In the previous section, we completed a discussion on period of trigonometric functions. In this section, we will see trigonometric equations.
Some basics about trigonometric equations can be written in 3 steps:
1. We are familiar with simple algebraic equation.
• Let us see an example:
♦ x+4 = 7 is an algebraic equation.
♦ Here x will take an unique value ‘3’
♦ We say that:
✰ x+4 = 7 is an algebraic equation.
✰ The solution for this equation is: x = 3.
2. In a similar way, there are trigonometric equations also.
• Let us see an example:
♦ sin x = $\frac{1}{2}$ is a trigonometric equation.
♦ Here x can take two values: $\frac{\pi}{6}\;\;\text{and}\;\;\frac{5\pi}{6}$ ♦ We say that:
✰ sin x = $\frac{1}{2}$ is a trigonometric equation.
✰ The solution for this equation is: $\mathbf{x = \frac{\pi}{6}\;\;\text{and}\;\;\frac{5\pi}{6}}$
◼ It seems that, we have successfully solved the given trigonometric equation. But there is a problem. It can be explained in two steps:
(i) In the previous section, we saw the periodicity of the sine function. Based on that, we can write:
• Given that: sin x = $\frac{1}{2}$
♦ x = $\frac{\pi}{6}$ will satisfy this equation. x = $\frac{5\pi}{6}$ will also satisfy this equation
♦ But $x=\left(2\pi+\frac{\pi}{6}\right)\;\; \text{and} \;\;x=\left(2\pi+\frac{5\pi}{6}\right)$ will also satisfy this equation.
♦ Similarly, $x=\left(4\pi+\frac{\pi}{6}\right)\;\; \text{and}\;\; x=\left(4\pi+\frac{5\pi}{6}\right)$ will also satisfy this equation.
♦ so on . . .
(ii) We see that, infinite number of solutions are possible for the trigonometric equation sin x = $\frac{1}{2}$. The problem is: Which one shall we choose?
3. Let us see another example:
♦ tan x = √3 is a trigonometric equation.
♦ Here x can take two values: $\frac{\pi}{3}\;\;\text{and}\;\;\frac{4\pi}{3}$
♦ We say that:
✰ tan x = √3 is a trigonometric equation.
✰ The solution for this equation is: $\mathbf{x = \frac{\pi}{3}\;\;\text{and}\;\;\frac{4\pi}{3}}$
◼
It seems that, we have successfully solved the given trigonometric
equation. But there is a problem. It can be explained in two steps:
(i) In the previous section, we saw the periodicity of the tangent function. Based on that, we can write:
• Given that: tan x = √3
♦ x = $\frac{\pi}{3}$ will satisfy this equation. x = $\frac{4\pi}{3}$ will also satisfy this equation
♦ But $x=\left(\pi+\frac{\pi}{3}\right)\;\; \text{and}
\;\;x=\left(\pi+\frac{4\pi}{3}\right)$ will also satisfy this
equation.
♦ Similarly, $x=\left(2\pi+\frac{\pi}{3}\right)\;\;
\text{and}\;\; x=\left(2\pi+\frac{4\pi}{3}\right)$ will also satisfy
this equation.
♦ Similarly, $x=\left(3\pi+\frac{\pi}{3}\right)\;\;
\text{and}\;\; x=\left(3\pi+\frac{4\pi}{3}\right)$ will also satisfy
this equation.
♦ so on . . .
(ii) We see that, infinite number of solutions are possible for the trigonometric equation tan x = √3. The problem is: Which one shall we choose?
• Equations involving sine, cosine and tangent will have this problem. They have infinite number of solutions.
• We know the reason: The values repeat after regular periods.
♦ The sine and cosine have a period of 2π.
♦ The tangent has a period of π.
(We have seen those details in the previous section)
• So what do we do?
The answer is: Give a general solution.
• The advantage of using a general solution is:
♦ All the possible solutions can be written in a single statement.
♦ From that statement, any solution can be obtained just by giving a suitable integer value.
• Three theorems will help us to write the general solution of any given trigonometric equation. So our next task is to learn about those theorems.
Theorem 1
For any real numbers x and y
sin x = sin y implies x = nπ + (-1)n × y, where n ∈ Z
• First we will write an explanation for this theorem. After that, we will see the proof. The explanation can be written in 3 steps:
1. Given that x and y are any two real numbers.
• Consider the case when:
♦ sine of x
♦ is equal to
♦ sine of y
• We often come across such situations in Scientific and Engineering problems.
