Question

Pn= cos^n(x)+ sin^n(x), find 6P2-4P3+3= ??
A)0 B)4 C)5 D)6

Answer 1

Answer:

answer is b

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Answer 2

Answer:

Step-by-step explanation:

Best Answer

2P₆ - 3P₄ + 1 = 2 ( cos⁶ x + sin⁶ x ) – ( 3 cos⁴x + sin⁴x ) + 1

RHS = 2 [ ( cos² x + sin² x )³ - 3 cos² x sin² x (cos² x + sin² x )] – 3 [ ( cos² x + sin² x )² - 2 cos² x sin² x ] + 1

=> 2 ( 1 - 3 cos² x sin² x ) – 3 ( 1 - 2 cos² x sin² x ) + 1

=> 0

Again for n ≥ 4, we have

P(n) - P(n-2) = cosⁿ x sinⁿ x - ( cosⁿ⁻² x + sinⁿ⁻²x )

=> cosⁿ⁻² x ( cos² x – 1) + sinⁿ⁻² x ( sin² x – 1)

=> - sin² x . cosⁿ⁻² x - cos² x . sinⁿ⁻² x

=> - sin² x cos² x - ( cosⁿ⁻⁴ x + sinⁿ⁻⁴ x )

=> - sin² x cos² x P(n-4)

Hence 6 P(10) - 15 P(8) + 10 P(6) - 1

= 6 ( P₁₀ - P₈ ) - 9 (P₈ - P₆ ) + ( P₆ - P₄ ) + P₄ - P₂

= - sin² x cos² x ( 6 P₆ - 9 P₄ + P₂ + P₀ )

= - 3 sin² x cos² x ( 2 P₆ - 3 P₄ ) - sin² x cos² x ( 1 + 2 )

= - 3 sin² x cos² x ( - 1 ) - 3 sin² x cos² x = 0

[ since ( 2 P₆ - 3 P₄ + 1 ) = 0 ]

Hence LHS = RHS ………………. Proved