Normal Form Example

It is necessary to practice how to give appropriate signs for sine and cosine terms in the normal form. We have seen an example in section 10.8. A similar example is given below:

Equation of a line is 3x - 5y - 15 = 0. Reduce this equation to normal form. Find the values of p and 𝜔.
Solution:
1. From the given equation, we get:
A = 3, B = -5 and C = -15
2. Calculation of p

$\begin{array}{ll} {}&{p} &{}={}& {\pm \frac{C}{\sqrt{A^2~+~B^2}}} &{} \\ {}&{} &{}={}& {\pm \frac{-15}{\sqrt{3^2~+~(-5)^2}}} &{} \\ {}&{} &{}={}& {\pm \frac{15}{\sqrt{34}}} &{} \\ \end{array}$

2. Calculation of cos 𝜔

$\begin{array}{ll} {}&{\cos \omega } &{}={}& {\pm \frac{A}{\sqrt{A^2~+~B^2}}} &{} \\ {}&{} &{}={}& {\pm \frac{3}{\sqrt{3^2~+~(-5)^2}}} &{} \\ {}&{} &{}={}& {\pm \frac{3}{\sqrt{34}}} &{} \\ \end{array}$

3. Calculation of sin 𝜔

$\begin{array}{ll} {}&{\sin \omega } &{}={}& {\pm \frac{B}{\sqrt{A^2~+~B^2}}} &{} \\ {}&{} &{}={}& {\pm \frac{-5}{\sqrt{3^2~+~(-5)^2}}} &{} \\ {}&{} &{}={}& {\pm \frac{5}{\sqrt{34}}} &{} \\ \end{array}$

4. Now we determine the signs:
(i) Slope of the line = $m=-\frac{A}{B}~=~-\frac{3}{-5}~=~\frac{3}{5}$
• This is +ve. So we move along the left arrow of the chart in fig.13.36.
• The chart is shown again below:

(ii) x-intercept of the line = $a=-\frac{C}{A}~=~-\frac{-15}{3}~=~5$
• This is +ve. So sin 𝜔 is -ve and cos 𝜔 is +ve.

5. So we can write:

$\begin{array}{ll} {}&{\sin \omega } &{}={}& {- \frac{5}{\sqrt{35}}} &{} \\ {}&{\cos \omega} &{}={}& {\frac{3}{\sqrt{35}}} &{} \\ \end{array}$

6.  The sign of p will be always +ve because, it is a distance.

7. Now we can write the normal form:

$\begin{array}{ll} {}&{x \cos \omega~+~y \sin \omega} &{}={}& {p} &{} \\ {\Rightarrow}&{x × \frac{3}{\sqrt{35}}~+~y × - \frac{5}{\sqrt{35}}} &{}={}& {\frac{15}{\sqrt{35}}} &{} \\ {\Rightarrow}&{ \frac{3x}{\sqrt{35}}~+~ - \frac{5y}{\sqrt{35}}} &{}={}& {\frac{15}{\sqrt{35}}} &{} \\ {\Rightarrow}&{\frac{3x}{\sqrt{35}}~-~ \frac{5y}{\sqrt{35}}} &{}={}& {\frac{15}{\sqrt{35}}} &{} \\ \end{array}$

8. Calculation of 𝜔:
(i) From (5), we have: tan 𝜔 = -5/3
• Using a scientific calculator, we get: 𝜔 = -59.04^o
(ii) In our present case, 𝜔 is in the fourth quadrant (sin is -ve and cos is +ve).
• -59.04^o indeed lies in the fourth quadrant. But we want the +ve angle. The +ve angle corresponding to -59.04^o is (360 - 59.04) = 300.96^o
• We get: tan 𝜔 = tan -59.04 = tan 300.96
(iii) So the value of 𝜔 is 300.96^o

9. So the normal form can be written as:
x cos 300.96^o + y sin 300.96^o = $\frac{15}{\sqrt{35}}$

• The actual plot is shown below: