Problem 27
Question
Exer. 1-38: Find all solutions of the equation. $$ \sqrt{3}+2 \sin \beta=0 $$
Step-by-Step Solution
Verified Answer
\( \beta = 4\pi/3 + 2k\pi \) and \( \beta = 5\pi/3 + 2k\pi \), \( k \in \mathbb{Z} \).
1Step 1: Isolate the Sine Function
Start by isolating \( \sin \beta \) on one side of the equation. The equation given is \( \sqrt{3} + 2 \sin \beta = 0 \). Subtract \( \sqrt{3} \) from both sides to isolate \( 2 \sin \beta \):\[ 2 \sin \beta = -\sqrt{3} \]
2Step 2: Solve for Sine
Divide both sides of the equation by 2 to find \( \sin \beta \):\[ \sin \beta = -\frac{\sqrt{3}}{2} \]
3Step 3: Determine General Solutions for Sin Beta
Recall that \( \sin \beta = -\frac{\sqrt{3}}{2} \) corresponds to the angles \( \beta = 4\pi/3 + 2k\pi \) and \( \beta = 5\pi/3 + 2k\pi \), where \( k \) is any integer. This is because these angles have a sine value of \(-\frac{\sqrt{3}}{2}\) on the unit circle.
Key Concepts
Sine FunctionUnit CircleGeneral Solutions
Sine Function
The sine function is a fundamental aspect of trigonometry. It is denoted by \( \sin \theta \) and describes the y-coordinate of a point on the unit circle corresponding to an angle \( \theta \). This function varies between -1 and 1 as it moves around the unit circle.
It's essential to grasp that the sine function is periodic, which means it repeats its values in regular intervals. Specifically, the period of the sine function is \( 2\pi \). This implies that sine returns to its starting point after every \( 2\pi \) radians.
For the equation \( \sin \beta = -\frac{\sqrt{3}}{2} \), knowing the specific interval where the sine of an angle equals this value is key. These specific angles play a crucial role in solving trigonometric equations.
It's essential to grasp that the sine function is periodic, which means it repeats its values in regular intervals. Specifically, the period of the sine function is \( 2\pi \). This implies that sine returns to its starting point after every \( 2\pi \) radians.
For the equation \( \sin \beta = -\frac{\sqrt{3}}{2} \), knowing the specific interval where the sine of an angle equals this value is key. These specific angles play a crucial role in solving trigonometric equations.
Unit Circle
The unit circle is an indispensable tool in trigonometry that helps us understand the behavior of trigonometric functions like sine. It is a circle with a radius of 1, centered at the origin on the coordinate plane.
Each point on the unit circle corresponds to a unique angle measured in radians from the positive x-axis. By definition, the sine of an angle is the y-coordinate of its corresponding point on the unit circle. Hence, understanding the unit circle helps us quickly find sine values for common angles.
For example, for \( \sin \beta = - \frac{\sqrt{3}}{2} \), the corresponding angles on the unit circle are \( 4\pi/3 \) and \( 5\pi/3 \). These angles occur in the third and fourth quadrants of the circle where sine values are negative.
Each point on the unit circle corresponds to a unique angle measured in radians from the positive x-axis. By definition, the sine of an angle is the y-coordinate of its corresponding point on the unit circle. Hence, understanding the unit circle helps us quickly find sine values for common angles.
For example, for \( \sin \beta = - \frac{\sqrt{3}}{2} \), the corresponding angles on the unit circle are \( 4\pi/3 \) and \( 5\pi/3 \). These angles occur in the third and fourth quadrants of the circle where sine values are negative.
General Solutions
General solutions in trigonometry refer to all possible angles that satisfy a given trigonometric equation. Since trigonometric functions are periodic, they have infinitely many solutions. To express all these solutions, we use a general formula that incorporates the periodic nature of the circle.
In the case of \( \sin \beta = -\frac{\sqrt{3}}{2} \), the general solution is derived from the specific solutions \( \beta = 4\pi/3 \) and \( 5\pi/3 \). To account for the full set of solutions given the periodicity of sine, we add \( 2k\pi \), where \( k \) is any integer, to these specific angles:
In the case of \( \sin \beta = -\frac{\sqrt{3}}{2} \), the general solution is derived from the specific solutions \( \beta = 4\pi/3 \) and \( 5\pi/3 \). To account for the full set of solutions given the periodicity of sine, we add \( 2k\pi \), where \( k \) is any integer, to these specific angles:
- \( \beta = 4\pi/3 + 2k\pi \)
- \( \beta = 5\pi/3 + 2k\pi \)
Other exercises in this chapter
Problem 26
Verify the identity. $$ \tan \theta+\cot \theta=2 \csc 2 \theta $$
View solution Problem 27
Exer. 23-30: Write the expression as an algebraic expression in \(x\) for \(x>0\). $$ \sin \left(2 \sin ^{-1} x\right) $$
View solution Problem 27
Exer. 1-50: Verify the identity. $$ \left(\sin ^{2} \theta+\cos ^{2} \theta\right)^{3}=1 $$
View solution Problem 27
Exer. 25-36: Verify the reduction formula. $$ \sin \left(x-\frac{5 \pi}{2}\right)=-\cos x $$
View solution