How do you use inverse trigonometric functions to find the solutions of the equation that are in the interval 0 2π)?
To find solutions to a trigonometric equation start by taking the inverse trig function (like inverse sin, inverse cosine, inverse tangent) of both sides of the equation and then set up reference angles to find the rest of the answers.
This inverse method gives the one answer on the interval for which each inverse trigonometric function is defined:
#sin^-1 -> [-pi/2,pi/2]#
#cos^-1 -> [0, pi]#
#tan^-1 -> (-pi/2, pi/2)#
To find the rest of the answers (answers often come in pairs unless a max or min) using reference triangles, draw your reference triangle on the set of axes in the proper quadrant/location. Remember negative angles go in quadrant IV.
For the other answer to a sine equation, reflect the reference angle over the y-axis. (Remember that sine is the y-coordinate of the unit circle and you need to find the angle with the matching y-coordinate.) A short cut is to subtract your answer from
For the other answer to a cosine equation, reflect the reference angle over the x-axis. (Remember that the cosine is the x-coordinate on the unit circle and you need to find the angle with the matching x-coordinate. A short cut is to subtract your answer from
For tangent, since the period of tangent is pi, add pi to your first answer.
In the case of sine if your first angle is negative, add
Remember exact answers can be found from famous ratios
Two examples are given below.
The section referenced in the heading is chapter 8 section 1 of Advanced Mathematics by Brown. And the answers in this written example are given in degrees. Conversion is easy.