Unizor - Trigonometry - Systems of Equations
Channel: Zor Shekhtman on UNIZOR Education
Date: 2013-11-14

A system of equations with more than one unknown involved in some trigonometric functions is usually called a system of trigonometric equations.
As for any recommendations about how to solve such a system, the first one usually is a method of substitution. It involves finding a solution for one unknown in terms of others from one of the equations and substitute it in all other equations, thus reducing the number of unknown variables and the number of equations.
Other methods are much less general and usually require some ingenuity.
It is doubtful you will have to solve a system of trigonometric equations in your practical life, but these problems present an excellent opportunity to develop your creativity.

Let's consider a simple system of two equations with two unknowns
a·sin(x) + b·cos(y) = c
x + y = d

The plan to solve it is to express one variable in terms of another using the second equation, substitute this value into the first equation and solve one trigonometric equation with one variable. Then we will derive the value of a second variable.

So, the first step is:
y = d − x
Then the first equation after the substitution looks like this:
a·sin(x) + b·cos(d−x) = c
Using a formula for cos(d−x), we get:
a·sin(x) + b·cos(d)·cos(x) +
+ b·sin(d)·sin(x) = c
Combining two terms with sin, we obtain:
[a+b·sin(d)]·sin(x) +
+ [b·cos(d)]·cos(x) = c
The left side of this equation represents a linear combination of sin(x) and cos(x). This, as we know is a sinusoidal function, the form of which is determined by coefficients for our trigonometric functions.

Let me remind this procedure for the transformation of the expression P·cos(x)+Q·sin(x).
Set R = √(P²+Q²).
Then our expression can be transformed into
R·[(P/R)·cos(x)+(Q/R)·sin(x)]
There exists one and only one angle φ from 0 to 2π such that
cos(φ)=P/R and sin(φ)=Q/R. It can be found using an inverse trigonometric functions.
Here R is always positive. Depending on the sign of P and Q, the angle φ is located in one of four quadrant. Let angle α=arccos(|P|/R). Using this angle α we can find all possible values of the angle φ.
If both P and Q are non-negative then φ=α. If P is non-negative, but Q is negative then φ=−α. For negative P and non-negative Q φ=π−α. Finally, for both P and Q negative φ=π+α.

Therefore, using thus defined angle φ, our expression is equal to
R·[cos(φ)·cos(x)+sin(φ·sin(x)]
= R·cos(x−φ)

Now we can use this technique for
P = b·cos(d) and
Q = a+b·sin(d).
Then R = b²·cos²(d) +
+ a²+2ab·sin(d)+b²sin²(d) =
= a²+2ab·sin(d)+b².
In terms of a constant R and an angle φ determined above our equation looks like
R·cos(x−φ) = c
Solution to this equation exists if and only if c/R is between −1 and 1 and equals in this case to
x = ±arccos(c/R) + φ + 2π·N
where N is any integer number.

The above example illustrates the basic substitution method for solving a system of trigonometric equations. More specific techniques can only be addressed in concrete problems included into this topic.