Problem 39
Question
Find the given trigonometric function value. Do not use a calculator. $$ \sin 135^{\circ} $$
Step-by-Step Solution
Verified Answer
\(\sin 135^{\circ} = \frac{\sqrt{2}}{2}\)
1Step 1: Recognize the Reference Angle
To solve for \(\sin 135^{\circ}\), we start by finding its reference angle. Since 135° is in the second quadrant, we subtract it from 180°: 180° - 135° = 45°. So, the reference angle is 45°.
2Step 2: Determine the Trigonometric Sign
In the second quadrant, the sine function is positive. Therefore, \(\sin 135^{\circ} = \sin 45^{\circ}\), and it remains positive.
3Step 3: Use Known Angle Values
We know that \(\sin 45^{\circ} = \frac{\sqrt{2}}{2}\). Hence, by using the reference angle from Step 1, \(\sin 135^{\circ} = \frac{\sqrt{2}}{2}\).
Key Concepts
Reference AngleSine FunctionAngle SubtractionSecond QuadrantUnit Circle
Reference Angle
The reference angle is a fundamental concept in trigonometry that helps simplify the calculation of trigonometric functions for angles greater than 90 degrees. The reference angle is the acute angle formed by the terminal side of the given angle and the x-axis.
To find the reference angle for an angle in degrees:
- If the angle is in the second quadrant, subtract it from 180°.
- If the angle is in the third quadrant, subtract 180° from it.
- If the angle is in the fourth quadrant, subtract it from 360°.
Sine Function
The sine function, denoted as \( \sin \theta \), is one of the primary trigonometric functions and relates an angle in a right triangle to the ratio of the length of the opposite side over the length of the hypotenuse. For angles on the unit circle, the sine value is the y-coordinate of the point where the terminal side of the angle intersects the circle.Key properties of the sine function include:
- Range: The sine value is always between -1 and 1.
- Periodicity: The sine function repeats every 360° or \(2\pi\) radians.
- Symmetry: It is an odd function, meaning \(\sin(-\theta) = -\sin \theta\).
Angle Subtraction
Angle subtraction is a method used to determine a reference angle when dealing with angles outside the first quadrant. This process involves subtracting your angle from 90°, 180°, 270°, or 360° depending on the quadrant the angle lies in.For any angle \( \theta \) located in the second quadrant, like 135°, the subtraction process involves taking the difference between 180° and \( \theta \):\[\text{Reference Angle} = 180^{\circ} - \theta\]This subtraction helps to identify the equivalent acute angle, making it easier to find trigonometric values as these are often memorized in the first quadrant. It ensures that despite the angle's initial size or location, you can always revert to a simpler angle that shares the same trigonometric values, just adjusted for sign depending on the quadrant.
Second Quadrant
In the unit circle, the second quadrant is defined by angles ranging from 90° to 180°. This specific section of the circle impacts the sign of the trigonometric functions, specifically noting that:
- Sine values are positive.
- Cosine and tangent values are negative.
Unit Circle
The unit circle is a fundamental concept in trigonometry, consisting of a circle with a radius of one unit centered at the origin of the coordinate plane. It simplifies understanding of trigonometric functions since any angle's sine or cosine can be identified as the y or x coordinate, respectively, of the point where the terminal side of the angle intersects the circle.Key features include:
- The radius is always 1.
- Angles are typically measured from the positive x-axis.
- Sine corresponds to the y-coordinate, while cosine corresponds to the x-coordinate.
Other exercises in this chapter
Problem 39
Write the given expression as an algebraic expression in \(x\). $$ \csc (\arctan x) $$
View solution Problem 39
In Problems \(33-40,\) convert the given angle from radians to degrees. $$ 3.1 $$
View solution Problem 39
Use a half-angle formula to find the exact value of the given trigonometric function. Do not use a calculator. $$ \cos 67.5^{\circ} $$
View solution Problem 40
If \(x_{1}+x_{2}+x_{3}=\pi\), then show that \(\sin 2 x_{1}+\sin 2 x_{2}+\sin 2 x_{3}=4 \sin x_{1} \sin x_{2} \sin x_{3}\)
View solution