Trignometric identities question
Asked by
| 5th Oct, 2008,
09:10: PM
taking Lcm and solve we get sin2Φ(1-sin2θ)2 + sin4θcos2Φ = sin2Φ (1-sin2Φ )
further solving in terma of sinΦ and sinθ we get sin4θ + sin4Φ = 2sin2Φ sin2θ which gives (sin2θ - sin2Φ)2 =0 =====> sin θ = sinΦ ===> θ = Φ
so putting θ = Φ in second eqn. we get sin2θ + cos2θ = 1 which is true
Answered by
| 16th Dec, 2008,
10:04: PM
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