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Trigonometric identities

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Derivations of trigonometric Identities

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Trigonometric identities

  1. 1. Trigonometric Identities by SBR www.harekrishnahub.com
  2. 2. www.harekrishnahub.com We have the following trigonometric identities: 𝑠𝑖𝑛2 πœƒ + π‘π‘œπ‘ 2 πœƒ = 1 1 + π‘‘π‘Žπ‘›2 πœƒ = 𝑠𝑒𝑐2 πœƒ 1 + π‘π‘œπ‘‘2 πœƒ = π‘π‘œπ‘ π‘’π‘2 πœƒ Let us prove one by one. Note: 𝑠𝑖𝑛 πœƒ 2 , π‘π‘œπ‘  πœƒ 2 , etc., are written as 𝑠𝑖𝑛2 πœƒand π‘π‘œπ‘ 2 πœƒ etc. and read as sine squared ΞΈ
  3. 3. www.harekrishnahub.com Consider a right angle triangle ABC as shown below. Let ∠𝐴𝐡𝐢 = 90Β° and ∠𝐢𝐴𝐡 = πœƒ 𝐴𝐡 = π‘₯ , 𝐡𝐢 = 𝑦 , 𝐴𝐢 = π‘Ÿ
  4. 4. www.harekrishnahub.com we have, 𝒙 𝟐 + π’š 𝟐 = 𝒓 𝟐 Dividing the above equation throughout by π‘Ÿ2 , we get 𝒙 𝟐 𝒓 𝟐 + π’š 𝟐 𝒓 𝟐 = 𝒓 𝟐 𝒓 𝟐 𝒙 𝒓 𝟐 + π’š 𝒓 𝟐 = 𝟏 But π‘₯ π‘Ÿ = π‘π‘œπ‘  πœƒ and 𝑦 π‘Ÿ = 𝑠𝑖𝑛 πœƒ ∴ π‘π‘œπ‘  πœƒ 2 + 𝑠𝑖𝑛 πœƒ 2 = 1 ∴ we have the identity π’”π’Šπ’ 𝟐 𝜽 + 𝒄𝒐𝒔 𝟐 𝜽 = 𝟏
  5. 5. www.harekrishnahub.com we have, 𝒙 𝟐 + π’š 𝟐 = 𝒓 𝟐 Dividing the above equation throughout by π‘₯2 (for 𝒙 β‰  𝟎) , we get 𝒙 𝟐 𝒙 𝟐 + π’š 𝟐 𝒙 𝟐 = 𝒓 𝟐 𝒙 𝟐 𝟏 + π’š 𝒙 𝟐 = 𝒓 𝒙 𝟐 But π‘Ÿ π‘₯ = 𝑠𝑒𝑐 πœƒ and 𝑦 π‘₯ = π‘‘π‘Žπ‘› πœƒ ∴ we have 𝟏 + 𝒕𝒂𝒏 𝜽 𝟐 = 𝒔𝒆𝒄 𝜽 𝟐 ∴ we have the identity 1 + 𝒕𝒂𝒏2 𝜽 = 𝒔𝒆𝒄2 𝜽
  6. 6. www.harekrishnahub.com we have, 𝒙 𝟐 + π’š 𝟐 = 𝒓 𝟐 Dividing the above equation throughout by 𝑦2(for 𝒙 𝟐 π’š 𝟐 + π’š 𝟐 π’š 𝟐 = 𝒓 𝟐 π’š 𝟐 𝒙 π’š 𝟐 + 𝟏 = 𝒓 π’š 𝟐 𝒓 π’š = 𝒄𝒐𝒔𝒆𝒄 𝜽 𝒙 π’š = 𝒄𝒐𝒕 𝜽 ∴ we have 𝒄𝒐𝒕 𝜽 𝟐 + 𝟏 = 𝒄𝒐𝒔𝒆𝒄 𝜽 𝟐 ∴ we have the identity 1 + 𝒄𝒐𝒕2 𝜽 = 𝒄𝒐𝒔𝒆𝒄2 𝜽

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