<< Chapter < Page Chapter >> Page >

x = 2 π - θ = 2 π - π 3 = 5 π 3

Problem : Find angles in [0,2π], if

cot x = 1 3

Solution : Considering only the magnitude of numerical value, we have :

cot θ = 1 3 = cot π 3

Thus, required acute angle is π/3. Now, cotangent function is positive in first and third quadrants. Looking at the value diagram, the angle in third quadrant is :

x = π + θ = π + π 3 = 4 π 3

Hence angles are π/3 and 4π/3.

Negative angles

When we consider angle as a real number entity, we need to express angles as negative angles as well. The corresponding negative angle (y) is obtained as :

y = x - 2 π

Thus, negative angles corresponding to 4π/3 and 5π/3 are :

y = 4 π 3 - 2 π = - 2 π 3 y = 5 π 3 - 2 π = - π 3

We can also find negative angle values using a separate negative value diagram (see figure). We draw negative value diagram by demarking quadrants with corresponding angles and writing angle values for negative values. We deduct “2π” from the relation for positive value diagram.

Trigonometric value diagram

Trigonometric value diagram for negative angles

Let us consider sinx = -√3/2 again. The acute angle in first quadrant is π/3. Sine is negative in third and fourth quadrants. The angles in these quadrants are :

y = - π + θ = - π + π 3 = - 2 π 3 y = - θ = - π 3

Trigonometric equations

Zeroes of sine and cosine functions

Trigonometric equations are formed by equating trigonometric functions to zero. The solutions of these equations are :

1 : sin x = 0 x = n π ; n Z

2 : cos x = 0 x = 2 n + 1 π 2 ; n Z

Definition of other trigonometric functions

We define other trigonometric functions in the light of zeroes of sine and cosine as listed above :

tan x = sin x cos x ; x 2 n + 1 π 2 ; n Z cot x = cos x sin x ; x n π ; n Z cosec x = 1 sin x ; x n π ; n Z sec x = 1 cos x ; x 2 n + 1 π 2 ; n Z

Trigonometric equations

Trigonometric function can be used to any other values as well. Solutions of such equations are given here without deduction for reference purpose. Solutions of three equations involving sine, cosine and tangent functions are listed here :

1. Sine equation

sin x = a = sin y

x = n π + - 1 n y ; n Z

2. Cosine equation

cos x = a = cos y

x = 2 n π ± y ; n Z

3. Tangent equation

tan x = a = tan y

x = n π + y ; n Z

In order to understand the working with trigonometric equation, let us consider an equation :

sin x = - 3 2

As worked out earlier, -√3/2 is sine value of two angles in the interval [0, π]. Important question here is to know which angle should be used in the solution set. Here,

sin 4 π 3 = sin 5 π 3 = - 3 2

We can write general solution using either of two values.

x = n π + - 1 n 4 π 3 ; n Z x = n π + - 1 n 5 π 3 ; n Z

The solution sets appear to be different, but are same on expansion. Conventionally, however, we use the smaller of two angles which lie in the interval [0, π]. In order to check that two series are indeed same, let us expand series from n=-4 to n=4,

For x = n π + - 1 n 4 π 3 ; n Z

- 4 π + 4 π 3 = - 8 π 3 , - 3 π - 4 π 3 = - 13 π 3 , - 2 π + 4 π 3 = - 2 π 3 , - π - 4 π 3 = - 7 π 3 ,

0 + 4 π / 3 = 4 π 3 , π - 4 π 3 = - π 3 , 2 π + 4 π 3 = 10 π 3 , 3 π - 4 π 3 = 5 π 3 , 4 π + 4 π 3 = 16 π 3

Arranging in increasing order :

- 13 π 3 , - 8 π 3 , - 7 π 3 , - 2 π 3 , - π 3 , 4 π 3 , 5 π 3 , 10 π 3 , 16 π 3

For x = n π + - 1 n 5 π 3 ; n Z

- 4 π + 5 π 3 = - 7 π 3 , - 3 π - 5 π 3 = - 14 π 3 , - 2 π + 5 π 3 = - π 3 , - π - 5 π 3 = - 8 π 3 ,

0 + 5 π 3 = 5 π 3 , π - 5 π 3 = - 2 π 3 , 2 π + 5 π 3 = 11 π 3 , 3 π - 5 π 3 = 4 π 3 , 4 π + 5 π 3 = 17 π 3

Arranging in increasing order :

- 14 π 3 , - 8 π 3 , - 7 π 3 , - 2 π 3 , - π 3 , 4 π 3 , 5 π 3 , 11 π 3 , 17 π 3

We see that there are common terms. There are, however, certain terms which do not appear in other series. We can though find those missing terms by evaluating some more values. For example, if we put n = 6 in the second series, then we get the missing term -13π/3. Also, putting n=5,7, we get 10π/3 and 16π/3. Thus, all missing terms in second series are obtained. Similarly, we can compute few more values in first series to find missing terms. We, therefore, conclude that both these series are equal.

