Question

guys help me to solve this...
(i) If sin 0 + sin2 0 = 1, prove that cos0 + cos4 = 1
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Answer 1

sinA + sin²A = 1

sinA = 1 - sin²A

sinA = cos²A ... (i) (sin²A + cos²A = 1)

now,

RHS

= 1

LHS

= cos²A + cos⁴A

= cos²A + (cos²A)²

= cos²A + (sinA)² ... (from i)

= cos²A + sin²A

= 1

= RHS

LHS = RHS

Hence, cos²A + cos⁴A = 1

Answer 2

Step-by-step explanation:

Given Equation is sinθ + sin²θ = 1

[∵ sin²θ + cos²θ = 1 → sin²θ = 1 - cos²θ]

=> sinθ + 1 - cos²θ = 1

⇒ sinθ - cos²θ = 1 - 1

⇒ sinθ - cos²θ = 0

⇒ sinθ = cos²θ

On Squaring both sides, we get

(sinθ)² = (cos²θ)²

⇒ sin²θ = cos⁴θ

⇒ 1 - cos²θ = cos⁴θ

⇒ cos²θ + cos⁴θ = 1

Hope it helps!