• For example: sin 30 is equal to sin 1830
♦ [sin 1830 = sin (5 × 360 + 30) = sin 30]
• Another example: sin 30 is equal to sin 2670
♦ [sin 2670 = sin (7 × 360 + 150) = sin 150)
♦ [sin 150 = sin (180 - 30) = sin 30 (using identity 9.d)]
2. When sin x = sin y, there will be a definite relationship between x and y.
• Theorem 1 gives us this relation: x = nπ + (-1)n × y, where n ∈ Z
• This relation can be written in words. We can write it in 4 steps:
(i) First, π is to be multiplied by an integer n. The product is nπ
(ii) Then y is to be added to or subtracted from nπ
• Addition/subtraction will depend on whether n is odd or even
♦ If n is odd, (-1)n will be -1. Then y will be subtracted from nπ
♦ If n is even, (-1)n will be +1. Then y will be added to nπ
(iii) When steps (i) and (ii) are completed, we will get x
(iv) Each x and y, which satisfy the relation sin x = sin y will have a unique value of 'n'.
3. Let us check this relation:
• In the equation sin 30 = sin 1830, x = 30 and y = 1830
♦ Then the RHS of the relation will be: n × 180 + (-1)n × 1830
♦ If we put n = -10, the RHS will become:
♦ $-10 \times 180 + {(-1)^{-10}}\times 1820$
♦ = $-10 \times 180 + \frac{1}{(-1)^{10}}\times 1820$
♦ = $-1800 + \frac{1}{1}\times 1830\;=\;30$ = LHS
♦ So when x = 30 and y = 1830, the unique value of n is -10
• In the equation sin 30 = sin 2670, x = 30 and y = 2670
♦ Then the RHS of the relation will be: n × 180 + (-1)n × 2670
♦ If we put n = 15, the RHS will become:
♦ $15 \times 180 + {(-1)^{15}}\times 2670$
♦ = $15 \times 180 + -1\times2670$
♦ = 2700 - 2670 = 30 = LHS
♦ So when x = 30 and y = 2760, the unique value of n is 15
(At present, we do not have to worry about how 'n' is calculated. All we need to know is that, there will be an unique 'n' for each case)
Now we have a basic idea about the theorem. Let us write the proof. It can be written in 7 steps:
1. If sin x = sin y, we can write: sin x - sin y = 0
To this, we can apply identity 20.d. We get:
$\begin{eqnarray}
&{}&\sin x \;=\; \sin y \nonumber \\
&\Rightarrow & \sin x - \sin y = 0 \nonumber \\
&\Rightarrow & 2\cos \left( \frac{x+y}{2}\right)\,\sin \left( \frac{x-y}{2}\right) = 0 \nonumber \\
&{}&\text{(Applying identity 20.d)} \nonumber \
\end{eqnarray}$
2. If $2\cos \left( \frac{x+y}{2}\right)\,\sin \left( \frac{x-y}{2}\right) = 0$,
Either $\cos \left( \frac{x+y}{2}\right) = 0$ or $\sin \left( \frac{x-y}{2}\right) = 0$
• That means:
If sin x = sin y,
either $\cos \left( \frac{x+y}{2}\right) = 0$ or $\sin \left( \frac{x-y}{2}\right) = 0$
3. Let us check this.
(i) We have seen that sin 30 = sin 1830. Here x = 30 and y = 1830
• So (x+y)/ 2 = (30 + 1830)/2 = 1860/2 = 930
♦ Then cos (x+y)/2 = cos 930 = cos (2 × 360+210) = cos 210
♦ = cos (180+30) = -cos 30 [using identity 9.e]
♦ = -0.8660
• Also (x-y)/ 2 = (30 - 1830)/2 = -1800/2 = -900
♦ Then sin (x-y)/2 = sin (-900) = -sin 900 = -sin (2 × 360+180)
♦ = -sin 180 = 0
(ii) We have seen that sin 30 = sin 2670. Here x = 30 and y = 2670
• So (x+y)/ 2 = (30 + 2670)/2 = 2700/2 = 1350
♦ Then cos (x+y)/2 = cos 1350 = cos (3 × 360+270) = cos 270 = 0
• Also (x-y)/ 2 = (30 - 2670)/2 = -2640/2 = -1320
♦ Then sin (x-y)/2 = sin (-1320) = -sin 1320 = -sin (3 × 360+240)
♦ = -sin 240 = -sin (180+60) = - (-sin 60) [using identity 9.f]
♦ = -(-0.8660) = 0.86600
(iii) Let us compare the results:
• In step (i), we see that:
When sin 30 = sin 1830, sin (x-y)/2 becomes zero.
