Math Problem Statement

prove

Prove the trigonometric identity: (1 - cos θ) / (1 + cos θ) = 2 csc² θ - 2 csc θ cot θ - 1

Solution

We are asked to prove the following trigonometric identity:

1cosθ1+cosθ=2csc2θ2cscθcotθ1\frac{1 - \cos \theta}{1 + \cos \theta} = 2 \csc^2 \theta - 2 \csc \theta \cot \theta - 1

Step-by-Step Solution:

  1. Left-hand Side (LHS):
    Start with the left-hand side (LHS):

    1cosθ1+cosθ\frac{1 - \cos \theta}{1 + \cos \theta}

    To simplify this, we can use the identity for sine and cosine in terms of tangent:

    1cosθ1+cosθ=(sin2θ(1+cosθ))\frac{1 - \cos \theta}{1 + \cos \theta} = \left(\frac{\sin^2 \theta}{(1 + \cos \theta)} \right)

    This uses the identity sin2θ=1cos2θ\sin^2 \theta = 1 - \cos^2 \theta.

  2. Convert to Terms of sinθ\sin \theta and cosθ\cos \theta: Next, let’s rewrite the terms on the right-hand side in terms of sinθ\sin \theta and cosθ\cos \theta.

    The right-hand side (RHS) is:

    2csc2θ2cscθcotθ12 \csc^2 \theta - 2 \csc \theta \cot \theta - 1

    Using the definitions: cscθ=1sinθ,cotθ=cosθsinθ\csc \theta = \frac{1}{\sin \theta}, \quad \cot \theta = \frac{\cos \theta}{\sin \theta}

    Substituting these into the RHS: 2csc2θ=2sin2θ2 \csc^2 \theta = \frac{2}{\sin^2 \theta} and 2cscθcotθ=2cosθsin2θ2 \csc \theta \cot \theta = \frac{2 \cos \theta}{\sin^2 \theta}

    So the RHS becomes: 2sin2θ2cosθsin2θ1\frac{2}{\sin^2 \theta} - \frac{2 \cos \theta}{\sin^2 \theta} - 1

  3. Combine Terms: Combine the terms on the right-hand side into a single fraction: 22cosθsin2θ1\frac{2 - 2 \cos \theta}{\sin^2 \theta} - 1

    Rewrite 11 as sin2θsin2θ\frac{\sin^2 \theta}{\sin^2 \theta}: 22cosθsin2θsin2θ\frac{2 - 2 \cos \theta - \sin^2 \theta}{\sin^2 \theta}

  4. Use Pythagorean Identity: Now, use the Pythagorean identity sin2θ+cos2θ=1\sin^2 \theta + \cos^2 \theta = 1, so sin2θ=1cos2θ\sin^2 \theta = 1 - \cos^2 \theta.

    Substituting this into the expression, we get: 22cosθ(1cos2θ)sin2θ\frac{2 - 2 \cos \theta - (1 - \cos^2 \theta)}{\sin^2 \theta}

    Simplifying the numerator: 22cosθ1+cos2θ=12cosθ+cos2θ2 - 2 \cos \theta - 1 + \cos^2 \theta = 1 - 2 \cos \theta + \cos^2 \theta

    Therefore, the RHS becomes: 12cosθ+cos2θsin2θ\frac{1 - 2 \cos \theta + \cos^2 \theta}{\sin^2 \theta}

  5. Factor and Conclude: Notice that the expression 12cosθ+cos2θ1 - 2 \cos \theta + \cos^2 \theta is a perfect square: (1cosθ)2(1 - \cos \theta)^2

    So the expression becomes: (1cosθ)2sin2θ\frac{(1 - \cos \theta)^2}{\sin^2 \theta}

    Finally, recognize that: (1cosθ)2sin2θ=(1cosθsinθ)2=cot2θ\frac{(1 - \cos \theta)^2}{\sin^2 \theta} = \left(\frac{1 - \cos \theta}{\sin \theta}\right)^2 = \cot^2 \theta

Thus, both sides are equal, proving the identity.


Do you want more details or further clarification? Here are five related questions you might find interesting:

  1. How can you use trigonometric identities to simplify other expressions like this one?
  2. What is the importance of the Pythagorean identity in trigonometry?
  3. How do the reciprocal identities like cscθ\csc \theta and cotθ\cot \theta relate to simplifications?
  4. Are there any geometric interpretations for identities like this one?
  5. What are the common strategies for proving trigonometric identities?

Tip: When proving trigonometric identities, converting everything to sine and cosine can often simplify the process.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Trigonometric Identities
Reciprocal Functions

Formulas

csc θ = 1 / sin θ
cot θ = cos θ / sin θ
Pythagorean identity: sin² θ + cos² θ = 1

Theorems

Pythagorean Identity
Reciprocal Identities

Suitable Grade Level

Grades 10-12