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Trigonometric identities
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- 2. www.harekrishnahub.com We have thefollowing trigonometric identities: π ππ2 π + πππ 2 π = 1 1 + π‘ππ2 π = π ππ2 π 1 + πππ‘2 π = πππ ππ2 π Let us prove one by one. Note: π ππ π 2 , πππ π 2 , etc., are written as π ππ2 πand πππ 2 π etc. and read as sine squared ΞΈ
- 3. www.harekrishnahub.com Consider a rightangle triangle ABC as shown below. Let β π΄π΅πΆ = 90Β° and β πΆπ΄π΅ = π π΄π΅ = π₯ , π΅πΆ = π¦ , π΄πΆ = π
- 4. www.harekrishnahub.com we have, ππ + π π = π π Dividing the above equation throughout by π2 , we get π π π π + π π π π = π π π π π π π + π π π = π But π₯ π = πππ π and π¦ π = π ππ π β΄ πππ π 2 + π ππ π 2 = 1 β΄ we have the identity πππ π π½ + πππ π π½ = π
- 5. www.harekrishnahub.com we have, ππ + π π = π π Dividing the above equation throughout by π₯2 (for π β π) , we get π π π π + π π π π = π π π π π + π π π = π π π But π π₯ = π ππ π and π¦ π₯ = π‘ππ π β΄ we have π + πππ π½ π = πππ π½ π β΄ we have the identity 1 + πππ2 π½ = πππ2 π½
- 6. www.harekrishnahub.com we have, ππ + π π = π π Dividing the above equation throughout by π¦2(for π π π π + π π π π = π π π π π π π + π = π π π π π = πππππ π½ π π = πππ π½ β΄ we have πππ π½ π + π = πππππ π½ π β΄ we have the identity 1 + πππ2 π½ = πππππ2 π½