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Trigonometric Equations

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Trigonometry is the study of angles and measurements of a right angled triangle. The main three trigonometric functions are taken as ratios of the right angles triangle. The main three trigonometric functions are sine, cosec and tangent function. These functions also have reciprocals named cosine, secant and cotangent respectively.
Right Angle Triangle
For the sake of simplicity we can take, opposite = p, adjacent = b and hypotenuse = h.

For the right angled triangle given above the trigonometric functions will be as given:

1) $sin\theta$ = $\frac{p}{h}$

2) $cos\theta$ = $\frac{b}{h}$

3) $tan\theta$ = $\frac{p}{b}$

4) $csc\theta$ = $\frac{h}{p}$

5) $sec\theta$ = $\frac{h}{b}$

6) $cot\theta$ = $\frac{b}{p}$

An equation which contains trigonometric functions is known as trigonometric equations.
These equations can be solved using the trigonometric identities. These equations are used in astronomy to measure the distance of celestial bodies. They are applied in many real life problems including navigation, architecture, graphics, medical sciences, and so on. 

Trigonometric Identities  and Equations

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Following are the basic identities used in trigonometry:
1) Negative angle identities:

sin($-\theta$) = -sin$\theta$.

Similarly, cos($-\theta$) = -cos$\theta$, tan($-\theta$) = -tan$\theta$, cosec($-\theta$) = -cosec$\theta$, sec($-\theta$) = -sec$\theta$, cot($-\theta$) = -cot$\theta$.

2) Quotient identities:

$tan \theta$ = $\frac{sin\theta}{cos\theta}$

$cot \theta$ = $\frac{cos\theta}{sin\theta}$

3) Pythagorean identities:

$sin^{2}\theta +cos^{2}\theta$ =1

$tan^{2}\theta$ +1=$sec^{2}\theta $

$cot^{2}\theta$ +1=$cosec^{2}\theta $

4) Cosine of sum or difference identities:

$cos(A + B)$ = $cos\ A\ cos\ B - sin\ A\ sin\  B$

$cos(A - B)$ = $cos\ A\ cos\ B + sin\ A\ sin\  B$

5) Sine of sum or difference identities:

$sin(A + B)$ = $sin\ A\ cos\ B + cos\ A\ sin\ B$

$sin(A - B)$ = $sin\ A\ cos\ B$ - $cos\ A\  sin\ B$

6) Tangent sum or difference identities:

$tan(A + B)$ = $\frac{tan\ A +tan\ B}{1-tan\ A tan\ B}$

$tan(A - B)$ = $\frac{tan\ A -tan\ B}{1+tan\ A tan\ B}$

7)  Cofunction identities:

$sin(90 - \theta)$ = $cos \theta$

$cos(90 - \theta)$ = $sin \theta$

$tan(90 - \theta)$ = $cot \theta$

$cot(90 - \theta)$ = tan $\theta$

$sec(90 - \theta)$ = $csc \theta$

csc(90 - $\theta$) = sec$\theta$

8) Half angle identities:

$co$ $s\frac{A}{2}$=$\pm$ $\sqrt{\frac{1+cosA}{2}}$

$sin$\$frac{A}{2}$=$\pm$ $\sqrt{\frac{1-cosA}{2}}$

$tan\frac{A}{2}$=$\pm$ $\sqrt{\frac{1+cosA}{1-cosA}}$

9) Double angle identities:

$sin\ 2A$= 2sinAcosA

$cos\ 2A$ = $cos^{2}A$ - $sin^{2}A$

$tan\ 2A$ = $\frac{2tanA}{1-tan^{2}A}$

There are many other identities which can be derived from these basic identities. Applying these identities, trigonometric equations can be solved.

Steps

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Trigonometry is defined as its own entity which constitutes the relation between angles and sides of a triangle. It is a branch of mathematics and defines various Relations which can be formed by the possible angles and sides of the triangle. Trigonometric equations are the equations which consist of various trigonometric Functions. In these equations either the value of the sides or the angles are to be determined.

To solve a trigonometric equation following tips can be useful:
1) Check if the equation has linear or quadratic form. If it is a linear equation with only one trigonometric function it can be solved easily.

2) If more than one trigonometric functions are present, Rearrange the equation to have right hand side as zero and factor it to solve.

3) If is a quadratic equation, factor it to solve.

4) If quadratic equation is not being factored easily apply the quadratic formula.

5) In any given equations, try to apply trigonometric identities to get a simpler equation.

There are four quadrants defined as first, second, third and fourth quadrant.
The first quadrant is positive for all trigonometric functions and lies between 0 to 900. The second quadrant lies between 900 and 1800 and is positive for sine and cosec functions. The third quadrant lies from 180 to 270 and is positive for tan and cot functions. And the last quadrant lies between 270 and 360 which are positive for cos and sec functions.

