Verifying Trigonometric Identities

Complementary Angle 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.

Even-Odd Identities

  • Based on whether trigonometric functions are even or odd.

  • Sine of negative theta: sin⁡(−θ)=−sin⁡(θ)\sin(-\theta) = -\sin(\theta)

    • If theta is in quadrant one, negative theta is in quadrant four.

    • sin⁡(−θ)\sin(-\theta) is the negative of sin⁡(θ)\sin(\theta).

  • Cosine of negative theta: cos⁡(−θ)=cos⁡(θ)\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.

Verifying Trigonometric Identities

  • 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.

Hints for Verifying Identities

  • 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.

Verifying Trig Identities Examples

Example 1

  • Prove: sec⁡2(θ)−1sec⁡2(θ)=sin⁡2(θ)\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(θ)+1=sec⁡2(θ)\tan^2(\theta) + 1 = \sec^2(\theta)

  • Rearrange: tan⁡2(θ)=sec⁡2(θ)−1\tan^2(\theta) = \sec^2(\theta) - 1

  • Substitute: tan⁡2(θ)sec⁡2(θ)\frac{\tan^2(\theta)}{\sec^2(\theta)}

  • Express tangent in terms of sine and cosine: tan⁡2(θ)=sin⁡2(θ)cos⁡2(θ)\tan^2(\theta) = \frac{\sin^2(\theta)}{\cos^2(\theta)}

  • Rewrite the expression: sin⁡2(θ)cos⁡2(θ)1cos⁡2(θ)\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:
    sin⁡2(θ)cos⁡2(θ)⋅cos⁡2(θ)=sin⁡2(θ)\frac{\sin^2(\theta)}{\cos^2(\theta)} \cdot \cos^2(\theta) = \sin^2(\theta)

  • Result: sin⁡2(θ)=sin⁡2(θ)\sin^2(\theta) = \sin^2(\theta), Verified.

Example 2

  • Prove: cot⁡(θ)⋅csc⁡(θ)=cos⁡(θ)\cot(\theta) \cdot \csc(\theta) = \cos(\theta)

  • Start with the left side.

  • Rewrite cotangent as one over tangent: cot⁡(θ)=1tan⁡(θ)\cot(\theta) = \frac{1}{\tan(\theta)}

  • Rewrite cosecant as the inverse of sine: csc⁡(θ)=1sin⁡(θ)\csc(\theta) = \frac{1}{\sin(\theta)}

  • Substitute: 1tan⁡(θ)⋅1sin⁡(θ)\frac{1}{\tan(\theta)} \cdot \frac{1}{\sin(\theta)}

  • Rewrite the expression: 1tan⁡(θ)⋅1sin⁡(θ)=sin⁡(θ)tan⁡(θ)\frac{1}{\tan(\theta)} \cdot \frac{1}{\sin(\theta)} = \frac{\sin(\theta)}{\tan(\theta)}

  • Express tangent in terms of sine and cosine: tan⁡(θ)=sin⁡(θ)cos⁡(θ)\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}

  • Substitute: sin⁡(θ)sin⁡(θ)cos⁡(θ)\frac{\sin(\theta)}{\frac{\sin(\theta)}{\cos(\theta)}}

  • Dividing by a fraction is the same as multiplying by its reciprocal:
    sin⁡(θ)sin⁡(θ)cos⁡(θ)=sin⁡(θ)⋅cos⁡(θ)sin⁡(θ)\frac{\sin(\theta)}{\frac{\sin(\theta)}{\cos(\theta)}} = \sin(\theta) \cdot \frac{\cos(\theta)}{\sin(\theta)}

  • Cancel out the sin⁡(θ)\sin(\theta) terms.

  • Result: cos⁡(θ)=cos⁡(θ)\cos(\theta) = \cos(\theta), Verified.

Example 3

  • Prove: 11−sin⁡(θ)=sec⁡(θ)+tan⁡(θ)\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: 11−sin⁡(θ)⋅1+sin⁡(θ)1+sin⁡(θ)\frac{1}{1 - \sin(\theta)} \cdot \frac{1 + \sin(\theta)}{1 + \sin(\theta)}

  • Multiply out the fractions.

  • Simplify denominator and rewrite it: 1+sin⁡(θ)1−sin⁡2(θ)=cos⁡(θ)+cos⁡(θ)sin⁡(θ)cos⁡2(θ)\frac{1 + \sin(\theta)}{1 - \sin^2(\theta)} = \frac{\cos(\theta) + \cos(\theta)\sin(\theta)}{\cos^2(\theta)}

  • Since cos⁡2(θ)=1−sin⁡2(θ)\cos^2(\theta) = 1 - \sin^2(\theta)

  • Break into two fractions: cos⁡(θ)cos⁡2(θ)+cos⁡(θ)sin⁡(θ)cos⁡2(θ)\frac{\cos(\theta)}{\cos^2(\theta)} + \frac{\cos(\theta)\sin(\theta)}{\cos^2(\theta)}

  • Simplify: 1cos⁡(θ)+sin⁡(θ)cos⁡(θ)\frac{1}{\cos(\theta)} + \frac{\sin(\theta)}{\cos(\theta)}

  • Replace with trig functions

  • sec⁡(θ)+tan⁡(θ)=sec⁡(θ)+tan⁡(θ)\sec(\theta) + \tan(\theta) = \sec(\theta) + \tan(\theta), Verified.

Example 4

  • Prove: tan⁡2(y)+1−tan⁡2(y)=1\tan^2(y)+1-\tan^2(y) = 1

  • Start with the left side.

  • Use Pythagorean identities to simplify expression

  • Substitute for sec⁡2(y)=tan⁡2(y)+1\sec^2(y) = \tan^2(y) + 1

  • tan⁡2(y)+1−tan⁡2(y)=sec⁡2(y)−tan⁡2(y)\tan^2(y)+1-\tan^2(y) = \sec^2(y)-\tan^2(y)

  • sec⁡2(y)−tan⁡2(y)=1\sec^2(y)-\tan^2(y) = 1

  • 1=11=1, Verified.