An equation which contains Trigonometric Functions is known as trigonometric equations.
Solving Trigonometric EquationsBack to Top
Sine, cosine, sec, cosec, tan, cot. Sine function is defined as the Ratio of perpendicular and hypotenuse. Cosine function is defined as the ratio of base and hypotenuse. Sec is the ratio of hypotenuse and base. Tan is the ratio of perpendicular and base and cot function is defined as the ratio of base and perpendicular.
Solving Trigonometric Equations is quiet important task in Trigonometry, as various Trigonometric Functions may not be defined. We can solve trigonometric equations by simply solving it using basic solving techniques.
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:
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.
Trigonometric Form of Complex NumbersBack to Top
z = r (cos α + i sin α),
where the value of α є Arg(z). here 'r' denotes the iota and the value of 'i' is √(-1), so we can easily find the absolute value of 'z'. This is the complex numbers trigonometric form.
= | x + iy | = √ (x² + y²). Complex number can be represented by 'z'.
Here we have to find the value of 'α'. We know that the value of 'α' is not unique, but is always find modulo 2π. In complex numbers the value, arg (z) always lie in the interval [0, 2π]. 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
= z / |z| = cos α + I sin α, so the value of α can be found from...
= tan α = 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 ∏/ 2 or 3∏ / 2, it is totally depends on the sign y. We know that in mathematics 0 is only complex number that is not associated with any of the argument.
So we can write it as:
= arg (z) = ∏ / 2, and the value of y > 0.
= arg (z) = 3∏ / 2, and the value of y < 0.
This is all about trigonometric form of a complex number.
Trigonometric Equations of Quadratic TypeBack to Top
X = (-B + √ (B2 – 4AC)) / 2A,
AND X = (-B - √ (B2 – 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: sin2X – 4 sin 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:
Sin X = - (-4) + √ (– 4)2 – 4 * 1 * (-1))) / 2 * 1 = (4 + √ (16 +4)) / 2 = 2 + √5,
AND Sin X = - (-4) - √ (– 4)2 – 4 * 1 * (-1))) / 2 * 1 = (4 - √ (16 +4)) / 2 = 2 - √5,
Let us consider an example of equation which has to be solved using Factorization method: 2sin2 X + sin X – 1. Where, 'X' belongs to [0, 2pi]. To solve for 'X' we follow these steps:
(2sin X – 1) + (sin X + 1) = 0 (by factorization),
2sin X – 1 = 0 or sin X + 1 = 0,
Sin X = 1 / 2 or sin X = - 1,
Taking inverse trigonometric function of value we get values for 'X' as:
X = sin-1 (1 / 2) and X = sin-1 (- 1),
X = ∏ / 6 or 5 ∏ / 6 and X = 3∏/ 2.