Math, asked by divyanshi6331, 1 year ago

prove that, sin^4A+2sin^2A.cos^2A=1-cos^4A​

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Answered by abhi178
3

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 Anonymous
5

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)

 \boxed{\sf sin^{2}\theta + cos^{2}\theta = 1}

(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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