quadrantoo cosec(x) < sec(x)oo 1/ sin(x) < 1 / cos(x)oo now we have sin(x).cos(x) > 0oo cos(x) < sin(x)

The solution set is ( pi/4, pi/2 ).

o Second quadrant

Now we have cos(x) < 0 and sin(x) > 0. There are no solutions.

o Third quadrantoo cosec(x) < sec(x)oo 1/ sin(x) < 1 / cos(x)oo now we have sin(x).cos(x) > 0


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-pi/2 + 2k pi < x - pi/4 < pi/2 + 2k pi and sin (x) not0

-pi/4 + 2k pi < x < 3pi/4 + 2k pi and sin (x) not 0

The solution set V is the union of the open intervals

V = { U (-pi/4 + 2k pi , 3pi/4 + 2k pi ) | k in Z } \ { kpi | k in Z }

k

Cyclometric equations

All equations are solved using the same method. We replace the equation successively by a

necessary condition. This means that the values that we find are not necessary solutions.

Afterwards, we have to test these values against the initial equation. The false or parasitic

values must be deleted.

Example 1

arcsin(2x) = pi/4 + arcsin(x)

/ arcsin(2x) = a| arcsin(x) = b (1)\ a = pi/4 + b

/ sin(a) = 2x

=> | sin(b) = x\ a = pi/4 + b

=> / sin(pi/4 + b) = 2x\ sin(b) = x

sum formulas

=> / cos(b) + sin(b) = 2x.sqrt(2)\ sin(b) = x

=> / cos(b) = 2x.sqrt(2) - x\ sin(b) = x

=> (2x.sqrt(2) - x)2 + x2 = 1

=> ....

=> x = +0.4798 or x = -0.4798

We test these values against the initial equation. The only solution is 0.4798.

The other x-value is false or parasitic.


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Example 2

arctan(x+1) = 3.arctan(x-1)

/ arctan(x+1) = a| arctan(x-1) = b\ a = 3 b

=> / tan(a) = x + 1| tan(b) = x - 1\ a = 3 b

=> / tan(3b) = x+ 1\ tan(b) = x - 1

3 tan(b) - tan3(b)but tan(3b) = --------------------

1 - 3.tan2(b)

3(x-1) - (x-1)3

=> x+1 = --------------------1 - 3 (x-1)2

=> (x+1) (1 - 3 (x-1)2) = 3(x-1) - (x-1)3

=> ...

=> x = 0 or x = sqrt(2) or x = -sqrt(2)We test these values against the initial equation. The only solution is sqrt(2).The other x-values are false or parasitic.

Example 3

arctan(x) + arctan(2x) = pi/4

/ arctan(x) = a

| arctan(2x) = b\ a + b = pi/4

=> / x = tan(a)| 2x = tan(b)\ a + b = pi/4

=> / x = tan(a)\ 2x = tan(pi/4-a)


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1 - tan(a)but tan(pi/4-a) = ---------------- since tan(pi/4) = 1

1 + tan(a)

1 - x=> 2x = ----------

1 + x

=> ...

=> x = (-3+sqrt(17))/4 or x = (-3-sqrt(17))/4

We test these values against the initial equation. The only solution is (-3+sqrt(17))/4.

The other x-value is false or parasitic.

Example 4

arctan( (x+1)/(x+2) ) - arctan ( (x-1)/(x-2) ) = arccos( 3/sqrt(13) )

/ arctan( (x+1)/(x+2) ) = a | arctan( (x-1)/(x-2) ) = b

| arccos( 3/sqrt(13) ) = c\ a - b = c

/ tan(a) = (x+1)/(x+2)=> | tan(b) = (x-1)/(x-2)

| cos(c) = 3/sqrt(13)

\ a - b = c

tan(a) - tan(b)but tan(a-b) = ------------------ and after calculation

one finds1 + tan(a) tan(b)

-2 x/ tan(c) = ----------------

=> | 2 x2 - 5|\ cos(c) = 3/sqrt(13)

From the last equation it follows1 + tan2(c) = 1/cos2(c) = 13/9 => tan2(c) = 4/9There are now two cases

First case :-2 x

/ tan(c) = ----------------=> | 2 x2 - 5

|


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\ tan(c) = 2/3

=> ....

=> x = 1 or x = -5/2

Second case:

-2 x/ tan(c) = ----------------

=> | 2 x2 - 5|\ tan(c) = - 2/3

=> ....

=> x = -1 or x = 5/2

We test these values against the initial equation. The only solutions are 1 and -5/2.

The other x-values are false or parasitic.

Calculation with inverse trigonometric functions

Example

________| 2

\| 1 - pShow that cot(arcsin(p)) = -----------

p

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

So, cos(b) = sqrt( 1 - p2) and________

| 2\| 1 - p

cot(arcsin(p)) = cot(b) = ---------

p

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