Problem : Find solution of equation :

2 cos 2 x + 3 sin x = 0

Solution : Our objective here is to covert equation to linear form. Here, we can not convert sine term to cosine term, but we can convert cos 2 x in terms of sin 2 x .

2 1 - sin 2 x + 3 sin x = 0 2 - 2 sin 2 x + 3 sin x = 0 2 sin 2 x 3 sin x 2 = 0

It is a quadratic equation in sinx. Factoring, we have :

2 sin 2 x + sin x 4 sin x 2 = 0 sin x 2 sin x + 1 2 2 sin x + 1 = 0 2 sin x + 1 sin x 2 = 0

Either, sinx=-1/2 or sinx = 2. But sinx can not be equal to 2. hence,

sin x = - 1 2 = sin π + π 6 = sin 7 π 6 x = n π + - 1 n 7 π 6 ; n Z

Note : We shall not work with any other examples here as purpose of this module is only to introduce general concepts of angles, identities and equations. These topics are part of separate detailed study.

Trigonometric identities

Reciprocal identities

Reciprocals are defined for values of x for which trigonometric function in the denominator is not zero.

sin x = 1 cosec x ; cos x = 1 sec x ; tan x = 1 cot x ; cosec x = 1 sin x ; sec x = 1 cos x ; cot x = 1 tan x

Negative angle identities

cos - x = cos x ; sin - x = - sin x ; tan - x = - tan x

Pythagorean identities

cos 2 x + sin 2 x = 1 ; 1 + tan 2 x = sec 2 x ; 1 + cot 2 x = cosec 2 x

Sum/difference identities

sin x ± y = sin x cos y ± sin y cos x cos x ± y = cos x cos y sin x sin y tan x ± y = tan s x ± tan y / 1 tan x tan y ; x,y and (x+y) are not odd multiple of π/2 cot x ± y = cot x cot y 1 / cot y ± cot x ; x,y and (x+y) are not odd multiple of π/2

Double angle identities

sin 2 x = 2 sin x cos x = 2 tan x 1 + tan 2 x cos 2 x = cos 2 x - sin 2 x = 2 cos 2 x - 1 = 1 - 2 sin 2 x = 1 - tan 2 x 1 + tan 2 x tan 2 x = 2 tan x 1 - tan 2 x cot 2 x = cot 2 x - 1 2 cot x

Triple angle identities

sin 3 x = 3 sin x 4 sin 3 x cos 3 x = 4 cos 3 x 3 cos x tan 3 x = 3 tan x tan 3 x 1 - 3 tan 2 x cot 3 x = 3 cot x cot 3 x 1 - 3 cot 2 x

Power reduction identities

sin 2 x = 1 - cos 2 x 2 cos 2 x = 1 + cos 2 x 2 sin 3 x = 3 sin x sin 3 x 4 cos 3 x = cos 3 x + 3 cos x 4

Product to sum identities

2 sin x cos y = sin x + y + sin x - y 2 cos x sin y = sin x + y - sin x - y 2 cos x cos y = cos x + y + cos x - y 2 sin x sin y = - cos x + y + cos x - y = cos x - y - cos x + y

Sum to product identities

sin x + sin y = 2 sin x + y 2 cos x - y 2 sin x - sin y = 2 cos x + y 2 sin x - y 2 cos x + cos y = 2 cos x + y 2 cos x - y 2 cos x - cos y = - 2 sin x + y 2 sin x - y 2 = 2 sin x + y 2 sin y - x 2

Half angle identities

sin x 2 = ± { 1 - cos x 2 } cos x 2 = ± { 1 + cos x 2 } tan x 2 = cosec x cot x = ± { 1 cos x 1 + cos x } = sin x 1 + cos x = 1 cos x sin x cot x 2 = cosec x + cot x = ± { 1 + cos x 1 cos x } = sin x 1 cos x = 1 + cos x sin x

Questions & Answers

a perfect square v²+2v+_
Dearan Reply
kkk nice
Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
y=10×
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
is it 3×y ?
Joan Reply
J, combine like terms 7x-4y
Bridget Reply
im not good at math so would this help me
Rachael Reply
how did I we'll learn this
Noor Reply
f(x)= 2|x+5| find f(-6)
Prince Reply
f(n)= 2n + 1
Samantha Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
Period of sin^6 3x+ cos^6 3x
Sneha Reply
Period of sin^6 3x+ cos^6 3x
Sneha Reply

Get the best Algebra and trigonometry course in your pocket!





Source:  OpenStax, Functions. OpenStax CNX. Sep 23, 2008 Download for free at http://cnx.org/content/col10464/1.64
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Functions' conversation and receive update notifications?

Ask