Verifying Trigonometric Identities

The sine of one part of an angle (in degrees) equals the cosine of the other part.
This concept extends to other trigonometric functions:
- Tangent and cotangent.
- Secant and cosecant.

Based on whether trigonometric functions are even or odd.
Sine of negative theta: \sin(-\theta) = -\sin(\theta)
- If theta is in quadrant one, negative theta is in quadrant four.
- \sin(-\theta) is the negative of \sin(\theta).
Cosine of negative theta: \cos(-\theta) = \cos(\theta)
- Cosine of negative theta and cosine of theta share the same x-value on the unit circle.
- Cosine is an even function.

Goal: Prove that a given trigonometric equation is valid.
Important Assumption: We are trying to prove both sides of the equation are equal.
Cannot solve the equation by performing the same operations on both sides, as we do not yet know they are equal.

Number One Hint: Start with the more complicated side of the equation.
- The goal is to manipulate the more complicated side until it looks like the simpler side.
Other helpful hints:
- Combine multiple fractions into one fraction.
- Factor expressions if possible.
- Rewrite everything in terms of sine and cosine.
- Multiply by a fraction equal to one to change the appearance of the expression.
Utilize algebraic techniques from previous lessons and knowledge of trigonometric identities.
General Rule of Thumb: Always try something; trial and error is key.

Prove: \frac{\sec^2(\theta) - 1}{\sec^2(\theta)} = \sin^2(\theta)
Start with the left side (more complicated).
Recognize the presence of squared terms, suggesting the use of Pythagorean identities.
Pythagorean identity: \tan^2(\theta) + 1 = \sec^2(\theta)
Rearrange: \tan^2(\theta) = \sec^2(\theta) - 1
Substitute: \frac{\tan^2(\theta)}{\sec^2(\theta)}
Express tangent in terms of sine and cosine: \tan^2(\theta) = \frac{\sin^2(\theta)}{\cos^2(\theta)}
Rewrite the expression: \frac{\frac{\sin^2(\theta)}{\cos^2(\theta)}}{\frac{1}{\cos^2(\theta)}}
Dividing by a fraction is the same as multiplying by its reciprocal:
\frac{\sin^2(\theta)}{\cos^2(\theta)} \cdot \cos^2(\theta) = \sin^2(\theta)
Result: \sin^2(\theta) = \sin^2(\theta), Verified.

Prove: \cot(\theta) \cdot \csc(\theta) = \cos(\theta)
Start with the left side.
Rewrite cotangent as one over tangent: \cot(\theta) = \frac{1}{\tan(\theta)}
Rewrite cosecant as the inverse of sine: \csc(\theta) = \frac{1}{\sin(\theta)}
Substitute: \frac{1}{\tan(\theta)} \cdot \frac{1}{\sin(\theta)}
Rewrite the expression: \frac{1}{\tan(\theta)} \cdot \frac{1}{\sin(\theta)} = \frac{\sin(\theta)}{\tan(\theta)}
Express tangent in terms of sine and cosine: \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}
Substitute: \frac{\sin(\theta)}{\frac{\sin(\theta)}{\cos(\theta)}}
Dividing by a fraction is the same as multiplying by its reciprocal:
\frac{\sin(\theta)}{\frac{\sin(\theta)}{\cos(\theta)}} = \sin(\theta) \cdot \frac{\cos(\theta)}{\sin(\theta)}
Cancel out the \sin(\theta) terms.
Result: \cos(\theta) = \cos(\theta), Verified.

Prove: \frac{1}{1 - \sin(\theta)} = \sec(\theta) + \tan(\theta)
Start with the left side.
Multiply the left side by a fraction equal to one to simplify the denominator: \frac{1}{1 - \sin(\theta)} \cdot \frac{1 + \sin(\theta)}{1 + \sin(\theta)}
Multiply out the fractions.
Simplify denominator and rewrite it: \frac{1 + \sin(\theta)}{1 - \sin^2(\theta)} = \frac{\cos(\theta) + \cos(\theta)\sin(\theta)}{\cos^2(\theta)}
Since \cos^2(\theta) = 1 - \sin^2(\theta)
Break into two fractions: \frac{\cos(\theta)}{\cos^2(\theta)} + \frac{\cos(\theta)\sin(\theta)}{\cos^2(\theta)}
Simplify: \frac{1}{\cos(\theta)} + \frac{\sin(\theta)}{\cos(\theta)}
Replace with trig functions
\sec(\theta) + \tan(\theta) = \sec(\theta) + \tan(\theta), Verified.