Problem 103
Question
Verify that equation is an identity. \(\sin ^{2} \alpha \sec ^{2} \alpha+\sin ^{2} \alpha \csc ^{2} \alpha=\sec ^{2} \alpha\)
Step-by-Step Solution
Verified Answer
The identity is verified: both sides simplify to \( \sec^2 \alpha \).
1Step 1: Simplify the identity
Start by rewriting the given equation: \( \sin^2 \alpha \sec^2 \alpha + \sin^2 \alpha \csc^2 \alpha = \sec^2 \alpha \). We aim to simplify both sides of the equation to verify the identity.
2Step 2: Express Reciprocal Trigonometric Functions
Rewrite \( \sec^2 \alpha \) and \( \csc^2 \alpha \) in terms of \( \sin \alpha \) and \( \cos \alpha \). We know that \( \sec^2 \alpha = \frac{1}{\cos^2 \alpha} \) and \( \csc^2 \alpha = \frac{1}{\sin^2 \alpha} \). Substitute these into the left side of the identity.
3Step 3: Simplify the Left Side
Substitute the reciprocal expressions: \( \sin^2 \alpha \frac{1}{\cos^2 \alpha} + \sin^2 \alpha \frac{1}{\sin^2 \alpha} \). This simplifies to \( \frac{\sin^2 \alpha}{\cos^2 \alpha} + 1 \).
4Step 4: Find a Common Denominator
The expression \( \frac{\sin^2 \alpha}{\cos^2 \alpha} + 1 \) requires a common denominator to combine the terms. Rewrite \( 1 \) as \( \frac{\cos^2 \alpha}{\cos^2 \alpha} \). The left side becomes \( \frac{\sin^2 \alpha + \cos^2 \alpha}{\cos^2 \alpha} \).
5Step 5: Use the Pythagorean Identity
Apply the Pythagorean identity \( \sin^2 \alpha + \cos^2 \alpha = 1 \). Therefore, the expression \( \frac{\sin^2 \alpha + \cos^2 \alpha}{\cos^2 \alpha} \) simplifies to \( \frac{1}{\cos^2 \alpha} \), which is \( \sec^2 \alpha \).
6Step 6: Verify Identity
Now both the left side and the right side of the equation simplify to \( \sec^2 \alpha \). Since they are equal, the identity \( \sin^2 \alpha \sec^2 \alpha + \sin^2 \alpha \csc^2 \alpha = \sec^2 \alpha \) is verified.
Key Concepts
Pythagorean IdentityReciprocal Trigonometric FunctionsSimplifying Trigonometric Expressions
Pythagorean Identity
When dealing with trigonometric identities, one of the most fundamental tools is the Pythagorean Identity. This identity states that for any angle \( \alpha \), the equation \( \sin^2 \alpha + \cos^2 \alpha = 1 \) holds true. This relation is intrinsic to the unit circle, where the square of the sine and cosine of an angle sum to one. This means that if you know one of these values, you can directly compute the other. For example:
- If \( \sin^2 \alpha = 0.4 \), then \( \cos^2 \alpha = 0.6 \).
- This can be solved as \( \cos \alpha = \sqrt{0.6} \) (consider the quadrant for sign), and vice versa for sine.
Reciprocal Trigonometric Functions
Reciprocal trigonometric functions are inverses of the basic trigonometry functions, and they can be helpful in transforming and simplifying expressions. The three primary reciprocal functions are:
- Cosecant (\( \csc \alpha = \frac{1}{\sin \alpha} \))
- Secant (\( \sec \alpha = \frac{1}{\cos \alpha} \))
- Cotangent (\( \cot \alpha = \frac{1}{\tan \alpha} \))
Simplifying Trigonometric Expressions
Simplifying trigonometric expressions is an essential skill in solving trigonometric identities and equations. The process generally involves:
- Using identities like the Pythagorean Identity: They help in reducing expressions to simpler forms.
- Applying reciprocal identities: They convert expressions into easier formats by fracturing them into sine, cosine, or tangent terms.
- Finding common denominators: This helps in combining fractions into single expressions, like what was done in our exercise, where we needed a common denominator to combine two terms.
Other exercises in this chapter
Problem 102
Verify that equation is an identity. \(\frac{1-\cos x}{1+\cos x}=\csc ^{2} x-2 \csc x \cot x+\cot ^{2} x\)
View solution Problem 103
Write each expression as an algebraic expression in \(u, u>0\). $$\sin (\arccos u)$$
View solution Problem 104
Write each expression as an algebraic expression in \(u, u>0\). $$\tan (\arccos u)$$
View solution Problem 104
Verify that equation is an identity. \((\sec \alpha+\csc \alpha)(\cos \alpha-\sin \alpha)=\cot \alpha-\tan \alpha\)
View solution