CodesJava

Easy learning with example program codes

Solution of trigonometric equations

The General solution of trigonometric equations of the type sin y = sin a, cos y = cos a and tan y = tan a

General Solution: The entire set of the values belonging to the unknown angle satisfies the equation. Also, it has all the specific solutions along with the principal solutions.

General Solutions of a Trigonometric Equation

From the below-given diagram, we can observe that sin(π -θ) = sin θ & cos ( -θ) = cos θ.

We use the above description for finding the solutions of a few trigonometric equations.

Now, Solve sin(x) = y for x.

http://mathsfirst.massey.ac.nz/Trig/images/singraph.gif

Case 1: When the condition is -1≤y≤ 1, i.e, y’s value is between

(-1 & 1), so, there’s the solution.

sin(x) = y is the set of all the possible solutions which is:

x = sin-1(y) + 2 & x = −sin-1(y) + (2k+1) π, Where “k” can be taken as any integer; i.e, solutions for x is having sin-1(y) in addition to all the even multiples of the value π, together with the minus sin-1(y) & all the odd multiples regarding π.

Case 2: When -1 > y or y > 1, i.e, y’s value is very much large or very much small for the possible solution.

There’re no solutions.

Solving Cos (x) = y for x.

http://mathsfirst.massey.ac.nz/Trig/images/cosgraph.gif

Case 1: -1≤y≤ 1

cos(x) = y is the set of all the solutions which is equal to

x = ± cos-1(y) + 2,

Where “k” can be taken as any integer;

Case 2: When, -1 > y or y > 1

There exist no solutions.

Solving tan(x) = y for x.

http://mathsfirst.massey.ac.nz/Trig/images/tanxgraph.gif

Tan(x) = y equation is the set of all solution to:

x = tan-1(y) + ,

Where the value of “k” can be taken as “any integer”

Example

Find the general solutions of 2 sinx = 1.

The equation is equivalent to Sinx =

Now, Sin ⁻¹ =

Hence the general solution is

x = ±π/4 + 2, where the value of k can be any integer.

Trigonometric Equations with their General solutions

Trigonometrical EquationGeneral Solution
Sin θ = 0 θ = nπ
Cos θ = 0 θ = nπ + π/2
Tan θ = 0 θ = nπ
Sin θ = 1 θ = 2nπ + π/2
Cos θ = 1 θ = 2nπ
Sin θ = a θ = nπ ± (-1)a
Cos θ = a θ = 2nπ ± a
Tan θ = a θ = nπ ± a
Sin²θ = Sin²a θ = nπ ± a
Cos²θ = Cos²a θ = nπ ± a
Tan²θ= Tan²a θ = nπ ± a
Sin θ = Sin a

Cos θ = Cos a

θ = nπ ± a
Sin θ = Sin a

Tan θ = Tan a

θ = nπ ± a
Tan θ = Tan a

Cos θ = Cos a

θ = nπ ± a

 

 

Please follow and like us:
error
Posted in A2   









Copyright © 2019 CodesJava DMCA.com Protection Status