Problem 48
Question
Verify that each of the following is an identity. \(\sin ^{2} \theta+\tan ^{2} \theta=\left(1-\cos ^{2} \theta\right)+\frac{\sec ^{2} \theta}{\csc ^{2} \theta}\)
Step-by-Step Solution
Verified Answer
The expression is an identity as both sides are equal.
1Step 1: Rewrite Known Identities
First, let's rewrite the known trigonometric identities. We know that \( \sin^2 \theta = 1 - \cos^2 \theta \) and \( \sec^2 \theta = 1 + \tan^2 \theta \). We also know \( \csc \theta = \frac{1}{\sin \theta} \), so \( \csc^2 \theta = \frac{1}{\sin^2 \theta} \).
2Step 2: Simplify Right Side
Substitute the identity \( \csc^2 \theta = \frac{1}{\sin^2 \theta} \) into the expression \( \frac{\sec^2 \theta}{\csc^2 \theta} \). Therefore, \( \frac{\sec^2 \theta}{\csc^2 \theta} = \sec^2 \theta \cdot \sin^2 \theta = (1 + \tan^2 \theta)\sin^2 \theta \).
3Step 3: Combine and Simplify
By substituting \( \sec^2 \theta = 1 + \tan^2 \theta \), the term \( \frac{\sec^2 \theta}{\csc^2 \theta} \) becomes \( (1 + \tan^2 \theta) \sin^2 \theta = \sin^2 \theta + \sin^2 \theta \tan^2 \theta \).
4Step 4: Compare Both Sides
Now let's compare both sides:- Left side: \( \sin^2 \theta + \tan^2 \theta \).- Right side: \( (1 - \cos^2 \theta) + \sin^2 \theta + \tan^2 \theta \sin^2 \theta \).Based on the identity \( \sin^2 \theta = 1 - \cos^2 \theta \), we combine terms on the right:\( 1 - \cos^2 \theta + \sin^2 \theta = \sin^2 \theta + \tan^2 \theta \).Thus, both sides are equal.
Key Concepts
Sine and CosineTangent and SecantCosecant
Sine and Cosine
The sine and cosine functions are fundamental in trigonometry, representing the coordinates of a point on the unit circle. The relationship \(\sin^2 \theta + \cos^2 \theta = 1\) is known as the Pythagorean identity. This identity illustrates that the square of the sine function added to the square of the cosine function equals 1.
This identity is crucial because it helps relate the two functions directly and it serves as the foundation for various other trigonometric identities.
This identity is crucial because it helps relate the two functions directly and it serves as the foundation for various other trigonometric identities.
- Sine (\(\sin \theta\)): Represents the y-coordinate of a point on the unit circle and is opposite the angle \(\theta\).
- Cosine (\(\cos \theta\)): Represents the x-coordinate of a point on the unit circle and is adjacent to the angle \(\theta\).
Tangent and Secant
Tangent and secant are two other important trigonometric functions. The tangent of an angle \(\theta\) is defined as the ratio of the sine to the cosine, while the secant is the reciprocal of the cosine.
- Tangent (\(\tan \theta = \frac{\sin \theta}{\cos \theta}\)) : This function helps to relate angles and side ratios in a right triangle.
- Secant (\(\sec \theta = \frac{1}{\cos \theta}\)) : It is the reciprocal of the cosine function and highlights angles where cosine approaches zero.
Cosecant
Cosecant is often less familiar than sine or cosine but is equally important in trigonometry. It is the reciprocal of the sine function.
In our verified identity, converting \(\frac{\sec^2 \theta}{\csc^2 \theta}\) to \(\sec^2 \theta \cdot \sin^2 \theta\) shows how these transformations help in reconciling both sides of an identity.
- Cosecant (\(\csc \theta = \frac{1}{\sin \theta}\)) : Used when dealing with scenarios where the sine function approaches zero.
In our verified identity, converting \(\frac{\sec^2 \theta}{\csc^2 \theta}\) to \(\sec^2 \theta \cdot \sin^2 \theta\) shows how these transformations help in reconciling both sides of an identity.
Other exercises in this chapter
Problem 48
Find the exact value of each expression. \(\cos 150^{\circ}\)
View solution Problem 48
State the amplitude, period, and phase shift of each function. Then graph each function. $$ y=3 \cos \left(\theta+\frac{\pi}{2}\right) $$
View solution Problem 48
State the vertical shift, equation of the midline, amplitude, and period for each function. Then graph the function. (Lesson 142\()\) \(y=\sin \theta-1\)
View solution Problem 48
Solve each equation. $$ \operatorname{Arccos} \frac{\sqrt{2}}{2}=x $$
View solution