Evaluate sin(theta)> 0

Solution:

Step 1:

Isolate $\theta$ by applying the inverse sine (arcsin) to both sides: $\theta >{\sin }^{-1}(0)$

Step 2:

Resolve the right-hand side of the inequality.

Step 2.1:

Since ${\sin }^{-1}(0)=0$, we have $\theta >0$.

Step 3:

Identify the quadrants where sine is positive, which are the first and second quadrants.

Step 4:

For the second quadrant solution, calculate $\theta$ by subtracting the reference angle from $\pi$: $\theta =\pi -0$

Step 5:

Determine the sine function's period.

Step 5.1:

The general period formula is $T=\frac{2\pi }{|b|}$.

Step 5.2:

Insert $b=1$ into the period formula: $T=\frac{2\pi }{|1|}$.

Step 5.3:

The absolute value of $1$ is $1$, so $T=\frac{2\pi }{1}$.

Step 5.4:

Simplify to find the period: $T=2\pi$.

Step 6:

Since the sine function repeats every $2\pi$, the general solutions are $\theta =2\pi n$ and $\theta =\pi +2\pi n$ for any integer $n$.

Step 7:

Combine these solutions into a single expression: $\theta =\pi n$ for any integer $n$.

Step 8:

Create test intervals using the roots: $(0,\pi )$ and $(\pi ,2\pi )$.

Step 9:

Test values from each interval in the original inequality to see if they satisfy the condition.

Step 9.1:

For the interval $(0,\pi )$, choose a test value such as $\theta =2$.

Step 9.1.1:

Substitute $\theta =2$ into the inequality: $\sin (2)>0$.

Step 9.1.2:

Evaluate to find that $\sin (2)\approx 0.90929742>0$, indicating the inequality holds true in this interval.

Step 9.2:

For the interval $(\pi ,2\pi )$, choose a test value such as $\theta =5$.

Step 9.2.1:

Substitute $\theta =5$ into the inequality: $\sin (5)>0$.

Step 9.2.2:

Evaluate to find that $\sin (5)\approx -0.95892427$ which is not greater than $0$, indicating the inequality does not hold true in this interval.

Step 10:

The solution set consists of the intervals where the inequality is true: $(0+2\pi n,\pi +2\pi n)$ for any integer $n$.

Knowledge Notes:

The problem at hand involves determining the values of $\theta$ for which $\sin (\theta )>0$. This is a trigonometric inequality problem, and the solution requires knowledge of the following points:

1. Inverse Trigonometric Functions: The inverse sine function, ${\sin }^{-1}(x)$ or $arcsin(x)$, returns the angle whose sine is $x$. It is used to isolate the variable within a trigonometric function.

2. Trigonometric Function Properties: The sine function is positive in the first and second quadrants of the unit circle, corresponding to angles between $0$ and $\pi$.

3. Periodicity of Trigonometric Functions: The sine function has a period of $2\pi$, meaning that its values repeat every $2\pi$ radians.

4. Absolute Value: The absolute value of a number, denoted $|a|$, is the non-negative value of $a$ without regard to its sign. It represents the distance from zero on the real number line.

5. Testing Intervals: To solve inequalities, it is often necessary to test values within different intervals to determine where the inequality holds true.

6. General Solutions for Trigonometric Equations: For periodic functions like sine, the general solution can be expressed in terms of all angles that satisfy the equation, typically involving an integer multiple of the period.

7. Combining Solutions: When multiple solutions exist, they can often be combined or expressed in a more compact form.

8. Interval Notation: In mathematics, intervals are used to represent a range of values. Open intervals, such as $(a,b)$, exclude the endpoints, while closed intervals, such as $[a,b]$, include them.