prove that, sin^4A+2sin^2A.cos^2A=1-cos^4A
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we know, sin²x + cos²x = 1
so, sin²θ = 1 - cos²θ .........(1)
so, sin⁴θ = (sin²θ)² = (1 - cos²θ)²
= 1 + cos⁴θ - 2cos²θ .......(2)
LHS = sin⁴θ + 2sin²θ.cos²θ
from equation (2),
= 1 + cos⁴θ - 2cos²θ + 2sin²θ.cos²θ
= 1 + cos⁴θ - 2cos²θ(1 - sin²θ)
= 1 + cos⁴θ - 2cos²θ.cos²θ [ from equation (1), ]
= 1 + cos⁴θ - 2cos⁴θ
= 1 - cos⁴θ = RHS
hence, sin⁴θ + 2sin²θ.cos²θ = 1 - cos⁴θ
Answered by
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Answer:
To Prove : sin⁴A + cos⁴A = 1 - 2sin²A × cos²A
Solution : sin⁴A + cos⁴A can be expressed as;
α² + β² = (α + β)² - 2αβ
(sin²A)² + (cos²A)² = (sin²A + cos²A)² - 2(sin²A)(cos²A)
(sin²A)² + (cos²A)² = (1)² - 2(sin²A)(cos²A)
(sin²A)² + (cos²A)² = 1 - 2 × sin²A × cos²A
Hence Proved.
Identities used in the Solution:
α² + β² = (α + β)² - 2αβ
sin²θ + cos²θ = 1
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