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Trigonometric EquationsTrigonometric Identities : Double Angles : Trigonometric Graphs : Equations |
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In this essay we will combine the Trigonometric Function into equations that can be solved.
We begin by reminding ourselves of the trigonometric relations:
In addition, there are relations called double angles:
Because Sin2X + Cos2X = 1, this last relation can also be written as:
Sines are periodic. They oscilate between 1 and -1 over 360o (2 π Radians) begining and ending at 0.
Below is the graph of Y = Cos X. This is similar but in a different phase.
Cosines also oscilate between 1 and -1 over 360o (2 π Radians) begining and ending with 1.
The table below summarises the information for both Sines and Cosines between 0o and 360o (0 to 2 π Radians). This information will be used when solving trigonometric equations.
Angle (o) |
Angle (Rad) |
Sine | Cosine |
---|---|---|---|
Sin X = 1 / 2
Using the table, it is easy to see that X has two values in the required range. They are:
X = 30o and X = 150o.
Cos X + 1 / 2 = 0
Re-arranging the equation (to get Cos X on one side and the numbers on the other side) gives:
Cos X = -1 / 2
Using the table, we can see that X has two values in the required range. They are:
X = 120o and X = 240o.
Cos X Tan X = 1 / √2
Using the identity to replace Tan X gives:
Cos X (Sin X / Cos X) = 1 / √2
The Cosines cancel out to give:
Sin X = 1 / √2
This gives two values of X:
X = 45o and X = 135o.
2 Cos 2X + 1 = 0
Re-arrange the equation:
Cos 2X = - 1 / 2
Therefore 2X = 120o and 240o which gives:
X = 60o and X = 120o.
Sin X - Cos 2X = 0
Using the double angle identity, replace Cos 2X by (1 - 2 Sin2X):
Sin X - (1 - 2 Sin2X) = 0
Sin X - 1 + 2 Sin2X = 0
which re-arranges to a quadratic equation in Sin2X:
2 Sin2X + Sin X - 1 = 0
This can be solved by factorising:
(2 Sin X - 1)(Sin X + 1) = 0
This equation gives 0 if either 2 Sin X - 1 = 0 or Sin X + 1 = 0. In other words:
Sin X = 1 / 2 or Sin X = -1
The first equation gives two vales (X = 30o, X = 150o), the second equation gives one value (270o). Thus the solution of the original equation is:
X = 30o, X = 150o and X = 270o.
2 Cos2X + Sin 2X = 0
Using the double angle identity, the Sin 2X can be replaced by 2 Sin X Cos X:
2 Cos2X + 2 Sin X Cos X = 0
2 Cos X is common to both terms so this can be re-written:
2 Cos X ( Cos X + Sin X) = 0
This equation gives 0 if either 2 Cos X = 0 or Cos X + Sin X = 0. In other words:
Cos X = 0 or Sin X = -Cos X
The first equation gives two vales (X = 90o, X = 270o). The second equation also gives two values (135o and 315o - check these figures in the table). Thus the solution of the original equation is:
X = 90o, X = 135o, X = 270o and 315o.
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