Chapter 7.3, Problem 43AYU

To determine

To solve: To find general formula of trigonometric equation $\sin \frac{\theta }{2}=-\frac{\sqrt{3}}{2}$ as mentioned in the problem and provide the six solution.

Answer to Problem 43AYU

General formula of trigonometric equation $\sin \frac{\theta }{2}=-\frac{\sqrt{3}}{2}$ is

  $\langle \theta =\frac{8\pi }{3}+4n\pi ,\theta =\frac{10\pi }{3}+4n\pi \rangle$ Where n is any integer and the six solutions are $\frac{8\pi }{3},\frac{10\pi }{3},\frac{20\pi }{3},\frac{22\pi }{3},\frac{32\pi }{3},\frac{34\pi }{3}$ .

Explanation of Solution

Given information:

The mentioned trigonometric equation is $\sin \frac{\theta }{2}=-\frac{\sqrt{3}}{2}$ .

Formula used:

For general formula of sine function we use $\sin \theta =\sin \alpha \Rightarrow \theta =n\pi +{\left(-1\right)}^n\alpha$ .

Also the trigonometric formula $\sin \left(2\pi +\theta \right)=\sin \theta$ , $\sin \left(\pi -\theta \right)=\sin \theta$ .

Calculation:

Consider the mentioned trigonometric expression $\sin \frac{\theta }{2}=-\frac{\sqrt{3}}{2}$ .

Recall that for general formula of sine function is $\sin \theta =\sin \alpha \Rightarrow \theta =n\pi +{\left(-1\right)}^n\alpha$ .

So, for general formula of trigonometric function is

  $\sin \text{}\frac{\theta }{2}=-\frac{\sqrt{3}}{2}\Rightarrow \sin \frac{\theta }{2}=-\sin \frac{\pi }{3}\Rightarrow \frac{\theta }{2}=n\pi -{\left(-1\right)}^n\frac{\pi }{3}\Rightarrow \theta =2n\pi -{\left(-1\right)}^n\frac{\pi }{3}$ Where n belongs to even integer.

Therefore $\theta =\frac{8\pi }{3}+4n\pi$ is the general formula where n is even integer.

Recall the trigonometric formula $\sin \left(\pi -\theta \right)=\sin \theta$ .

Therefore $\sin \left(-\frac{\pi }{3}\right)=\sin \left(\pi +\frac{\pi }{3}\right)=\sin \frac{4\pi }{3}$ .

Therefore the general formula is $\theta =\frac{10\pi }{3}+4n\pi$ where n is odd integer.

So the general formula of trigonometric function $\sin \frac{\theta }{2}=-\frac{\sqrt{3}}{2}$ is $\langle \theta =\frac{8\pi }{3}+4n\pi ,\theta =\frac{10\pi }{3}+4n\pi \rangle$ where n is even integer and odd integer respectively.

Now if we enter the value of n $=1,2,3$ in the general formula mentioned above.We will have the solutions of trigonometric functions $\sin \frac{\theta }{2}=-\frac{\sqrt{3}}{2}$ that is $\frac{8\pi }{3},\frac{10\pi }{3},\frac{20\pi }{3},\frac{22\pi }{3},\frac{32\pi }{3},\frac{34\pi }{3}$

So the six solutions of trigonometric functions $\sin \frac{\theta }{2}=-\frac{\sqrt{3}}{2}$ is $\frac{8\pi }{3},\frac{10\pi }{3},\frac{20\pi }{3},\frac{22\pi }{3},\frac{32\pi }{3},\frac{34\pi }{3}$ .