Matzic|

We have learned the way to find roots of a quadrant equation. Now, we are going to learn the mathematical method to

SinΘ = C

Where C is a constant value which may be anything between 0 to 1. Let us assume the corresponding value of the constant value would be Sinα. So, we can write above trigonometric equation as

SinΘ = Sinα

Now we can write this as an equation

SinΘ – Sinα = 0

As we already know that

SinC – SinD = 2.Cos[(C + D)/2].Sin[(C – D)/2]

Now, we can simplify trigonometric equation by using this formula

2.Cos[(Θ + α)/2].Sin[(Θ – α)/2] = 0

We get two equations now.

Cos[(Θ + α)/2] = 0 And Sin[(Θ – α)/2] = 0

Solution : 1

Cos[(Θ + α)/2] = 0

Trigonometry function Cosine gives zero for π/2, 3π/2, 5π/2, 7π/2, …. ∞. The principle value is π/2 for this equation. Similarly, we have to think for general solution. Generally, we can write (2n + 1)π/2 as general solution of this equation where n = 0, 1, 2, 3, 4, …. ∞.

[(Θ + α)/2] = Cos^–1(0)

» (Θ + α)/2 = (2n + 1)π/2

» Θ = (2n + 1)π – α

Here we have to use one trick to get complete solution.

Θ = (2n + 1)π + (–1)²ⁿ⁺¹α

Both solutions gives same value.

Solution : 2

Sin[(Θ – α)/2] = 0

Trigonometry function Sine gives zero value for 0, π, 2π, 3π, 4π, … ∞. The principle value is zero for this equation but we have to find the general solution. Generally, we can tell nπ as general solution. So, we can write it as

[(Θ – α)/2] = Sin^–1(0)

» (Θ – α)/2 = nπ

» Θ = 2nπ + α

Just like above solution. We have to apply same technique here to solve this.

Θ = 2nπ + (–1)²ⁿα

By considering both solutions, we can write a general solution for Sine function

Θ = nπ + (–1)ⁿα

This general solution satisfies both solutions.