• In step (ii), we see that:
When sin 30 = sin 3670, cos (x+y)/2 becomes zero.
(iv) So the result that we obtained in (2) is confirmed.
4. Consider the first result obtained in (2): $\cos \left( \frac{x+y}{2}\right) = 0$
It can be analyzed in 4 steps:
(i) We see that: cosine of (x+y)/2 is zero.
• We know the situations when cosine becomes zero:
The angle must be odd multiple of $\frac{\pi}{2}$
• So we can write:
$\text{If}\;\cos \left( \frac{x+y}{2}\right) = 0,\;\;\text{Then}\; \left( \frac{x+y}{2}\right) = (2n+1)\frac{\pi}{2}$, Where n ∈ Z
(ii) If $\frac{x+y}{2} = (2n+1)\frac{\pi}{2}$, then (x+y) = (2n+1)π
Thus we get: x = (2n+1)π - y
(iii) We see a -ve sign before 'y'. That means, 'y' is to be multiplied by '-1'
• We can write '-1' as: (-1)2n+1
♦ That means, '-1' is raised to the power (2n+1).
♦ We know that (2n+1) will be an odd integer. So (-1)2n+1 will always be '-1'.
• So (-1)2n+1 × y is same as -y.
• Thus the result in (ii) becomes: x = (2n+1)π + (-1)2n+1 y
(iv) So we can write:
If $\cos \left( \frac{x+y}{2}\right) = 0$, then x = (2n+1)π + (-1)2n+1 y
5. Consider the second result obtained in (2): $\sin \left( \frac{x-y}{2}\right) = 0$
It can be analyzed in 4 steps:
(i) We see that: sine of (x-y)/2 is zero.
• We know the situations when sine becomes zero:
The angle must be a multiple of π
• So we can write:
$\text{If}\;\sin \left( \frac{x-y}{2}\right) = 0,\;\;\text{Then}\;
\left( \frac{x-y}{2}\right) = n \pi$, Where n ∈ Z
(ii) If $\frac{x-y}{2} = n \pi$, then (x-y) = 2nπ
Thus we get: x = 2nπ + y
(iii) We see a +ve sign before 'y'. That means, 'y' is to be multiplied by '+1'
• We can write '+1' as: (-1)2n
♦ That means, '-1' is raised to the power (2n).
♦ We know that (2n) will be an even integer. So (-1)2n will always be '+1'.
• So (-1)2n × y is same as +y.
• Thus the result in (ii) becomes: x = 2nπ + (-1)2n y
(iv) So we can write:
If $\sin \left( \frac{x-y}{2}\right) = 0$, then x = 2nπ + (-1)2n y
6. Let us compare the results in (4) and (5). The comparison can be done in 5 steps:
(i) In (4), we have: x = (2n+1)π + (-1)2n+1 y
(ii) In (5), we have: x = 2nπ + (-1)2n y
(iii) Multiplication of π:
♦ In (i) we see that, π can be multiplied by any odd integer.
♦ In (ii) we see that, π can be multiplied by any even integer.
♦ Combining these, we get:
✰ π can be multiplied by any integer n
✰ We get: nπ
(iv) Power of '-1'
♦ In (i) we see that, -1 can be raised to any odd integer.
♦ In (ii) we see that, -1 can be raised to any even integer.
♦ Combining these, we get:
✰ -1 can be raised to any integer n
✰ We get: (-1)n
(v) So combining the results in (i) and (ii), we get: x = nπ + (-1)n y
7. Let us write a summary. It can be written in 4 steps:
(i) If sin x = sin y,
Then either $\cos \left( \frac{x+y}{2}\right) = 0$ or $\sin \left( \frac{x-y}{2}\right) = 0$
(ii) Then either x = (2n+1)π + (-1)2n+1 y or x = 2nπ + (-1)2n y
(iii) Combining the results in (ii), we get: x = nπ + (-1)n y
(iv) Thus Theorem 1 is proved.
In
the next
section, we will see theorem 2.
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