Chapter 7.3, Problem 42AYU

To determine

To solve: To find general formula of trigonometric equation $\sin 2\theta =-1$ as mentioned in the problem and provide the six solution.

Answer to Problem 42AYU

General formula of trigonometric equation $\sin 2\theta =-1$ is

  $\langle \theta =\frac{n\pi }{2}+{\left(-1\right)}^n\frac{3\pi }{4},\theta =\frac{n\pi }{2}+{\left(-1\right)}^n\frac{\pi }{4}\rangle$ Where n is any integer and the six solutions are $\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4},\frac{9\pi }{4},\frac{11\pi }{4}$ .

Explanation of Solution

Given information:

The mentioned trigonometric equation is $\sin 2\theta =-1$ .

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 $\sin 2\theta =-1$ .

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

  $\sin 2\theta =-1\Rightarrow \sin 2\theta =\sin \frac{3\pi }{2}\Rightarrow 2\theta =n\pi +{\left(-1\right)}^n\frac{3\pi }{2}\Rightarrow \theta =\frac{n\pi }{2}+{\left(-1\right)}^n\frac{3\pi }{4}$ Where n belongs to integer.

Therefore $\theta =\frac{n\pi }{2}+{\left(-1\right)}^n\frac{3\pi }{4}$ is the general formula where n is any integer.

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

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

Therefore the general formula is $\theta =\frac{n\pi }{2}+{\left(-1\right)}^n\frac{\pi }{4}$ where n is any integer.

So the general formula of trigonometric function $\sin 2\theta =-1$ is $\langle \theta =\frac{n\pi }{2}+{\left(-1\right)}^n\frac{3\pi }{4},\theta =\frac{n\pi }{2}+{\left(-1\right)}^n\frac{\pi }{4}\rangle$ where n is any integer.

Now if we enter the value of n $=1,2,3$ in the general formula, we will have the six solutions of trigonometric functions $\sin 2\theta =-1$ that is $\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4},\frac{9\pi }{4},\frac{11\pi }{4}$ .

So the six solutions of trigonometric functions $\sin 2\theta =-1$ is $\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4},\frac{9\pi }{4},\frac{11\pi }{4}$ .