text: Cos x + cos y formula
  
  Chapter 3 Class 11 Trigonometric Functions
  Concept wise
  
  ### Transcript
  
  Ex 3.3, 20 Prove that 〖sin x − 〗⁡sin⁡3x /(sin2⁡x − cos2⁡x ) = 2 sin x Solving L.H.S. 〖sin x −〗⁡sin⁡3x /(sin2⁡x − cos2⁡x ) We solve sin x – sin 3x & sin2 x – cos2 x separately sin x – sin 3x = 2 cos ((x + 3x)/2) sin((x − 3x)/2) = 2 cos (4𝑥/2) sin ((−2𝑥)/2) = 2 cos 2x sin (–x) sin2 x – cos2 x = – cos 2x We know that cos 2x = cos2 x – sin2 x cos 2x = – ( sin2 x – cos2 x) Thus, −( sin2 x – cos2 x) = cos 2x − ( sin2 x – cos2 x) = – cos 2x Now, sin⁡〖𝑥 − sin⁡3𝑥 〗/sin2⁡〖𝑥 − cos2⁡𝑥 〗 = 〖𝟐 𝒄𝒐𝒔 〗⁡〖𝟐𝒙 〖 𝒔𝒊𝒏〗⁡〖(−𝒙)〗 〗/〖−𝒄𝒐𝒔〗⁡𝟐𝒙 = 〖2 𝑐𝑜𝑠〗⁡〖2𝑥 〖(− sin〗⁡𝑥)〗/〖−𝑐𝑜𝑠〗⁡2𝑥 = 2sin x = R.H.S. Hence L.H.S. = R.H.S. Hence proved
explanation: The document is a transcript from a mathematics textbook, specifically from Chapter 3, Class 11 Trigonometric Functions. It contains a detailed solution to a trigonometric identity problem, proving that \(\frac{\sin x - \sin 3x}{\sin^2 x - \cos^2 x} = 2 \sin x\). The document includes the step-by-step solution, breaking down the problem into manageable parts and using trigonometric identities to simplify the expression.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: Cos x + cos y formula
  
  Chapter 3 Class 11 Trigonometric Functions
  Concept wise
  
  ### Transcript
  
  Ex 3.3, 20 Prove that 〖sin x − 〗⁡sin⁡3x /(sin2⁡x − cos2⁡x ) = 2 sin x Solving L.H.S. 〖sin x −〗⁡sin⁡3x /(sin2⁡x − cos2⁡x ) We solve sin x – sin 3x & sin2 x – cos2 x separately sin x – sin 3x = 2 cos ((x + 3x)/2) sin((x − 3x)/2) = 2 cos (4𝑥/2) sin ((−2𝑥)/2) = 2 cos 2x sin (–x) sin2 x – cos2 x = – cos 2x We know that cos 2x = cos2 x – sin2 x cos 2x = – ( sin2 x – cos2 x) Thus, −( sin2 x – cos2 x) = cos 2x − ( sin2 x – cos2 x) = – cos 2x Now, sin⁡〖𝑥 − sin⁡3𝑥 〗/sin2⁡〖𝑥 − cos2⁡𝑥 〗 = 〖𝟐 𝒄𝒐𝒔 〗⁡〖𝟐𝒙 〖 𝒔𝒊𝒏〗⁡〖(−𝒙)〗 〗/〖−𝒄𝒐𝒔〗⁡𝟐𝒙 = 〖2 𝑐𝑜𝑠〗⁡〖2𝑥 〖(− sin〗⁡𝑥)〗/〖−𝑐𝑜𝑠〗⁡2𝑥 = 2sin x = R.H.S. Hence L.H.S. = R.H.S. Hence proved
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a transcript from a mathematics textbook, specifically from Chapter 3, Class 11 Trigonometric Functions. It contains a detailed solution to a trigonometric identity problem, proving that \(\frac{\sin x - \sin 3x}{\sin^2 x - \cos^2 x} = 2 \sin x\). The document includes the step-by-step solution, breaking down the problem into manageable parts and using trigonometric identities to simplify the expression.
  
  Label: question-and-answer