Its figure is shown below:

  Quadrant of Circle
Let’s take an example for solving trigonometric equations:

Suppose we need to solve the equation 2 sin x - 2 = 0. For solving the equation firstly use simple mathematics and then apply logics considering quadrants. From the above
equation we get, sin x =1. And value of sin is 1 at 900.
 
Therefore value of x is equals to 1. Since it lies in first quadrant, therefore value will be positive as first quadrant gives positive values of all the functions. 

Factoring 

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We all must be aware of quadratic equations that they are in form AX2 + BX + C = 0, where 'A' is not equals to 0, then solutions for equation can be given as: 
X = $\frac{-B + \sqrt{B^2 - 4AC}}{ 2A}$,

and X = $\frac{-B + \sqrt{B^2 - 4AC}}{ 2A}$

Trigonometric equations of quadratic type consists of Trigonometric Functions instead of having “X” variable. Quadratic formula for such equations is applicable in same way.

For Example: suppose we have a trigonometric equation given in quadratic form as:
$sin^2X$ - 4sin X - 1 = 0. To find solution of this problem for all angles between 0 and 360 degrees we replace 'X' with sin X in Quadratic Formula:

Values of A, B and C in equation are: A = 1, B = - 4, and C = -1.

Substituting values in formula we get:

$sinX$= $\frac{-4\pm \sqrt{(-4)^{2}-4*1*(-1)}}{2*1}$ = $\frac{-4\pm \sqrt{20}}{2}$ =2 $\pm \sqrt{5}$.

Factoring of a trigonometric function can help in solving a quadratic trigonometric equation. For an equation like $sin^{2}A - 2 sinA + 1 = $, it can be easily factorized into $(sinA - 1)^{2} = 0$. It can be solved for sinA - 1 = 0 which gives sinA = 1. For an interval [0, $\pi$] the value of A will be $\frac{\pi}{2}$.

Trigonometric Form of Complex Numbers

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In Trigonometry the Complex Number can be represented in the trigonometric or in polar form. The trigonometric form of complex Numbers is given by:
z = r (cos α + i sin α).

where the value of α $\epsilon$ Arg(z). here 'r' denotes the iota and the value of 'i' is $\sqrt{(-1)}$, so we can easily find the absolute value of 'z'. This is the complex numbers trigonometric form.

= | x + iy | = $\sqrt{x^{2}+y^{2}}$. Complex number can be represented by 'z'.

Here we have to find the value of 'α'. We know that the value of $\alpha $ is not unique, but is always find modulo 2$\pi$. In complex numbers the value, arg (z) always lie in the interval [0,  2$\pi$]. Assume that , z = x + yi. Then we need to find the value of 'α'. As we know that the angle 'α' is formed with the x-axis by the radius vector of the Point (x, y) or

= $\frac{z}{|z|}$ = cos $\alpha $ + i sin $\alpha $, so the value of α can be found from...
= tan $\alpha $ = $\frac{y}{ x}$.
 
If we put '0' as the value of x then it is clearly an exceptional case, but in this case we can put the value of arg (z) either  $\frac{\pi}{2}$ or $\frac{3\pi}{2}$, it is totally depends on the sign of y. We know that in mathematics 0 is the only complex number that is not associated with any of the argument.

So we can write it as:   

= arg (z) =$\frac{\pi}{2}$, and the value of y > 0.

= arg (z) = $\frac{3\pi}{2}$, and the value of y < 0.

This is all about trigonometric form of a complex number.

Examples

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Example 1: Solve cos2A = cosA in the interval of [0, 2$\pi$].

Solution: cos2A = cos A

$\Rightarrow 2cos^{2}A-1=cosA$

$\Rightarrow 2cos^{2}A - cosA - 1 = 0$

$\Rightarrow (2cosA + 1)(cosA - 1) = 0$

$\Rightarrow cosA$ = $\frac{-1}{2}$, 1

$\Rightarrow A$ = 0, $\frac{2\pi}{3}$, $\frac{4\pi}{3}$

Example 2: Solve $sin\ A\  cos\ A$ = $sin\ A$ in the interval [0, $\pi$].

Solution: $sin\ A\ cos\ A$ = $sin\ A$

$\Rightarrow sin\ A\ cos\ A$ - $sin\ A$ = 0

$\Rightarrow sin\ A(cos\ A - 1)$ = 0

$\Rightarrow sin\ A$ = 0,$cos\ A$ = $\frac{1}{2}$

$\Rightarrow A = 0, \pi$, $\frac{\pi}{3}$, $\frac{5\pi}{3}$

Example 3: Solve cos $\frac{A}{2}$ = 1 in the interval [0, 2$\pi$].

Solution: In the interval [0, 2$\pi$], for the trigonometric equation cos$\frac{A}{2}$ = 1

$\frac{A}{2}$ = 0, 2$\pi$.

A = 0, 4$\pi$

Here, 4$\pi$ is not in the interval 0, 2$\pi$]. Hence, the angle A = 0 in [0, 2$\pi$].